I am totally new to python and I am trying to implement quicksort in it.
Could someone please help me complete my code?
I do not know how to concatenate the three arrays and print them.
def sort(array=[12,4,5,6,7,3,1,15]):
less = []
equal = []
greater = []
if len(array) > 1:
pivot = array[0]
for x in array:
if x < pivot:
less.append(x)
if x == pivot:
equal.append(x)
if x > pivot:
greater.append(x)
sort(less)
sort(pivot)
sort(greater)
def sort(array):
"""Sort the array by using quicksort."""
less = []
equal = []
greater = []
if len(array) > 1:
pivot = array[0]
for x in array:
if x < pivot:
less.append(x)
elif x == pivot:
equal.append(x)
elif x > pivot:
greater.append(x)
# Don't forget to return something!
return sort(less)+equal+sort(greater) # Just use the + operator to join lists
# Note that you want equal ^^^^^ not pivot
else: # You need to handle the part at the end of the recursion - when you only have one element in your array, just return the array.
return array
Quick sort without additional memory (in place)
Usage:
array = [97, 200, 100, 101, 211, 107]
quicksort(array)
print(array)
# array -> [97, 100, 101, 107, 200, 211]
def partition(array, begin, end):
pivot = begin
for i in range(begin+1, end+1):
if array[i] <= array[begin]:
pivot += 1
array[i], array[pivot] = array[pivot], array[i]
array[pivot], array[begin] = array[begin], array[pivot]
return pivot
def quicksort(array, begin=0, end=None):
if end is None:
end = len(array) - 1
def _quicksort(array, begin, end):
if begin >= end:
return
pivot = partition(array, begin, end)
_quicksort(array, begin, pivot-1)
_quicksort(array, pivot+1, end)
return _quicksort(array, begin, end)
There is another concise and beautiful version
def qsort(arr):
if len(arr) <= 1:
return arr
else:
return qsort([x for x in arr[1:] if x < arr[0]])
+ [arr[0]]
+ qsort([x for x in arr[1:] if x >= arr[0]])
Let me explain the above codes for details
pick the first element of array arr[0] as pivot
[arr[0]]
qsort those elements of array which are less than pivot with List Comprehension
qsort([x for x in arr[1:] if x < arr[0]])
qsort those elements of array which are larger than pivot with List Comprehension
qsort([x for x in arr[1:] if x >= arr[0]])
This answer is an in-place QuickSort for Python 2.x. My answer is an interpretation of the in-place solution from Rosetta Code which works for Python 3 too:
import random
def qsort(xs, fst, lst):
'''
Sort the range xs[fst, lst] in-place with vanilla QuickSort
:param xs: the list of numbers to sort
:param fst: the first index from xs to begin sorting from,
must be in the range [0, len(xs))
:param lst: the last index from xs to stop sorting at
must be in the range [fst, len(xs))
:return: nothing, the side effect is that xs[fst, lst] is sorted
'''
if fst >= lst:
return
i, j = fst, lst
pivot = xs[random.randint(fst, lst)]
while i <= j:
while xs[i] < pivot:
i += 1
while xs[j] > pivot:
j -= 1
if i <= j:
xs[i], xs[j] = xs[j], xs[i]
i, j = i + 1, j - 1
qsort(xs, fst, j)
qsort(xs, i, lst)
And if you are willing to forgo the in-place property, below is yet another version which better illustrates the basic ideas behind quicksort. Apart from readability, its other advantage is that it is stable (equal elements appear in the sorted list in the same order that they used to have in the unsorted list). This stability property does not hold with the less memory-hungry in-place implementation presented above.
def qsort(xs):
if not xs: return xs # empty sequence case
pivot = xs[random.choice(range(0, len(xs)))]
head = qsort([x for x in xs if x < pivot])
tail = qsort([x for x in xs if x > pivot])
return head + [x for x in xs if x == pivot] + tail
Quicksort with Python
In real life, we should always use the builtin sort provided by Python. However, understanding the quicksort algorithm is instructive.
My goal here is to break down the subject such that it is easily understood and replicable by the reader without having to return to reference materials.
The quicksort algorithm is essentially the following:
Select a pivot data point.
Move all data points less than (below) the pivot to a position below the pivot - move those greater than or equal to (above) the pivot to a position above it.
Apply the algorithm to the areas above and below the pivot
If the data are randomly distributed, selecting the first data point as the pivot is equivalent to a random selection.
Readable example:
First, let's look at a readable example that uses comments and variable names to point to intermediate values:
def quicksort(xs):
"""Given indexable and slicable iterable, return a sorted list"""
if xs: # if given list (or tuple) with one ordered item or more:
pivot = xs[0]
# below will be less than:
below = [i for i in xs[1:] if i < pivot]
# above will be greater than or equal to:
above = [i for i in xs[1:] if i >= pivot]
return quicksort(below) + [pivot] + quicksort(above)
else:
return xs # empty list
To restate the algorithm and code demonstrated here - we move values above the pivot to the right, and values below the pivot to the left, and then pass those partitions to same function to be further sorted.
Golfed:
This can be golfed to 88 characters:
q=lambda x:x and q([i for i in x[1:]if i<=x[0]])+[x[0]]+q([i for i in x[1:]if i>x[0]])
To see how we get there, first take our readable example, remove comments and docstrings, and find the pivot in-place:
def quicksort(xs):
if xs:
below = [i for i in xs[1:] if i < xs[0]]
above = [i for i in xs[1:] if i >= xs[0]]
return quicksort(below) + [xs[0]] + quicksort(above)
else:
return xs
Now find below and above, in-place:
def quicksort(xs):
if xs:
return (quicksort([i for i in xs[1:] if i < xs[0]] )
+ [xs[0]]
+ quicksort([i for i in xs[1:] if i >= xs[0]]))
else:
return xs
Now, knowing that and returns the prior element if false, else if it is true, it evaluates and returns the following element, we have:
def quicksort(xs):
return xs and (quicksort([i for i in xs[1:] if i < xs[0]] )
+ [xs[0]]
+ quicksort([i for i in xs[1:] if i >= xs[0]]))
Since lambdas return a single epression, and we have simplified to a single expression (even though it is getting more unreadable) we can now use a lambda:
quicksort = lambda xs: (quicksort([i for i in xs[1:] if i < xs[0]] )
+ [xs[0]]
+ quicksort([i for i in xs[1:] if i >= xs[0]]))
And to reduce to our example, shorten the function and variable names to one letter, and eliminate the whitespace that isn't required.
q=lambda x:x and q([i for i in x[1:]if i<=x[0]])+[x[0]]+q([i for i in x[1:]if i>x[0]])
Note that this lambda, like most code golfing, is rather bad style.
In-place Quicksort, using the Hoare Partitioning scheme
The prior implementation creates a lot of unnecessary extra lists. If we can do this in-place, we'll avoid wasting space.
The below implementation uses the Hoare partitioning scheme, which you can read more about on wikipedia (but we have apparently removed up to 4 redundant calculations per partition() call by using while-loop semantics instead of do-while and moving the narrowing steps to the end of the outer while loop.).
def quicksort(a_list):
"""Hoare partition scheme, see https://en.wikipedia.org/wiki/Quicksort"""
def _quicksort(a_list, low, high):
# must run partition on sections with 2 elements or more
if low < high:
p = partition(a_list, low, high)
_quicksort(a_list, low, p)
_quicksort(a_list, p+1, high)
def partition(a_list, low, high):
pivot = a_list[low]
while True:
while a_list[low] < pivot:
low += 1
while a_list[high] > pivot:
high -= 1
if low >= high:
return high
a_list[low], a_list[high] = a_list[high], a_list[low]
low += 1
high -= 1
_quicksort(a_list, 0, len(a_list)-1)
return a_list
Not sure if I tested it thoroughly enough:
def main():
assert quicksort([1]) == [1]
assert quicksort([1,2]) == [1,2]
assert quicksort([1,2,3]) == [1,2,3]
assert quicksort([1,2,3,4]) == [1,2,3,4]
assert quicksort([2,1,3,4]) == [1,2,3,4]
assert quicksort([1,3,2,4]) == [1,2,3,4]
assert quicksort([1,2,4,3]) == [1,2,3,4]
assert quicksort([2,1,1,1]) == [1,1,1,2]
assert quicksort([1,2,1,1]) == [1,1,1,2]
assert quicksort([1,1,2,1]) == [1,1,1,2]
assert quicksort([1,1,1,2]) == [1,1,1,2]
Conclusion
This algorithm is frequently taught in computer science courses and asked for on job interviews. It helps us think about recursion and divide-and-conquer.
Quicksort is not very practical in Python since our builtin timsort algorithm is quite efficient, and we have recursion limits. We would expect to sort lists in-place with list.sort or create new sorted lists with sorted - both of which take a key and reverse argument.
