I have come across an exercise in Python:
Read in some strings and put them into a queue
Sort the strings lexicographically into a new queue, but the original queue shouldn't be changed. I should write a function from scratch (e.g. the sorted function cannot be used)
The use of arrays is not allowed
I think I have managed to come up with a function for step 1, but I have been struggling with step 2 for hours. I would really appreciate any kind of help!
Here is my code snippet for step 1:
q1 = []
def DisplayQueue(queue):
for Item in queue:
print(Item)
def PushQueue(queue):
x = True
while x:
user_input = input("Please enter a string (for exit type: exit): ")
if user_input == "exit":
x = False
else:
queue.append(user_input)
return queue
queue = PushQueue(q1)
Sorting can be done in many ways (bubble sort, insert sort, quick sort radix sort), but I recommend you start with something simple, although not the fastest.
Create a queue (or list) to store the answer in.
Find the smallest element in the old data list. *
Remove that element from the old list. **
Add than element to the answer list.
Repeat from step 2 (if the data list is not empty).
The data list will get shorter and shorter for each turn,
and the elements will be added in increasing order to the answer list.
*) To find the smallest element in the data list (you could put that in a separate function), keep the value of the currently smallest value in a variable called x or something, and go through the data items one by one. If the data item is smaller than what you have in the variable x, then put the value of that data item into x.
Now, once you have gone through the whole data list, your variable x will contain the value of the smallest element in the list.
**) You can remove a value v from a list x with x.remove(v). It just removes the first occurence of that value.
Related
I'm very new to the world of programming, I've been trying to solve a specific python academic exercise but I ran into an obstacle.
The problem is that I need to generate a lucky numbers sequence, as in the user inputs a sequence [1,...,n] and these steps happen:
Every second element is removed
Every third element is removed
Every fourth element is removed
.
.
.
When it becomes impossible to remove more numbers, the numbers left in the list are "lucky".
This is my code:
def lucky(l):
index = 2
new_list = []
while(index<len(l)):
for i in range(len(l)):
if(i%index==0):
new_list.append(l[i])
index=index+1
return new_list
The while loop is to have the final condition when " it is impossible to remove more numbers". However with every iteration, the list gets shorter more and more, but I don't know how to do it.
My code works for the first condition when index=2(remove every 2nd element), then in the following loops it doesn't work because:
It is still limited by length of the original list.
new_list.append(l[i]) will just add more elements to the new_list, rather than updating it in its place.
I don't know how to update the list without creating multiple amounts of lists and with each iteration adding the new elements to a new list.
Any help is appreciated.
You could use del with appropriate list slicing (see the manual for more details) to update the list in-place:
def lucky(l):
interval = 2
while interval <= len(l):
del l[interval-1::interval]
interval += 1
I am not sure if I understand your question correctly, but you can remove items from your original list via del l[index], where index is the index of the element to be removed.
For more details on lists look here:
https://docs.python.org/3/tutorial/datastructures.html
import math
def lucky(l, index):
for i in range(math.floor(len(l)/index)):
del l[(i+1)*(index-1)]
Not sure if the code will work, as I cannot test it right now. But I think it should work somehow like that.
EDIT:
Tested it and the code works. If you want to run all three steps, just do:
l = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
lucky(l,2)
lucky(l,3)
lucky(l,4)
print(l)
>>>[1,3,7,13,15]
Currently, I have a list as such:
aList = [1,2,3,4,5]
and what I'm trying to do is to emulate 'aList[index] = item' function in python. the program prompts the user for input of negative index and the item to replace.
enter negative index: -2
enter item to replace: 7
this would give me:
[1,2,3,7,5]
since, when the index is negative, python starts counting from the back of the list. Here's my code:
aList = [1,2,3,4,5]
index = int(input("Enter index:"))
item = int(input("Enter item:"))
j = -1 #-1 means the counter should begin with -1 not 0
start = len(aList)-1 #i want python to start from the back of the list
while j<start:
if j == index:
aList[j] = item
j-=1
print(lst)
I'm getting an infinite loop because of the j-=1 and I'm wondering if I'm emulating it correctly?
I think first you need to clear you concept about array.
What is Array.?