There are many answers to this already, but I think this approach is the most clean implementation:
def quicksort(arr):
""" Quicksort a list
:type arr: list
:param arr: List to sort
:returns: list -- Sorted list
"""
if not arr:
return []
pivots = [x for x in arr if x == arr[0]]
lesser = quicksort([x for x in arr if x < arr[0]])
greater = quicksort([x for x in arr if x > arr[0]])
return lesser + pivots + greater
You can of course skip storing everything in variables and return them straight away like this:
def quicksort(arr):
""" Quicksort a list
:type arr: list
:param arr: List to sort
:returns: list -- Sorted list
"""
if not arr:
return []
return quicksort([x for x in arr if x < arr[0]]) \
+ [x for x in arr if x == arr[0]] \
+ quicksort([x for x in arr if x > arr[0]])
functional approach:
def qsort(lst):
if len(lst) < 2:
return lst
pivot = lst[0]
left = list(filter(lambda x: x <= pivot, lst[1:]))
right = list(filter(lambda x: x > pivot, lst[1:]))
return qsort(left) + [pivot] + qsort(right)
Easy implementation from grokking algorithms
def quicksort(arr):
if len(arr) < 2:
return arr #base case
else:
pivot = arr[0]
less = [i for i in arr[1:] if i <= pivot]
more = [i for i in arr[1:] if i > pivot]
return quicksort(less) + [pivot] + quicksort(more)
This is a version of the quicksort using Hoare partition scheme and with fewer swaps and local variables
def quicksort(array):
qsort(array, 0, len(array)-1)
def qsort(A, lo, hi):
if lo < hi:
p = partition(A, lo, hi)
qsort(A, lo, p)
qsort(A, p + 1, hi)
def partition(A, lo, hi):
pivot = A[lo]
i, j = lo-1, hi+1
while True:
i += 1
j -= 1
while(A[i] < pivot): i+= 1
while(A[j] > pivot ): j-= 1
if i >= j:
return j
A[i], A[j] = A[j], A[i]
test = [21, 4, 1, 3, 9, 20, 25, 6, 21, 14]
print quicksort(test)
functional programming aproach
smaller = lambda xs, y: filter(lambda x: x <= y, xs)
larger = lambda xs, y: filter(lambda x: x > y, xs)
qsort = lambda xs: qsort(smaller(xs[1:],xs[0])) + [xs[0]] + qsort(larger(xs[1:],xs[0])) if xs != [] else []
print qsort([3,1,4,2,5]) == [1,2,3,4,5]
Partition - Split an array by a pivot that smaller elements move to the left and greater elemets move to the right or vice versa. A pivot can be an random element from an array. To make this algorith we need to know what is begin and end index of an array and where is a pivot. Then set two auxiliary pointers L, R.
So we have an array user[...,begin,...,end,...]
The start position of L and R pointers
[...,begin,next,...,end,...]
R L
while L < end
1. If a user[pivot] > user[L] then move R by one and swap user[R] with user[L]
2. move L by one
After while swap user[R] with user[pivot]
Quick sort - Use the partition algorithm until every next part of the split by a pivot will have begin index greater or equals than end index.
def qsort(user, begin, end):
if begin >= end:
return
# partition
# pivot = begin
L = begin+1
R = begin
while L < end:
if user[begin] > user[L]:
R+=1
user[R], user[L] = user[L], user[R]
L+= 1
user[R], user[begin] = user[begin], user[R]
qsort(user, 0, R)
qsort(user, R+1, end)
tests = [
{'sample':[1],'answer':[1]},
{'sample':[3,9],'answer':[3,9]},
{'sample':[1,8,1],'answer':[1,1,8]},
{'sample':[7,5,5,1],'answer':[1,5,5,7]},
{'sample':[4,10,5,9,3],'answer':[3,4,5,9,10]},
{'sample':[6,6,3,8,7,7],'answer':[3,6,6,7,7,8]},
{'sample':[3,6,7,2,4,5,4],'answer':[2,3,4,4,5,6,7]},
{'sample':[1,5,6,1,9,0,7,4],'answer':[0,1,1,4,5,6,7,9]},
{'sample':[0,9,5,2,2,5,8,3,8],'answer':[0,2,2,3,5,5,8,8,9]},
{'sample':[2,5,3,3,2,0,9,0,0,7],'answer':[0,0,0,2,2,3,3,5,7,9]}
]
for test in tests:
sample = test['sample'][:]
answer = test['answer']
qsort(sample,0,len(sample))
print(sample == answer)
I think both answers here works ok for the list provided (which answer the original question), but would breaks if an array containing non unique values is passed. So for completeness, I would just point out the small error in each and explain how to fix them.
For example trying to sort the following array [12,4,5,6,7,3,1,15,1] (Note that 1 appears twice) with Brionius algorithm .. at some point will end up with the less array empty and the equal array with a pair of values (1,1) that can not be separated in the next iteration and the len() > 1...hence you'll end up with an infinite loop
You can fix it by either returning array if less is empty or better by not calling sort in your equal array, as in zangw answer
def sort(array=[12,4,5,6,7,3,1,15]):
less = []
equal = []
greater = []
if len(array) > 1:
pivot = array[0]
for x in array:
if x < pivot:
less.append(x)
elif x == pivot:
equal.append(x)
else: # if x > pivot
greater.append(x)
# Don't forget to return something!
return sort(less) + equal + sort(greater) # Just use the + operator to join lists
# Note that you want equal ^^^^^ not pivot
else: # You need to hande the part at the end of the recursion - when you only have one element in your array, just return the array.
return array
The fancier solution also breaks, but for a different cause, it is missing the return clause in the recursion line, which will cause at some point to return None and try to append it to a list ....
To fix it just add a return to that line
def qsort(arr):
if len(arr) <= 1:
return arr
else:
return qsort([x for x in arr[1:] if x<arr[0]]) + [arr[0]] + qsort([x for x in arr[1:] if x>=arr[0]])
Or if you prefer a one-liner that also illustrates the Python equivalent of C/C++ varags, lambda expressions, and if expressions:
qsort = lambda x=None, *xs: [] if x is None else qsort(*[a for a in xs if a<x]) + [x] + qsort(*[a for a in xs if a>=x])
A "true" in-place implementation [Algorithms 8.9, 8.11 from the Algorithm Design and Applications Book by Michael T. Goodrich and Roberto Tamassia]:
from random import randint
def partition (A, a, b):
p = randint(a,b)
# or mid point
# p = (a + b) / 2
piv = A[p]
# swap the pivot with the end of the array
A[p] = A[b]
A[b] = piv
i = a # left index (right movement ->)
j = b - 1 # right index (left movement <-)
while i <= j:
# move right if smaller/eq than/to piv
while A[i] <= piv and i <= j:
i += 1
# move left if greater/eq than/to piv
while A[j] >= piv and j >= i:
j -= 1
# indices stopped moving:
if i < j:
# swap
t = A[i]
A[i] = A[j]
A[j] = t
# place pivot back in the right place
# all values < pivot are to its left and
# all values > pivot are to its right
A[b] = A[i]
A[i] = piv
return i
def IpQuickSort (A, a, b):
while a < b:
p = partition(A, a, b) # p is pivot's location
#sort the smaller partition
if p - a < b - p:
IpQuickSort(A,a,p-1)
a = p + 1 # partition less than p is sorted
else:
IpQuickSort(A,p+1,b)
b = p - 1 # partition greater than p is sorted
def main():
A = [12,3,5,4,7,3,1,3]
print A
IpQuickSort(A,0,len(A)-1)
print A
if __name__ == "__main__": main()
def quick_sort(self, nums):
def helper(arr):
if len(arr) <= 1: return arr
#lwall is the index of the first element euqal to pivot
#rwall is the index of the first element greater than pivot
#so arr[lwall:rwall] is exactly the middle part equal to pivot after one round
lwall, rwall, pivot = 0, 0, 0
#choose rightmost as pivot
pivot = arr[-1]
for i, e in enumerate(arr):
if e < pivot:
#when element is less than pivot, shift the whole middle part to the right by 1
arr[i], arr[lwall] = arr[lwall], arr[i]
lwall += 1
arr[i], arr[rwall] = arr[rwall], arr[i]
rwall += 1
elif e == pivot:
#when element equals to pivot, middle part should increase by 1
arr[i], arr[rwall] = arr[rwall], arr[i]
rwall += 1
elif e > pivot: continue
return helper(arr[:lwall]) + arr[lwall:rwall] + helper(arr[rwall:])
return helper(nums)
I know many people have answered this question correctly and I enjoyed reading them. My answer is almost the same as zangw but I think the previous contributors did not do a good job of explaining visually how things actually work...so here is my attempt to help others that might visit this question/answers in the future about a simple solution for quicksort implementation.
How does it work ?
We basically select the first item as our pivot from our list and then we create two sub lists.
Our first sublist contains the items that are less than pivot
Our second sublist contains our items that are great than or equal to pivot
We then quick sort each of those and we combine them the first group + pivot + the second group to get the final result.
Here is an example along with visual to go with it ...
(pivot)9,11,2,0
average: n log of n
worse case: n^2
The code:
def quicksort(data):
if (len(data) < 2):
return data
else:
pivot = data[0] # pivot
#starting from element 1 to the end
rest = data[1:]
low = [each for each in rest if each < pivot]
high = [each for each in rest if each >= pivot]
return quicksort(low) + [pivot] + quicksort(high)
items=[9,11,2,0]
print(quicksort(items))
The algorithm contains two boundaries, one having elements less than the pivot (tracked by index "j") and the other having elements greater than the pivot (tracked by index "i").
In each iteration, a new element is processed by incrementing j.
Invariant:-
all elements between pivot and i are less than the pivot, and
all elements between i and j are greater than the pivot.
If the invariant is violated, ith and jth elements are swapped, and i
is incremented.
After all elements have been processed, and everything after the pivot
has been partitioned, the pivot element is swapped with the last element
smaller than it.
The pivot element will now be in its correct place in the sequence. The
elements before it will be less than it and the ones after it will be
greater than it, and they will be unsorted.
def quicksort(sequence, low, high):
if low < high:
pivot = partition(sequence, low, high)
quicksort(sequence, low, pivot - 1)
quicksort(sequence, pivot + 1, high)
def partition(sequence, low, high):
pivot = sequence[low]
i = low + 1
for j in range(low + 1, high + 1):
if sequence[j] < pivot:
sequence[j], sequence[i] = sequence[i], sequence[j]
i += 1
sequence[i-1], sequence[low] = sequence[low], sequence[i-1]
return i - 1
def main(sequence):
quicksort(sequence, 0, len(sequence) - 1)
return sequence
if __name__ == '__main__':
sequence = [-2, 0, 32, 1, 56, 99, -4]
print(main(sequence))
Selecting a pivot
A "good" pivot will result in two sub-sequences of roughly the same
size. Deterministically, a pivot element can either be selected in a
naive manner or by computing the median of the sequence.
A naive implementation of selecting a pivot will be the first or last
element. The worst-case runtime in this case will be when the input
sequence is already sorted or reverse sorted, as one of the subsequences
will be empty which will cause only one element to be removed per
recursive call.
A perfectly balanced split is achieved when the pivot is the median
element of the sequence. There are an equal number of elements greater
than it and less than it. This approach guarantees a better overall
running time, but is much more time-consuming.
A non-deterministic/random way of selecting the pivot would be to pick
an element uniformly at random. This is a simple and lightweight
approach that will minimize worst-case scenario and also lead to a
roughly balanced split. This will also provide a balance between the naive approach and the median approach of selecting the pivot.
def quicksort(array):
if len(array) < 2:
return array
else:
pivot = array[0]
less = [i for i in array[1:] if i <= pivot]
greater = [i for i in array[1:] if i > pivot]
return quicksort(less) + [pivot] + quicksort(greater)
def quick_sort(array):
return quick_sort([x for x in array[1:] if x < array[0]]) + [array[0]] \
+ quick_sort([x for x in array[1:] if x >= array[0]]) if array else []
def Partition(A,p,q):
i=p
x=A[i]
for j in range(p+1,q+1):
if A[j]<=x:
i=i+1
tmp=A[j]
A[j]=A[i]
A[i]=tmp
l=A[p]
A[p]=A[i]
A[i]=l
return i
def quickSort(A,p,q):
if p<q:
r=Partition(A,p,q)
quickSort(A,p,r-1)
quickSort(A,r+1,q)
return A
The algorithm has 4 simple steps:
Divide the array into 3 different parts: left, pivot and right, where pivot will have only one element. Let us choose this pivot element as the first element of array
Append elements to the respective part by comparing them to pivot element. (explanation in comments)
Recurse this algorithm till all elements in the array have been sorted
Finally, join left+pivot+right parts
Code for the algorithm in python:
def my_sort(A):
p=A[0] #determine pivot element.
left=[] #create left array
right=[] #create right array
for i in range(1,len(A)):
#if cur elem is less than pivot, add elem in left array
if A[i]< p:
left.append(A[i])
#the recurssion will occur only if the left array is atleast half the size of original array
if len(left)>1 and len(left)>=len(A)//2:
left=my_sort(left) #recursive call
elif A[i]>p:
right.append(A[i]) #if elem is greater than pivot, append it to right array
if len(right)>1 and len(right)>=len(A)//2: # recurssion will occur only if length of right array is atleast the size of original array
right=my_sort(right)
A=left+[p]+right #append all three part of the array into one and return it
return A
my_sort([12,4,5,6,7,3,1,15])
Carry on with this algorithm recursively with the left and right parts.
Another quicksort implementation:
# A = Array
# s = start index
# e = end index
# p = pivot index
# g = greater than pivot boundary index
def swap(A,i1,i2):
A[i1], A[i2] = A[i2], A[i1]
def partition(A,g,p):
# O(n) - just one for loop that visits each element once
for j in range(g,p):
if A[j] <= A[p]:
swap(A,j,g)
g += 1
swap(A,p,g)
return g
def _quicksort(A,s,e):
# Base case - we are sorting an array of size 1
if s >= e:
return
# Partition current array
p = partition(A,s,e)
_quicksort(A,s,p-1) # Left side of pivot
_quicksort(A,p+1,e) # Right side of pivot
# Wrapper function for the recursive one
def quicksort(A):
_quicksort(A,0,len(A)-1)
A = [3,1,4,1,5,9,2,6,5,3,5,8,9,7,9,3,2,3,-1]
print(A)
quicksort(A)
print(A)
For Version Python 3.x: a functional-style using operator module, primarily to improve readability.
from operator import ge as greater, lt as lesser
def qsort(L):
if len(L) <= 1: return L
pivot = L[0]
sublist = lambda op: [*filter(lambda num: op(num, pivot), L[1:])]
return qsort(sublist(lesser))+ [pivot] + qsort(sublist(greater))
and is tested as
print (qsort([3,1,4,2,5]) == [1,2,3,4,5])
Here's an easy implementation:-
def quicksort(array):
if len(array) < 2:
return array
else:
pivot= array[0]
less = [i for i in array[1:] if i <= pivot]
greater = [i for i in array[1:] if i > pivot]
return quicksort(less) + [pivot] + quicksort(greater)
print(quicksort([10, 5, 2, 3]))
My answer is very similar to the great one from #alisianoi . However, I believe there is a slight inefficiency in his code (see my comment), which I removed. Moreover, I added more explanation and was a bit more specific about the problem of duplicate (pivot) values.
def quicksort(nums, begin=0, end=None):
# Only at the beginning end=None. In this case set to len(nums)-1
if end is None: end = len(nums) - 1
# If list part is invalid or has only 1 element, do nothing
if begin>=end: return
# Pick random pivot
pivot = nums[random.randint(begin, end)]
# Initialize left and right pointers
left, right = begin, end
while left < right:
# Find first "wrong" value from left hand side, i.e. first value >= pivot
# Find first "wrong" value from right hand side, i.e. first value <= pivot
# Note: In the LAST while loop, both left and right will point to pivot!
while nums[left] < pivot: left += 1
while nums[right] > pivot: right -= 1
# Swap the "wrong" values
if left != right:
nums[left], nums[right] = nums[right], nums[left]
# Problem: loop can get stuck if pivot value exists more than once. Simply solve with...
if nums[left] == nums[right]:
assert nums[left]==pivot
left += 1
# Now, left and right both point to a pivot value.
# All values to its left are smaller (or equal in case of duplicate pivot values)
# All values to its right are larger.
assert left == right and nums[left] == pivot
quicksort(nums, begin, left - 1)
quicksort(nums, left + 1, end)
return
Without recursion:
def quicksort(nums, ranges=None):
if ranges is None:
ranges = [[0, len(nums) - 1]]
while ranges != []:
[start, end] = ranges[0]
ranges = ranges[1:]
if start >= end:
continue
pivot = nums[randint(start, end)]
left = start
right = end
while left < right:
while nums[left] < pivot:
left += 1
while nums[right] > pivot:
right -= 1
if left != right:
nums[left], nums[right] = nums[right], nums[left]
if nums[left] == nums[right]:
left += 1
ranges = [[start, left - 1], [left + 1, end]] + ranges
First we declare the first value in the array to be the
pivot_value and we also set the left and right marks
We create the first while loop, this while loop is there to tell
the partition process to run again if it doesn't satisfy the
necessary condition
then we apply the partition process
after both partition processes have ran, we check to see if it
satisfies the proper condition. If it does, we mark it as done,
if not we switch the left and right values and apply it again
Once its done switch the left and right values and return the
split_point
I am attaching the code below! This quicksort is a great learning tool because of the Location of the pivot value. Since it is in a constant place, you can walk through it multiple times and really get a hang of how it all works. In practice it is best to randomize the pivot to avoid O(N^2) runtime.
def quicksort10(alist):
quicksort_helper10(alist, 0, len(alist)-1)
def quicksort_helper10(alist, first, last):
""" """
if first < last:
split_point = partition10(alist, first, last)
quicksort_helper10(alist, first, split_point - 1)
quicksort_helper10(alist, split_point + 1, last)
def partition10(alist, first, last):
done = False
pivot_value = alist[first]
leftmark = first + 1
rightmark = last
while not done:
while leftmark <= rightmark and alist[leftmark] <= pivot_value:
leftmark = leftmark + 1
while leftmark <= rightmark and alist[rightmark] >= pivot_value:
rightmark = rightmark - 1
if leftmark > rightmark:
done = True
else:
temp = alist[leftmark]
alist[leftmark] = alist[rightmark]
alist[rightmark] = temp
temp = alist[first]
alist[first] = alist[rightmark]
alist[rightmark] = temp
return rightmark
def quick_sort(l):
if len(l) == 0:
return l
pivot = l[0]
pivots = [x for x in l if x == pivot]
smaller = quick_sort([x for x in l if x < pivot])
larger = quick_sort([x for x in l if x > pivot])
return smaller + pivots + larger
Full example with printed variables at partition step:
def partition(data, p, right):
print("\n==> Enter partition: p={}, right={}".format(p, right))
pivot = data[right]
print("pivot = data[{}] = {}".format(right, pivot))
i = p - 1 # this is a dangerous line
for j in range(p, right):
print("j: {}".format(j))
if data[j] <= pivot:
i = i + 1
print("new i: {}".format(i))
print("swap: {} <-> {}".format(data[i], data[j]))
data[i], data[j] = data[j], data[i]
print("swap2: {} <-> {}".format(data[i + 1], data[right]))
data[i + 1], data[right] = data[right], data[i + 1]
return i + 1
def quick_sort(data, left, right):
if left < right:
pivot = partition(data, left, right)
quick_sort(data, left, pivot - 1)
quick_sort(data, pivot + 1, right)
data = [2, 8, 7, 1, 3, 5, 6, 4]
print("Input array: {}".format(data))
quick_sort(data, 0, len(data) - 1)
print("Output array: {}".format(data))
def is_sorted(arr): #check if array is sorted
for i in range(len(arr) - 2):
if arr[i] > arr[i + 1]:
return False
return True
def qsort_in_place(arr, left, right): #arr - given array, #left - first element index, #right - last element index
if right - left < 1: #if we have empty or one element array - nothing to do
return
else:
left_point = left #set left pointer that points on element that is candidate to swap with element under right pointer or pivot element
right_point = right - 1 #set right pointer that is candidate to swap with element under left pointer
while left_point <= right_point: #while we have not checked all elements in the given array
swap_left = arr[left_point] >= arr[right] #True if we have to move that element after pivot
swap_right = arr[right_point] < arr[right] #True if we have to move that element before pivot
if swap_left and swap_right: #if both True we can swap elements under left and right pointers
arr[right_point], arr[left_point] = arr[left_point], arr[right_point]
left_point += 1
right_point -= 1
else: #if only one True we don`t have place for to swap it
if not swap_left: #if we dont need to swap it we move to next element
left_point += 1
if not swap_right: #if we dont need to swap it we move to prev element
right_point -= 1
arr[left_point], arr[right] = arr[right], arr[left_point] #swap left element with pivot
qsort_in_place(arr, left, left_point - 1) #execute qsort for left part of array (elements less than pivot)
qsort_in_place(arr, left_point + 1, right) #execute qsort for right part of array (elements most than pivot)
def main():
import random
arr = random.sample(range(1, 4000), 10) #generate random array
print(arr)
print(is_sorted(arr))
qsort_in_place(arr, 0, len(arr) - 1)
print(arr)
print(is_sorted(arr))
if __name__ == "__main__":
main()
This algorithm doesn't use recursive functions.