Arrays a kind of data structure that can store a fixed-size sequential collection of elements of the same type. An array is used to store a collection of data, but it is often more useful to think of an array as a collection of variables of the same type.
Instead of declaring individual variables, such as number0, number1, ..., and number99, you declare one array variable such as numbers and use numbers[0], numbers1, and ..., numbers[99] to represent individual variables. A specific element in an array is accessed by an index.
All arrays consist of contiguous memory locations. The lowest address corresponds to the first element and the highest address to the last element.
Array In Python
To define a list you simply write a comma separated list of items in square brackets:
myList=[1,2,3,4,5,6]
This looks like an array because you can use "slicing" notation to pick out an individual element - indexes start from 0. For example
print myList[2]
will display the third element, i.e. the value 3 in this case. Similarly to change the third element you can assign directly to it:
myList[2]=100
The slicing notation looks like array indexing but it is a lot more flexible. For example
myList[2:5]
is a sublist from the third element to the fifth i.e. from myList[2] to myList[4]. notice that the final element specified i.e. [5] is not included in the slice.
Also notice that you can leave out either of the start and end indexes and they will be assumed to have their maximum possible value. For example
myList[5:]
is the list from List[5] to the end of the list and
myList[:5]
is the list up to and not including myList[5] and
myList[:]
is the entire list.
List slicing is more or less the same as string slicing except that you can modify a slice. For example:
myList[0:2]=[0,1]
has the same effect as
myList[0]=0
myList[1]=1
Finally is it worth knowing that the list you assign to a slice doesn't have to be the same size as the slice - it simply replaces it even if it is a different size.
I am not sure why you need a loop, when you can just access the elements by index:
aList = [1,2,3,4,5]
index = int(input("Enter index:"))
item = int(input("Enter item:"))
aList[index] = item
Of course your loop is infinite...the variable 'j' keeps getting more negative and you're comparing it so 'start', which is the length of the list minus one.
Trying to create insertion sort but receive an error...
Don't really know why it is happening. It always tends to miss 37 aswell
numbers = [45,56,37,79,46,18,90,81,50]
def insertionSort(items):
Tsorted = []
Tsorted.append(items[0])
items.remove(items[0])
for i in range(0,len(items)):
print (Tsorted)
if items[i] > Tsorted[len(Tsorted)-1]:
Tsorted.append(items[i])
else:
Tsorted[len(Tsorted)-2] = items[i]
items.remove(items[i])
insertionSort(numbers)
Error:
if items[i] > Tsorted[len(Tsorted)-1]:
IndexError: list index out of range
First thing: you are removing items from the Array that you are iterating inside the loop here: items.remove(items[i]). This is generally not a good idea.
Second: This algorithm doesn't implement insertion sort, even if you fix the deletion issue. You should review the algorithm, e.g. here Insertion sort in Wikipedia. Thre is another loop required to find the right insertion place.
Third: in the else case, you are overwriting instead of inserting values.
You are removing elements from items over the course of the loop; thus, i may become a value that was a valid index in the original items, but is no longer in the shortened one.
If you need to remove elements from items, it looks like you should wait until the loop is finished.
That's because you're calling tems.remove(). Your code fails when i=4 and items=[37, 46, 90, 50].
So they already has no an element with index 4 but with 3 since indexing starts with 0.
range(0,len(items) will only be calculated the first time your code hits your for-loop, at which state len(list) = 8. Which means that you wlil iterate
for i in [0,1,2,3,4,5,6,7]
#Do stuff...
But at the same time you remove items from your list in each loop. So when hitting i = 4, you have iterated your loop 4 times and the length of you item-list is only 4, which means that items[4]
no longer exists.
here is a merge sort logic in python : (this is the first part, ignore the function merge()) The point in question is converting the recursive logic to a while loop.
Code courtesy: Rosettacode Merge Sort
def merge_sort(m):
if len(m) <= 1:
return m
middle = len(m) / 2
left = m[:middle]
right = m[middle:]
left = merge_sort(left)
right = merge_sort(right)
return list(merge(left, right))
Is it possible to make it a sort of dynamically in the while loop while each left and right array breaks into two, a sort of pointer keeps increasing based on the number of left and right arrays and breaking them until only single length sized list remains?
because every time the next split comes while going on both left- and right- side the array keeps breaking down till only single length list remains, so the number of left sided (left-left,left-right) and right sided (right-left,right-right) breaks will increase till it reaches a list of size 1 for all.