Let N be any list of numbers with len(N) > 0. Set K = [N] and execute the following program.
Note: This is a stable sorting algorithm.
def BinaryRip2Singletons(K, S):
K_L = []
K_P = [ [K[0][0]] ]
K_R = []
for i in range(1, len(K[0])):
if K[0][i] < K[0][0]:
K_L.append(K[0][i])
elif K[0][i] > K[0][0]:
K_R.append(K[0][i])
else:
K_P.append( [K[0][i]] )
K_new = [K_L]*bool(len(K_L)) + K_P + [K_R]*bool(len(K_R)) + K[1:]
while len(K_new) > 0:
if len(K_new[0]) == 1:
S.append(K_new[0][0])
K_new = K_new[1:]
else:
break
return K_new, S
N = [16, 19, 11, 15, 16, 10, 12, 14, 4, 10, 5, 2, 3, 4, 7, 1]
K = [ N ]
S = []
print('K =', K, 'S =', S)
while len(K) > 0:
K, S = BinaryRip2Singletons(K, S)
print('K =', K, 'S =', S)
PROGRAM OUTPUT:
K = [[16, 19, 11, 15, 16, 10, 12, 14, 4, 10, 5, 2, 3, 4, 7, 1]] S = []
K = [[11, 15, 10, 12, 14, 4, 10, 5, 2, 3, 4, 7, 1], [16], [16], [19]] S = []
K = [[10, 4, 10, 5, 2, 3, 4, 7, 1], [11], [15, 12, 14], [16], [16], [19]] S = []
K = [[4, 5, 2, 3, 4, 7, 1], [10], [10], [11], [15, 12, 14], [16], [16], [19]] S = []
K = [[2, 3, 1], [4], [4], [5, 7], [10], [10], [11], [15, 12, 14], [16], [16], [19]] S = []
K = [[5, 7], [10], [10], [11], [15, 12, 14], [16], [16], [19]] S = [1, 2, 3, 4, 4]
K = [[15, 12, 14], [16], [16], [19]] S = [1, 2, 3, 4, 4, 5, 7, 10, 10, 11]
K = [[12, 14], [15], [16], [16], [19]] S = [1, 2, 3, 4, 4, 5, 7, 10, 10, 11]
K = [] S = [1, 2, 3, 4, 4, 5, 7, 10, 10, 11, 12, 14, 15, 16, 16, 19]
Related
I would like to write a function to process a list of integers, best way is to show as an example:
input [0,1,2,3, -1,-2,-3, 0,1,2,3, -1,-2,-3] will return [6,-6,6,-6]
I have a draft here that will actually work:
def group_pos_neg_list(nums):
p_nums = []
# to determine if the first element >=0 or <0
# create pos_combined and neg_combined as a list to check the length in the future
if nums[0] >= 0:
pos_combined, neg_combined = [nums[0]], []
elif nums[0] < 0:
pos_combined, neg_combined = [], [nums[0]]
# loop over each element from position 1 to the end
# accumulate pos num and neg nums and set back to 0 if next element is different
index = 1
while index < len(nums):
if nums[index] >= 0 and nums[index-1] >= 0: # both posivite
pos_combined.append(nums[index])
index += 1
elif nums[index] < 0 and nums[index-1] < 0: # both negative
neg_combined.append(nums[index])
index += 1
else:
if len(pos_combined) > 0:
p_nums.append(sum(pos_combined))
pos_combined, neg_combined = [], [nums[index]]
elif len(neg_combined) > 0:
p_nums.append(sum(neg_combined))
pos_combined, neg_combined = [nums[index]], []
index += 1
# finish the last combined group
if len(pos_combined) > 0:
p_nums.append(sum(pos_combined))
elif len(neg_combined) > 0:
p_nums.append(sum(neg_combined))
return p_nums
But I am not quite happy with it, because it looks a bit complicate.
Especially that there is a repeating part of code:
if len(pos_combined) > 0:
p_nums.append(sum(pos_combined))
pos_combined, neg_combined = [], [nums[index]]
elif len(neg_combined) > 0:
p_nums.append(sum(neg_combined))
pos_combined, neg_combined = [nums[index]], []
I have to write this twice as the final group of integers will not be counted in the loop, so an extra step is needed.
Is there anyway to simplify this?
Using groupby
No need to make it that complex: we can first groupby the signum, and then we can calculate the sum, so:
from itertools import groupby
[sum(g) for _, g in groupby(data, lambda x: x >= 0)]
This then produces:
>>> from itertools import groupby
>>> data = [0,1,2,3, -1,-2,-3, 0,1,2,3, -1,-2,-3]
>>> [sum(g) for _, g in groupby(data, lambda x: x >= 0)]
[6, -6, 6, -6]
So groupby produces tuples with the "key" (the part we calculate with the lambda), and an iterable of the "burst" (a continuous subsequence of elements with the same key). We are only interested in the latter g, and then calculate sum(g) and add that to the list.
Custom made algorithm
We can also write our own version, by using:
swap_idx = [0]
swap_idx += [i+1 for i, (v1, v2) in enumerate(zip(data, data[1:]))
if (v1 >= 0) != (v2 >= 0)]
swap_idx.append(None)
our_sums = [sum(data[i:j]) for i, j in zip(swap_idx, swap_idx[1:])]
Here we first construct a list swap_idx that stores the indices where of the element where the signum changes. So for your sample code that is:
>>> swap_idx
[0, 4, 7, 11, None]
The 0 and None are added by the code explicitly. So now that we identified the points where the sign has changed, we can sum these subsequences together, with sum(data[i:j]). We thus use zip(swap_idx, swap_idx[1:]) to obtain two consecutive indices, and thus we can then sum that slice together.
More verbose version
The above is not very readable: yes it works, but it requires some reasoning. We can also produce a more verbose version, and make it even more generic, for example:
def groupby_aggregate(iterable, key=lambda x: x, aggregate=list):
itr = iter(iterable)
nx = next(itr)
kx = kxcur = key(nx)
current = [nx]
try:
while True:
nx = next(itr)
kx = key(nx)
if kx != kxcur:
yield aggregate(current)
current = [nx]
kxcur = kx
else:
current.append(nx)
except StopIteration:
yield aggregate(current)
We can then use it like:
list(groupby_aggregate(data, lambda x: x >= 0, sum))
You can use itertools.groupby, utilizing a key to group by all the values greater than or equal to zero:
import itertools
s = [0,1,2,3, -1,-2,-3, 0,1,2,3, -1,-2,-3]
new_s = [sum(b) for a, b in itertools.groupby(s, key=lambda x: x >=0)]
Output:
[6, -6, 6, -6]
Here is a way to do without any external imports, only using reduce():
def same_sign(a, b):
"""Returns True if a and b have the same sign"""
return (a*b>0) or (a>=0 and b>=0)
l = [0,1,2,3, -1,-2,-3, 0,1,2,3, -1,-2,-3]
reduce(
lambda x, y: (x+y if same_sign(x,y) else [x, y]) if not isinstance(x, list) else x[:-1] + [x[-1] + y] if same_sign(x[-1],y) else x + [y],
l
)
#[6, -6, 6, -6]
Explanation
This is a bit hard to explain, but I'll try.
From the docs calling reduce() will:
Apply function of two arguments cumulatively to the items of iterable, from left to right
In this case I take two values (x and y) from your list and do the following:
If x is not a list:
If x and y have the same sign (product >=0), sum them
Otherwise return a list [x, y]
If x is a list, only modify the last element of x.
If the signs match, add y.
Otherwise append a new element to the list x
Note
You probably shouldn't do it this way because the code is hard to read and understand. I just wanted to show that it was possible.
Update
A more readable version of the same code above:
def same_sign(a, b):
"""Returns True if a and b have the same sign"""
return (a*b>0) or (a>=0 and b>=0)
l = [0,1,2,3, -1,-2,-3, 0,1,2,3, -1,-2,-3]
def reducer(x, y):
if isinstance(x, list):
if same_sign(x[-1], y):
return x[:-1] + [x[-1] + y]
else:
return x + [y]
else:
if same_sign(x, y):
return x+y
else:
return [x, y]
reduce(reducer, l)
#[6, -6, 6, -6]
I want to define a recursive function can sort any list of ints:
def sort_l(l):
if l==[]:
return []
else:
if len(l)==1:
return [l[-1]]
elif l[0]<l[1]:
return [l[0]]+sort_l(l[1:])
else:
return sort_l(l[1:])+[l[0]]
Calling this function on a list [3, 1, 2,4,7,5,6,9,8] should give me:
[1,2,3,4,5,6,7,8,9]
But I get:
print(sort_l([3, 1, 2,4,7,5,6,9,8]))--> [1, 2, 4, 5, 6, 8, 9, 7, 3]
Please help me to fix the problem, actual code would be appreciated. Thanks!
The quick sort is recursive and easy to implement in Python:
def quick_sort(l):
if len(l) <= 1:
return l
else:
return quick_sort([e for e in l[1:] if e <= l[0]]) + [l[0]] +\
quick_sort([e for e in l[1:] if e > l[0]])
will give:
>>> quick_sort([3, 1, 2, 4, 7, 5, 6, 9, 8])
[1, 2, 3, 4, 5, 6, 7, 8, 9]
For this you would want to use merge sort. Essentially in a merge sort you recursively split the list in half until you have single elements and than build it back up in the correct order. merge sort on has a complexity of O(n log(n)) and is an extremely stable sorting method.
Here are some good in depth explanations and visuals for merge sorting:
https://www.youtube.com/watch?v=vxENKlcs2Tw
http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Merge_sort.html
def maximum(lis):
if len(lis) == 1:
return lis[0]
return maximum(lis[1:]) if lis[0] < lis[1] else maximum(lis[:1] + lis[2:])
def sorter(lis):
if len(lis) == 1:
return lis
x = maximum(lis)
lis.remove(x)
return sorter(lis) + [x]
with functional programming:
sor = lambda lis: lis if len(lis) == 1 else [lis.pop(lis.index(reduce(lambda x, y: x if x > y else y, lis)))] + sor(lis)
def quicksort(lst):
"Quicksort over a list-like sequence"
if len(lst) == 0:
return lst
pivot = lst[0]
pivots = [x for x in lst if x == pivot]
small = quicksort([x for x in lst if x < pivot])
large = quicksort([x for x in lst if x > pivot])
return small + pivots + large
Above is a more readable recursive implementation of Quick Sort Algorithm. Above piece of code is from book Functional programing in python by O'REILLY.
Above function will produce.
list=[9,8,7,6,5,4]
quicksort(list)
>>[4,5,6,7,8,9]
def sort(array, index = 0, bigNumber = 0):
if len(array) == index:
return array
elif bigNumber > array[index]:
array[index - 1] = array[index]
array[index] = bigNumber
bigNumber = array[0]
index = 0
else:
bigNumber = array[index]
return sort(array, (index + 1), bigNumber)
#sort an int list using recursion
global array
array=[5,3,8,4,2,6,1]
def sort1(array:[])->[]:
if len(array)==1:
return
temp=array[-1]
array.pop()
sort1(array)
sort2(array,temp)
def sort2(array:[],temp):
if len(array)==0 or temp>=array[-1]:
array.append(temp)
return
a=array[-1]
array.pop()
sort2(array,temp)
array.append(a)
sort1(array)
print(array)
Here i am explaining recursive approach to sort a list. we can follow "Induction Base-condition Hypothesis" recursion approach. so basically we consider our hypothesis here sort_l(nums) function which sorts for given list and Base condition will be found when we have singly number list available which is already sorted. Now in induction step, we insert the temp element (last element of list) in the correct position of given list.
example-
sort_l([1,5,0,2]) will make below recursively call
sort_l([1]) <-- 5 (here you need to insert 5 in correct position)
sort_l([1,5]) <-- 0 (here you need to insert 0 in correct position)
sort_l([0,1,5]) <-- 2 (here you need to insert 2 in correct position)
sort_l([0,1,5,2]) Finally it will be in sorted list.
====== Below is working code=======
def insert_element(nums, temp):
if len(nums) == 1:
if nums[0] > temp:
nums.insert(0, temp)
elif nums[0] < temp:
nums.append(temp)
else:
for i in range(len(nums)):
if nums[i] > temp:
nums.insert(i, temp)
break
if nums[-1] < temp:
nums.append(temp)
def sort_l(nums): ## hypothesis
if len(nums)==1: ## base condition
return nums
temp = nums[-1]
nums.pop()
sort_l(nums)
insert_element(nums, temp) ## induction
return nums
This is a complementary answer since both quicksort and complexity are already covered in previous answers. Although, I believe an easy-to-get sort function that covers Python's identity is missing*.
def sort(xs: list) -> list:
if not xs:
return xs
else:
xs.remove(num := min(xs))
return [num] + sort(xs)
[*] Python is a (slow) interpreted language, but it has become famous because of its readability and easiness to learn. It doesn't really "reward" its developers for using immutable objects nor it is a language that should be used for computation intensive applications
This is a recursive solution. For an explanation, refer to this video:
arr = [3,1,2,4,7,5,6,9,8]
def insert_fn(arr, temp): # Hypothesis
if len(arr) == 0 or arr[-1] <= temp: # Base - condition
arr.append(temp)
return arr
# Induction
val = arr[-1]
arr.pop()
insert_fn(arr, temp) # Call function on a smaller input.
arr.append(val) # Induction step
return arr
def sort_fn(arr): # Hypothesis
if len(arr) == 1: # Base - condition
return arr
# Induction
val = arr[-1]
arr.pop()
sort_fn(arr) # Call function on a smaller input.
insert_fn(arr, val) # Induction step
return arr
print(sort_fn(arr))
In class we are doing sorting algorithms and, although I understand them fine when talking about them and writing pseudocode, I am having problems writing actual code for them.
This is my attempt in Python:
mylist = [12, 5, 13, 8, 9, 65]
def bubble(badList):
length = len(badList) - 1
unsorted = True
while unsorted:
for element in range(0,length):
unsorted = False
if badList[element] > badList[element + 1]:
hold = badList[element + 1]
badList[element + 1] = badList[element]
badList[element] = hold
print badList
else:
unsorted = True
print bubble(mylist)
Now, this (as far as I can tell) sorts correctly, but once it finishes it just loops indefinitely.
How can this code be fixed so the function finishes properly and correctly sorts a list of any (reasonable) size?
P.S. I know I should not really have prints in a function and I should have a return, but I just have not done that yet as my code does not really work yet.
To explain why your script isn't working right now, I'll rename the variable unsorted to sorted.
At first, your list isn't yet sorted. Of course, we set sorted to False.
As soon as we start the while loop, we assume that the list is already sorted. The idea is this: as soon as we find two elements that are not in the right order, we set sorted back to False. sorted will remain True only if there were no elements in the wrong order.
sorted = False # We haven't started sorting yet
while not sorted:
sorted = True # Assume the list is now sorted
for element in range(0, length):
if badList[element] > badList[element + 1]:
sorted = False # We found two elements in the wrong order
hold = badList[element + 1]
badList[element + 1] = badList[element]
badList[element] = hold
# We went through the whole list. At this point, if there were no elements
# in the wrong order, sorted is still True. Otherwise, it's false, and the
# while loop executes again.
There are also minor little issues that would help the code be more efficient or readable.
In the for loop, you use the variable element. Technically, element is not an element; it's a number representing a list index. Also, it's quite long. In these cases, just use a temporary variable name, like i for "index".
for i in range(0, length):
The range command can also take just one argument (named stop). In that case, you get a list of all the integers from 0 to that argument.
for i in range(length):
The Python Style Guide recommends that variables be named in lowercase with underscores. This is a very minor nitpick for a little script like this; it's more to get you accustomed to what Python code most often resembles.
def bubble(bad_list):
To swap the values of two variables, write them as a tuple assignment. The right hand side gets evaluated as a tuple (say, (badList[i+1], badList[i]) is (3, 5)) and then gets assigned to the two variables on the left hand side ((badList[i], badList[i+1])).
bad_list[i], bad_list[i+1] = bad_list[i+1], bad_list[i]
Put it all together, and you get this:
my_list = [12, 5, 13, 8, 9, 65]
def bubble(bad_list):
length = len(bad_list) - 1
sorted = False
while not sorted:
sorted = True
for i in range(length):
if bad_list[i] > bad_list[i+1]:
sorted = False
bad_list[i], bad_list[i+1] = bad_list[i+1], bad_list[i]
bubble(my_list)
print my_list
(I removed your print statement too, by the way.)
The goal of bubble sort is to move the heavier items at the bottom in each round, while moving the lighter items up. In the inner loop, where you compare the elements, you don't have to iterate the whole list in each turn. The heaviest is already placed last. The swapped variable is an extra check so we can mark that the list is now sorted and avoid continuing with unnecessary calculations.
def bubble(badList):
length = len(badList)
for i in range(0,length):
swapped = False
for element in range(0, length-i-1):
if badList[element] > badList[element + 1]:
hold = badList[element + 1]
badList[element + 1] = badList[element]
badList[element] = hold
swapped = True
if not swapped: break
return badList
Your version 1, corrected:
def bubble(badList):
length = len(badList) - 1
unsorted = True
while unsorted:
unsorted = False
for element in range(0,length):
#unsorted = False
if badList[element] > badList[element + 1]:
hold = badList[element + 1]
badList[element + 1] = badList[element]
badList[element] = hold
unsorted = True
#print badList
#else:
#unsorted = True
return badList
This is what happens when you use variable name of negative meaning, you need to invert their values. The following would be easier to understand:
sorted = False
while not sorted:
...