One possible implementation might be this:
def merge_sort(m):
l = [[x] for x in m] # split each element to its own list
while len(l) > 1: # while there's merging to be done
for x in range(len(l) >> 1): # take the first len/2 lists
l[x] = merge(l[x], l.pop()) # and merge with the last len/2 lists
return l[0] if len(l) else []
Stack frames in the recursive version are used to store progressively smaller lists that need to be merged. You correctly identified that at the bottom of the stack, there's a one-element list for each element in whatever you're sorting. So, by starting from a series of one-element lists, we can iteratively build up larger, merged lists until we have a single, sorted list.
Reposted from alternative to recursion based merge sort logic at the request of a reader:
One way to eliminate recursion is to use a queue to manage the outstanding work. For example, using the built-in collections.deque:
from collections import deque
from heapq import merge
def merge_sorted(iterable):
"""Return a list consisting of the sorted elements of 'iterable'."""
queue = deque([i] for i in iterable)
if not queue:
return []
while len(queue) > 1:
queue.append(list(merge(queue.popleft(), queue.popleft())))
return queue[0]
It's said, that every recursive function can be written in a non-recursive manner, so the short answer is: yes, it's possible. The only solution I can think of is to use the stack-based approach. When recursive function invokes itself, it puts some context (its arguments and return address) on the inner stack, which isn't available for you. Basically, what you need to do in order to eliminate recursion is to write your own stack and every time when you would make a recursive call, put the arguments onto this stack.
For more information you can read this article, or refer to the section named 'Eliminating Recursion' in Robert Lafore's "Data Structures and Algorithms in Java" (although all the examples in this book are given in Java, it's pretty easy to grasp the main idea).
Going with Dan's solution above and taking the advice on pop, still I tried eliminating while and other not so pythonic approach. Here is a solution that I have suggested:
PS: l = len
My doubt on Dans solution is what if L.pop() and L[x] are same and a conflict is created, as in the case of an odd range after iterating over half of the length of L?
def merge_sort(m):
L = [[x] for x in m] # split each element to its own list
for x in xrange(l(L)):
if x > 0:
L[x] = merge(L[x-1], L[x])
return L[-1]
This can go on for all academic discussions but I got my answer to an alternative to recursive method.
In Python, I have 3 lists of floating-point numbers (angles), in the range 0-360, and the lists are not the same length. I need to find the triplet (with 1 number from each list) in which the numbers are the closest. (It's highly unlikely that any of the numbers will be identical, since this is real-world data.) I was thinking of using a simple lowest-standard-deviation method to measure agreement, but I'm not sure of a good way to implement this. I could loop through each list, comparing the standard deviation of every possible combination using nested for loops, and have a temporary variable save the indices of the triplet that agrees the best, but I was wondering if anyone had a better or more elegant way to do something like this. Thanks!
I wouldn't be surprised if there is an established algorithm for doing this, and if so, you should use it. But I don't know of one, so I'm going to speculate a little.
If I had to do it, the first thing I would try would be just to loop through all possible combinations of all the numbers and see how long it takes. If your data set is small enough, it's not worth the time to invent a clever algorithm. To demonstrate the setup, I'll include the sample code:
# setup
def distance(nplet):
'''Takes a pair or triplet (an "n-plet") as a list, and returns its distance.
A smaller return value means better agreement.'''
# your choice of implementation here. Example:
return variance(nplet)
# algorithm
def brute_force(*lists):
return min(itertools.product(*lists), key = distance)
For a large data set, I would try something like this: first create one triplet for each number in the first list, with its first entry set to that number. Then go through this list of partially-filled triplets and for each one, pick the number from the second list that is closest to the number from the first list and set that as the second member of the triplet. Then go through the list of triplets and for each one, pick the number from the third list that is closest to the first two numbers (as measured by your agreement metric). Finally, take the best of the bunch. This sample code demonstrates how you could try to keep the runtime linear in the length of the lists.