On the other hand, the logic of the algorithm is a little bit off. You need to check whether two elements swapped during the for loop. Here's how I would write it:
def bubble(values):
length = len(values) - 1
sorted = False
while not sorted:
sorted = True
for element in range(0,length):
if values[element] > values[element + 1]:
hold = values[element + 1]
values[element + 1] = values[element]
values[element] = hold
sorted = False
return values
Your use of the Unsorted variable is wrong; you want to have a variable that tells you if you have swapped two elements; if you have done that, you can exit your loop, otherwise, you need to loop again. To fix what you've got here, just put the "unsorted = false" in the body of your if case; remove your else case; and put "unsorted = true before your for loop.
def bubble_sort(l):
for passes_left in range(len(l)-1, 0, -1):
for index in range(passes_left):
if l[index] < l[index + 1]:
l[index], l[index + 1] = l[index + 1], l[index]
return l
#A very simple function, can be optimized (obviously) by decreasing the problem space of the 2nd array. But same O(n^2) complexity.
def bubble(arr):
l = len(arr)
for a in range(l):
for b in range(l-1):
if (arr[a] < arr[b]):
arr[a], arr[b] = arr[b], arr[a]
return arr
You've got a couple of errors in there. The first is in length, and the second is in your use of unsorted (as stated by McWafflestix). You probably also want to return the list if you're going to print it:
mylist = [12, 5, 13, 8, 9, 65]
def bubble(badList):
length = len(badList) - 2
unsorted = True
while unsorted:
for element in range(0,length):
unsorted = False
if badList[element] > badList[element + 1]:
hold = badList[element + 1]
badList[element + 1] = badList[element]
badList[element] = hold
print badList
unsorted = True
return badList
print bubble(mylist)
eta: You're right, the above is buggy as hell. My bad for not testing through some more examples.
def bubble2(badList):
swapped = True
length = len(badList) - 2
while swapped:
swapped = False
for i in range(0, length):
if badList[i] > badList[i + 1]:
# swap
hold = badList[i + 1]
badList[i + 1] = badList[i]
badList[i] = hold
swapped = True
return badList
I am a fresh fresh beginner, started to read about Python yesterday.
Inspired by your example I created something maybe more in the 80-ties style, but nevertheless it kinda works
lista1 = [12, 5, 13, 8, 9, 65]
i=0
while i < len(lista1)-1:
if lista1[i] > lista1[i+1]:
x = lista1[i]
lista1[i] = lista1[i+1]
lista1[i+1] = x
i=0
continue
else:
i+=1
print(lista1)
The problem with the original algorithm is that if you had a lower number further in the list, it would not bring it to the correct sorted position. The program needs to go back the the beginning each time to ensure that the numbers sort all the way through.
I simplified the code and it will now work for any list of numbers regardless of the list and even if there are repeating numbers. Here's the code
mylist = [9, 8, 5, 4, 12, 1, 7, 5, 2]
print mylist
def bubble(badList):
length = len(badList) - 1
element = 0
while element < length:
if badList[element] > badList[element + 1]:
hold = badList[element + 1]
badList[element + 1] = badList[element]
badList[element] = hold
element = 0
print badList
else:
element = element + 1
print bubble(mylist)
def bubble_sort(l):
exchanged = True
iteration = 0
n = len(l)
while(exchanged):
iteration += 1
exchanged = False
# Move the largest element to the end of the list
for i in range(n-1):
if l[i] > l[i+1]:
exchanged = True
l[i], l[i+1] = l[i+1], l[i]
n -= 1 # Largest element already towards the end
print 'Iterations: %s' %(iteration)
return l
def bubbleSort(alist):
if len(alist) <= 1:
return alist
for i in range(0,len(alist)):
print "i is :%d",i
for j in range(0,i):
print "j is:%d",j
print "alist[i] is :%d, alist[j] is :%d"%(alist[i],alist[j])
if alist[i] > alist[j]:
alist[i],alist[j] = alist[j],alist[i]
return alist
alist = [54,26,93,17,77,31,44,55,20,-23,-34,16,11,11,11]
print bubbleSort(alist)
def bubble_sort(a):
t = 0
sorted = False # sorted = False because we have not began to sort
while not sorted:
sorted = True # Assume sorted = True first, it will switch only there is any change
for key in range(1,len(a)):
if a[key-1] > a[key]:
sorted = False
t = a[key-1]; a[key-1] = a[key]; a[key] = t;
print a
A simpler example:
a = len(alist)-1
while a > 0:
for b in range(0,a):
#compare with the adjacent element
if alist[b]>=alist[b+1]:
#swap both elements
alist[b], alist[b+1] = alist[b+1], alist[b]
a-=1
This simply takes the elements from 0 to a(basically, all the unsorted elements in that round) and compares it with its adjacent element, and making a swap if it is greater than its adjacent element. At the end the round, the last element is sorted, and the process runs again without it, until all elements have been sorted.
There is no need for a condition whether sort is true or not.
Note that this algorithm takes into consideration the position of the numbers only when swapping, so repeated numbers will not affect it.
PS. I know it has been very long since this question was posted, but I just wanted to share this idea.
def bubble_sort(li):
l = len(li)
tmp = None
sorted_l = sorted(li)
while (li != sorted_l):
for ele in range(0,l-1):
if li[ele] > li[ele+1]:
tmp = li[ele+1]
li[ele+1] = li [ele]
li[ele] = tmp
return li
def bubbleSort ( arr ):
swapped = True
length = len ( arr )
j = 0
while swapped:
swapped = False
j += 1
for i in range ( length - j ):
if arr [ i ] > arr [ i + 1 ]:
# swap
tmp = arr [ i ]
arr [ i ] = arr [ i + 1]
arr [ i + 1 ] = tmp
swapped = True
if __name__ == '__main__':
# test list
a = [ 67, 45, 39, -1, -5, -44 ];
print ( a )
bubbleSort ( a )
print ( a )
def bubblesort(array):
for i in range(len(array)-1):
for j in range(len(array)-1-i):
if array[j] > array[j+1]:
array[j], array[j+1] = array[j+1], array[j]
return(array)
print(bubblesort([3,1,6,2,5,4]))
arr = [5,4,3,1,6,8,10,9] # array not sorted
for i in range(len(arr)):
for j in range(i, len(arr)):
if(arr[i] > arr[j]):
arr[i], arr[j] = arr[j], arr[i]
print (arr)
I consider adding my solution because ever solution here is having
greater time
greater space complexity
or doing too much operations
then is should be
So, here is my solution:
def countInversions(arr):
count = 0
n = len(arr)
for i in range(n):
_count = count
for j in range(0, n - i - 1):
if arr[j] > arr[j + 1]:
count += 1
arr[j], arr[j + 1] = arr[j + 1], arr[j]
if _count == count:
break
return count
If anyone is interested in a shorter implementation using a list comprehension:
def bubble_sort(lst: list) -> None:
[swap_items(lst, i, i+1) for left in range(len(lst)-1, 0, -1) for i in range(left) if lst[i] > lst[i+1]]
def swap_items(lst: list, pos1: int, pos2: int) -> None:
lst[pos1], lst[pos2] = lst[pos2], lst[pos1]
Here is a different variation of bubble sort without for loop. Basically you are considering the lastIndex of the array and slowly decrementing it until it first index of the array.
The algorithm will continue to move through the array like this until an entire pass is made without any swaps occurring.
The bubble is sort is basically Quadratic Time: O(n²) when it comes to performance.
class BubbleSort:
def __init__(self, arr):
self.arr = arr;
def bubbleSort(self):
count = 0;
lastIndex = len(self.arr) - 1;
while(count < lastIndex):
if(self.arr[count] > self.arr[count + 1]):
self.swap(count)
count = count + 1;
if(count == lastIndex):
count = 0;
lastIndex = lastIndex - 1;
def swap(self, count):
temp = self.arr[count];
self.arr[count] = self.arr[count + 1];
self.arr[count + 1] = temp;
arr = [9, 1, 5, 3, 8, 2]
p1 = BubbleSort(arr)
print(p1.bubbleSort())
def bubblesort(L,s):
if s >-1 :
bubblesort(L,s-1)
for i in range(len(L)-1-s):
if L[i]>L[i+1]:
temp = L[i+1]
L[i+1] = L[i]
L[i] = temp
return L
Nlist = [3,50,7,1,8,11,9,0,-1,5]
print(bubblesort(Nlist,len(Nlist)))
Answers provided by the-fury and Martin Cote fixed the problem of the infinite loop, but my code would still not work correctly (for a larger list, it would not sort correctly.). I ended up ditching the unsorted variable and used a counter instead.
def bubble(badList):
length = len(badList) - 1
n = 0
while n < len(badList):
for element in range(0,length):
if badList[element] > badList[element + 1]:
hold = badList[element + 1]
badList[element + 1] = badList[element]
badList[element] = hold
n = 0
else:
n += 1
return badList
if __name__ == '__main__':
mylist = [90, 10, 2, 76, 17, 66, 57, 23, 57, 99]
print bubble(mylist)
If anyone could provide any pointers on how to improve my code in the comments, it would be much appreciated.
Try this
a = int(input("Enter Limit"))
val = []
for z in range(0,a):
b = int(input("Enter Number in List"))
val.append(b)
for y in range(0,len(val)):
for x in range(0,len(val)-1):
if val[x]>val[x+1]:
t = val[x]
val[x] = val[x+1]
val[x+1] = t
print(val)
idk if this might help you after 9 years...
its a simple bubble sort program
l=[1,6,3,7,5,9,8,2,4,10]
for i in range(1,len(l)):
for j in range (i+1,len(l)):
if l[i]>l[j]:
l[i],l[j]=l[j],l[i]
def merge_bubble(arr):
k = len(arr)
while k>2:
for i in range(0,k-1):
for j in range(0,k-1):
if arr[j] > arr[j+1]:
arr[j],arr[j+1] = arr[j+1],arr[j]
return arr
break
else:
if arr[0] > arr[1]:
arr[0],arr[1] = arr[1],arr[0]
return arr
def bubble_sort(l):
for i in range(len(l) -1):
for j in range(len(l)-i-1):
if l[j] > l[j+1]:
l[j],l[j+1] = l[j+1], l[j]
return l
def bubble_sorted(arr:list):
while True:
for i in range(0,len(arr)-1):
count = 0
if arr[i] > arr[i+1]:
count += 1
arr[i], arr[i+1] = arr[i+1], arr[i]
if count == 0:
break
return arr
arr = [30,20,80,40,50,10,60,70,90]
print(bubble_sorted(arr))
#[20, 30, 40, 50, 10, 60, 70, 80, 90]
def bubbleSort(a):
def swap(x, y):
temp = a[x]
a[x] = a[y]
a[y] = temp
#outer loop
for j in range(len(a)):
#slicing to the center, inner loop, python style
for i in range(j, len(a) - j):
#find the min index and swap
if a[i] < a[j]:
swap(j, i)
#find the max index and swap
if a[i] > a[len(a) - j - 1]:
swap(len(a) - j - 1, i)
return a
Okay. I give up. I've been trying to implement the median of medians algorithm but I am continually given the wrong result. I know there is a lot of code below, but I can't find my error, and each chunk of code has a fairly process design. Quicksort is what I use to sort the medians I get from the median of medians pivot selection. Should be a straightforward quicksort implementation. getMean simply returns the mean of a given list.
getPivot is what I use to select the pivot. It runs through the list, taking 5 integers at a time, finding the mean of those integers, placing that mean into a list c, then finding the median of c. That is the pivot I use for the dSelect algorithm.
The dSelect algorithm is simple in theory. The base case returns an item when the list is 1 item long. Otherwise, much like in quickSort, I iterate over the list. If the number I am currently on, j, is less than the pivot, I move it to the left of the list, i, and increment i. If it is larger, I move it to the right of the list, i + 1, and do not increment i. After this loops through the entire list, I should have the pivot in its proper index, and print statements indicate that I do. At this point, I recurse to the left or the right depending on whether the pivot is greater than or less than the position I am trying to find.
I am not sure what other print statements to test at this point, so I'm turning to anyone dedicated enough to take a stab at this code. I know there are related topics, I know I could do more print statements, but believe me, I've tried. What should be a simple algo has got me quite stumped.
def quickSort(m, left, right):
if right - left <= 1:
return m
pivot = m[left]
i = left + 1
j = left + 1
for j in range(j, right):
if m[j] < pivot:
m[j], m[i] = m[i], m[j]
i += 1
elif m[j] == pivot:
m[j], m[i] = m[i], m[j]
print m
m[left], m[i-1] = m[i-1], m[left]
m = quickSort(m, left, i-1)
m = quickSort(m, i, right)
print m
return m
def getMedian(list):
length = len(list)
if length <= 1:
return list[0]
elif length % 2 == 0:
i = length/2
return list[i]
else:
i = (length + 1)/2
return list[i]
def getPivot(m):
c = []
i = 0
while i <= len(m) - 1:
tempList = []
j = 0
while j < 5 and i <= len(m) - 1:
tempList.append(m[i])
i = i + 1
j = j + 1
tempList = quickSort(tempList, 0, len(tempList) - 1)
c.append(getMedian(tempList))
c = quickSort(c, 0, len(c) - 1)
medianOfMedians = getMedian(c)
return medianOfMedians
def dSelect(m, position):
pivot = getPivot(m)
i = 0
j = 0
if len(m) <= 1:
return m[0]
for j in range(0, len(m)):
if m[j] < pivot:
m[j], m[i] = m[i], m[j]
i += 1
elif m[j] == pivot:
m[j], m[i] = m[i], m[j]
print "i: " + str(i)
print "j: " + str(j)
print "index of pivot: " + str(m.index(pivot))
print "pivot: " + str(pivot) + " list: " + str(m)
if m.index(pivot) == position:
return pivot
elif m.index(pivot) > position:
return dSelect(m[0:i], position)
else:
return dSelect(m[i:], position - i)
The biggest issue is with this line here:
i = (length + 1)/2
if list = [1, 2, 3, 4, 5, 6, 7] the answer should be 4 which is list[3]. Your version looks like the following:
i = (7 + 1) / 2
and so i is equal to 4 instead of 3. Similar problem with the even length list part although that shouldn't be as big of an issue.
In class we are doing sorting algorithms and, although I understand them fine when talking about them and writing pseudocode, I am having problems writing actual code for them.
This is my attempt in Python:
mylist = [12, 5, 13, 8, 9, 65]
def bubble(badList):
length = len(badList) - 1
unsorted = True
while unsorted:
for element in range(0,length):
unsorted = False
if badList[element] > badList[element + 1]:
hold = badList[element + 1]
badList[element + 1] = badList[element]
badList[element] = hold
print badList
else:
unsorted = True
print bubble(mylist)
Now, this (as far as I can tell) sorts correctly, but once it finishes it just loops indefinitely.
How can this code be fixed so the function finishes properly and correctly sorts a list of any (reasonable) size?
P.S. I know I should not really have prints in a function and I should have a return, but I just have not done that yet as my code does not really work yet.
To explain why your script isn't working right now, I'll rename the variable unsorted to sorted.
At first, your list isn't yet sorted. Of course, we set sorted to False.
As soon as we start the while loop, we assume that the list is already sorted. The idea is this: as soon as we find two elements that are not in the right order, we set sorted back to False. sorted will remain True only if there were no elements in the wrong order.
sorted = False # We haven't started sorting yet
while not sorted:
sorted = True # Assume the list is now sorted
for element in range(0, length):
if badList[element] > badList[element + 1]:
sorted = False # We found two elements in the wrong order
hold = badList[element + 1]
badList[element + 1] = badList[element]
badList[element] = hold
# We went through the whole list. At this point, if there were no elements
# in the wrong order, sorted is still True. Otherwise, it's false, and the
# while loop executes again.
There are also minor little issues that would help the code be more efficient or readable.
In the for loop, you use the variable element. Technically, element is not an element; it's a number representing a list index. Also, it's quite long. In these cases, just use a temporary variable name, like i for "index".
for i in range(0, length):
The range command can also take just one argument (named stop). In that case, you get a list of all the integers from 0 to that argument.
for i in range(length):
The Python Style Guide recommends that variables be named in lowercase with underscores. This is a very minor nitpick for a little script like this; it's more to get you accustomed to what Python code most often resembles.
def bubble(bad_list):
To swap the values of two variables, write them as a tuple assignment. The right hand side gets evaluated as a tuple (say, (badList[i+1], badList[i]) is (3, 5)) and then gets assigned to the two variables on the left hand side ((badList[i], badList[i+1])).
bad_list[i], bad_list[i+1] = bad_list[i+1], bad_list[i]
Put it all together, and you get this:
my_list = [12, 5, 13, 8, 9, 65]
def bubble(bad_list):
length = len(bad_list) - 1
sorted = False
while not sorted:
sorted = True
for i in range(length):
if bad_list[i] > bad_list[i+1]:
sorted = False
bad_list[i], bad_list[i+1] = bad_list[i+1], bad_list[i]
bubble(my_list)
print my_list
(I removed your print statement too, by the way.)
The goal of bubble sort is to move the heavier items at the bottom in each round, while moving the lighter items up. In the inner loop, where you compare the elements, you don't have to iterate the whole list in each turn. The heaviest is already placed last. The swapped variable is an extra check so we can mark that the list is now sorted and avoid continuing with unnecessary calculations.
def bubble(badList):
length = len(badList)
for i in range(0,length):
swapped = False
for element in range(0, length-i-1):
if badList[element] > badList[element + 1]:
hold = badList[element + 1]
badList[element + 1] = badList[element]
badList[element] = hold
swapped = True
if not swapped: break
return badList
Your version 1, corrected:
def bubble(badList):
length = len(badList) - 1
unsorted = True
while unsorted:
unsorted = False
for element in range(0,length):
#unsorted = False
if badList[element] > badList[element + 1]:
hold = badList[element + 1]
badList[element + 1] = badList[element]
badList[element] = hold
unsorted = True
#print badList
#else:
#unsorted = True
return badList
This is what happens when you use variable name of negative meaning, you need to invert their values. The following would be easier to understand:
sorted = False
while not sorted:
...
On the other hand, the logic of the algorithm is a little bit off. You need to check whether two elements swapped during the for loop. Here's how I would write it:
def bubble(values):
length = len(values) - 1
sorted = False
while not sorted:
sorted = True
for element in range(0,length):
if values[element] > values[element + 1]:
hold = values[element + 1]
values[element + 1] = values[element]
values[element] = hold
sorted = False
return values
Your use of the Unsorted variable is wrong; you want to have a variable that tells you if you have swapped two elements; if you have done that, you can exit your loop, otherwise, you need to loop again. To fix what you've got here, just put the "unsorted = false" in the body of your if case; remove your else case; and put "unsorted = true before your for loop.
def bubble_sort(l):
for passes_left in range(len(l)-1, 0, -1):
for index in range(passes_left):
if l[index] < l[index + 1]:
l[index], l[index + 1] = l[index + 1], l[index]
return l
#A very simple function, can be optimized (obviously) by decreasing the problem space of the 2nd array. But same O(n^2) complexity.
def bubble(arr):
l = len(arr)
for a in range(l):
for b in range(l-1):
if (arr[a] < arr[b]):
arr[a], arr[b] = arr[b], arr[a]
return arr
You've got a couple of errors in there. The first is in length, and the second is in your use of unsorted (as stated by McWafflestix). You probably also want to return the list if you're going to print it:
mylist = [12, 5, 13, 8, 9, 65]
def bubble(badList):
length = len(badList) - 2
unsorted = True
while unsorted:
for element in range(0,length):
unsorted = False
if badList[element] > badList[element + 1]:
hold = badList[element + 1]
badList[element + 1] = badList[element]
badList[element] = hold
print badList
unsorted = True
return badList
print bubble(mylist)
eta: You're right, the above is buggy as hell. My bad for not testing through some more examples.
def bubble2(badList):
swapped = True
length = len(badList) - 2
while swapped:
swapped = False
for i in range(0, length):
if badList[i] > badList[i + 1]:
# swap
hold = badList[i + 1]
badList[i + 1] = badList[i]
badList[i] = hold
swapped = True
return badList
I am a fresh fresh beginner, started to read about Python yesterday.