def item_selection(listA, listB, listC):
# make the list of partially-filled triplets
triplets = [[a] for a in listA]
iT = 0
iB = 0
while iT < len(triplets):
# make iB the index of a value in listB closes to triplets[iT][0]
while iB < len(listB) and listB[iB] < triplets[iT][0]:
iB += 1
if iB == 0:
triplets[iT].append(listB[0])
elif iB == len(listB)
triplets[iT].append(listB[-1])
else:
# look at the values in listB just below and just above triplets[iT][0]
# and add the closer one as the second member of the triplet
dist_lower = distance([triplets[iT][0], listB[iB]])
dist_upper = distance([triplets[iT][0], listB[iB + 1]])
if dist_lower < dist_upper:
triplets[iT].append(listB[iB])
elif dist_lower > dist_upper:
triplets[iT].append(listB[iB + 1])
else:
# if they are equidistant, add both
triplets[iT].append(listB[iB])
iT += 1
triplets[iT:iT] = [triplets[iT-1][0], listB[iB + 1]]
iT += 1
# then another loop while iT < len(triplets) to add in the numbers from listC
return min(triplets, key = distance)
The thing is, I can imagine situations where this wouldn't actually find the best triplet, for instance if a number from the first list is close to one from the second list but not at all close to anything in the third list. So something you could try is to run this algorithm for all 6 possible orderings of the lists. I can't think of a specific situation where that would fail to find the best triplet, but one might still exist. In any case the algorithm will still be O(N) if you use a clever implementation, assuming the lists are sorted.
def symmetrized_item_selection(listA, listB, listC):
best_results = []
for ordering in itertools.permutations([listA, listB, listC]):
best_results.extend(item_selection(*ordering))
return min(best_results, key = distance)
Another option might be to compute all possible pairs of numbers between list 1 and list 2, between list 1 and list 3, and between list 2 and list 3. Then sort all three lists of pairs together, from best to worst agreement between the two numbers. Starting with the closest pair, go through the list pair by pair and any time you encounter a pair which shares a number with one you've already seen, merge them into a triplet. For a suitable measure of agreement, once you find your first triplet, that will give you a maximum pair distance that you need to iterate up to, and once you get up to it, you just choose the closest triplet of the ones you've found. I think that should consistently find the best possible triplet, but it will be O(N^2 log N) because of the requirement for sorting the lists of pairs.
def pair_sorting(listA, listB, listC):
# make all possible pairs of values from two lists
# each pair has the structure ((number, origin_list),(number, origin_list))
# so we know which lists the numbers came from
all_pairs = []
all_pairs += [((nA,0), (nB,1)) for (nA,nB) in itertools.product(listA,listB)]
all_pairs += [((nA,0), (nC,2)) for (nA,nC) in itertools.product(listA,listC)]
all_pairs += [((nB,1), (nC,2)) for (nB,nC) in itertools.product(listB,listC)]
all_pairs.sort(key = lambda p: distance(p[0][0], p[1][0]))
# make a dict to track which (number, origin_list)s we've already seen
pairs_by_number_and_list = collections.defaultdict(list)
min_distance = INFINITY
min_triplet = None
# start with the closest pair
for pair in all_pairs:
# for the first value of the current pair, see if we've seen that particular
# (number, origin_list) combination before
for pair2 in pairs_by_number_and_list[pair[0]]:
# if so, that means the current pair shares its first value with
# another pair, so put the 3 unique values together to make a triplet
this_triplet = (pair[1][0], pair2[0][0], pair2[1][0])
# check if the triplet agrees more than the previous best triplet
this_distance = distance(this_triplet)
if this_distance < min_distance:
min_triplet = this_triplet
min_distance = this_distance
# do the same thing but checking the second element of the current pair
for pair2 in pairs_by_number_and_list[pair[1]]:
this_triplet = (pair[0][0], pair2[0][0], pair2[1][0])
this_distance = distance(this_triplet)
if this_distance < min_distance:
min_triplet = this_triplet
min_distance = this_distance
# finally, add the current pair to the list of pairs we've seen
pairs_by_number_and_list[pair[0]].append(pair)
pairs_by_number_and_list[pair[1]].append(pair)
return min_triplet
N.B. I've written all the code samples in this answer out a little more explicitly than you'd do it in practice to help you to understand how they work. But when doing it for real, you'd use more list comprehensions and such things.
N.B.2. No guarantees that the code works :-P but it should get the rough idea across.