Inspired by your example I created something maybe more in the 80-ties style, but nevertheless it kinda works
lista1 = [12, 5, 13, 8, 9, 65]
i=0
while i < len(lista1)-1:
if lista1[i] > lista1[i+1]:
x = lista1[i]
lista1[i] = lista1[i+1]
lista1[i+1] = x
i=0
continue
else:
i+=1
print(lista1)
The problem with the original algorithm is that if you had a lower number further in the list, it would not bring it to the correct sorted position. The program needs to go back the the beginning each time to ensure that the numbers sort all the way through.
I simplified the code and it will now work for any list of numbers regardless of the list and even if there are repeating numbers. Here's the code
mylist = [9, 8, 5, 4, 12, 1, 7, 5, 2]
print mylist
def bubble(badList):
length = len(badList) - 1
element = 0
while element < length:
if badList[element] > badList[element + 1]:
hold = badList[element + 1]
badList[element + 1] = badList[element]
badList[element] = hold
element = 0
print badList
else:
element = element + 1
print bubble(mylist)
def bubble_sort(l):
exchanged = True
iteration = 0
n = len(l)
while(exchanged):
iteration += 1
exchanged = False
# Move the largest element to the end of the list
for i in range(n-1):
if l[i] > l[i+1]:
exchanged = True
l[i], l[i+1] = l[i+1], l[i]
n -= 1 # Largest element already towards the end
print 'Iterations: %s' %(iteration)
return l
def bubbleSort(alist):
if len(alist) <= 1:
return alist
for i in range(0,len(alist)):
print "i is :%d",i
for j in range(0,i):
print "j is:%d",j
print "alist[i] is :%d, alist[j] is :%d"%(alist[i],alist[j])
if alist[i] > alist[j]:
alist[i],alist[j] = alist[j],alist[i]
return alist
alist = [54,26,93,17,77,31,44,55,20,-23,-34,16,11,11,11]
print bubbleSort(alist)
def bubble_sort(a):
t = 0
sorted = False # sorted = False because we have not began to sort
while not sorted:
sorted = True # Assume sorted = True first, it will switch only there is any change
for key in range(1,len(a)):
if a[key-1] > a[key]:
sorted = False
t = a[key-1]; a[key-1] = a[key]; a[key] = t;
print a
A simpler example:
a = len(alist)-1
while a > 0:
for b in range(0,a):
#compare with the adjacent element
if alist[b]>=alist[b+1]:
#swap both elements
alist[b], alist[b+1] = alist[b+1], alist[b]
a-=1
This simply takes the elements from 0 to a(basically, all the unsorted elements in that round) and compares it with its adjacent element, and making a swap if it is greater than its adjacent element. At the end the round, the last element is sorted, and the process runs again without it, until all elements have been sorted.
There is no need for a condition whether sort is true or not.
Note that this algorithm takes into consideration the position of the numbers only when swapping, so repeated numbers will not affect it.
PS. I know it has been very long since this question was posted, but I just wanted to share this idea.
def bubble_sort(li):
l = len(li)
tmp = None
sorted_l = sorted(li)
while (li != sorted_l):
for ele in range(0,l-1):
if li[ele] > li[ele+1]:
tmp = li[ele+1]
li[ele+1] = li [ele]
li[ele] = tmp
return li
def bubbleSort ( arr ):
swapped = True
length = len ( arr )
j = 0
while swapped:
swapped = False
j += 1
for i in range ( length - j ):
if arr [ i ] > arr [ i + 1 ]:
# swap
tmp = arr [ i ]
arr [ i ] = arr [ i + 1]
arr [ i + 1 ] = tmp
swapped = True
if __name__ == '__main__':
# test list
a = [ 67, 45, 39, -1, -5, -44 ];
print ( a )
bubbleSort ( a )
print ( a )
def bubblesort(array):
for i in range(len(array)-1):
for j in range(len(array)-1-i):
if array[j] > array[j+1]:
array[j], array[j+1] = array[j+1], array[j]
return(array)
print(bubblesort([3,1,6,2,5,4]))
arr = [5,4,3,1,6,8,10,9] # array not sorted
for i in range(len(arr)):
for j in range(i, len(arr)):
if(arr[i] > arr[j]):
arr[i], arr[j] = arr[j], arr[i]
print (arr)
I consider adding my solution because ever solution here is having
greater time
greater space complexity
or doing too much operations
then is should be
So, here is my solution:
def countInversions(arr):
count = 0
n = len(arr)
for i in range(n):
_count = count
for j in range(0, n - i - 1):
if arr[j] > arr[j + 1]:
count += 1
arr[j], arr[j + 1] = arr[j + 1], arr[j]
if _count == count:
break
return count
If anyone is interested in a shorter implementation using a list comprehension:
def bubble_sort(lst: list) -> None:
[swap_items(lst, i, i+1) for left in range(len(lst)-1, 0, -1) for i in range(left) if lst[i] > lst[i+1]]
def swap_items(lst: list, pos1: int, pos2: int) -> None:
lst[pos1], lst[pos2] = lst[pos2], lst[pos1]
Here is a different variation of bubble sort without for loop. Basically you are considering the lastIndex of the array and slowly decrementing it until it first index of the array.
The algorithm will continue to move through the array like this until an entire pass is made without any swaps occurring.
The bubble is sort is basically Quadratic Time: O(n²) when it comes to performance.
class BubbleSort:
def __init__(self, arr):
self.arr = arr;
def bubbleSort(self):
count = 0;
lastIndex = len(self.arr) - 1;
while(count < lastIndex):
if(self.arr[count] > self.arr[count + 1]):
self.swap(count)
count = count + 1;
if(count == lastIndex):
count = 0;
lastIndex = lastIndex - 1;
def swap(self, count):
temp = self.arr[count];
self.arr[count] = self.arr[count + 1];
self.arr[count + 1] = temp;
arr = [9, 1, 5, 3, 8, 2]
p1 = BubbleSort(arr)
print(p1.bubbleSort())
def bubblesort(L,s):
if s >-1 :
bubblesort(L,s-1)
for i in range(len(L)-1-s):
if L[i]>L[i+1]:
temp = L[i+1]
L[i+1] = L[i]
L[i] = temp
return L
Nlist = [3,50,7,1,8,11,9,0,-1,5]
print(bubblesort(Nlist,len(Nlist)))
Answers provided by the-fury and Martin Cote fixed the problem of the infinite loop, but my code would still not work correctly (for a larger list, it would not sort correctly.). I ended up ditching the unsorted variable and used a counter instead.
def bubble(badList):
length = len(badList) - 1
n = 0
while n < len(badList):
for element in range(0,length):
if badList[element] > badList[element + 1]:
hold = badList[element + 1]
badList[element + 1] = badList[element]
badList[element] = hold
n = 0
else:
n += 1
return badList
if __name__ == '__main__':
mylist = [90, 10, 2, 76, 17, 66, 57, 23, 57, 99]
print bubble(mylist)
If anyone could provide any pointers on how to improve my code in the comments, it would be much appreciated.
Try this
a = int(input("Enter Limit"))
val = []
for z in range(0,a):
b = int(input("Enter Number in List"))
val.append(b)
for y in range(0,len(val)):
for x in range(0,len(val)-1):
if val[x]>val[x+1]:
t = val[x]
val[x] = val[x+1]
val[x+1] = t
print(val)
idk if this might help you after 9 years...
its a simple bubble sort program
l=[1,6,3,7,5,9,8,2,4,10]
for i in range(1,len(l)):
for j in range (i+1,len(l)):
if l[i]>l[j]:
l[i],l[j]=l[j],l[i]
def merge_bubble(arr):
k = len(arr)
while k>2:
for i in range(0,k-1):
for j in range(0,k-1):
if arr[j] > arr[j+1]:
arr[j],arr[j+1] = arr[j+1],arr[j]
return arr
break
else:
if arr[0] > arr[1]:
arr[0],arr[1] = arr[1],arr[0]
return arr
def bubble_sort(l):
for i in range(len(l) -1):
for j in range(len(l)-i-1):
if l[j] > l[j+1]:
l[j],l[j+1] = l[j+1], l[j]
return l
def bubble_sorted(arr:list):
while True:
for i in range(0,len(arr)-1):
count = 0
if arr[i] > arr[i+1]:
count += 1
arr[i], arr[i+1] = arr[i+1], arr[i]
if count == 0:
break
return arr
arr = [30,20,80,40,50,10,60,70,90]
print(bubble_sorted(arr))
#[20, 30, 40, 50, 10, 60, 70, 80, 90]
def bubbleSort(a):
def swap(x, y):
temp = a[x]
a[x] = a[y]
a[y] = temp
#outer loop
for j in range(len(a)):
#slicing to the center, inner loop, python style
for i in range(j, len(a) - j):
#find the min index and swap
if a[i] < a[j]:
swap(j, i)
#find the max index and swap
if a[i] > a[len(a) - j - 1]:
swap(len(a) - j - 1, i)
return a