I am new to programming and am trying to understand how to "think" more programmatically in Python. I wrote a function that returns the largest element in an array without using max:
def max_userdefined(array):
"""
Finds largest value of nonempty array of numbers in O(n)
:param: list/tuple of values
:return: max value
"""
try:
assert isinstance(array, list or tuple)
result = array[0]
for element in array:
if element > result:
result = element
return result
except IndexError:
return 'Please pass in a nonempty array'
except AssertionError:
return 'Please pass in a list or tuple'
except TypeError:
return 'Please make sure elements of array are all floats or ints'
How can I improve the above code? Open to any criticism or if anyone can recommend a good Pythonic style guide.
def max_userdefined(array):
myMax = l[0]
for num in l:
if myMax < num:
myMax = num
return myMax
print max_userdefined ([1,2,3,4,5])
Output
5
Or You can do something like following by using sorted.
def max_userdefined(array):
return sorted(array)[-1]
>>> def umax(array):
global biggest
for num in array:
if num > biggest:
biggest = num
>>> biggest = int()
>>> nums = (1, 9, 6, 4, 5, 3, 2, 7, 0)
>>> umax(nums)
>>> biggest
9
For returning the greatest number from a list the simple algorithm followed is compare two numbers and keep the larger number in a variable, then print the variable OR you can use python shortcut using sort.
Using user defined function:
def greatest_ele_list(listobj):
greatest=listobj[0]
for i in range(len(listobj)):
if listobj[i]>greatest:
greatest=listobj[i]
else:
pass
return greatest
print greatest_ele_list([4,5,3,8,1])
Using sort:
def greatest_ele_list(listobj):
listobj2=listobj.sort()
return listobj2[-1]
print greatest_ele_list([4,5,3,8,1])
Output: >>> 8
For example
def find_all_occurrences(a,b):
'''
>>>find_all_occurrences('wswwswwwswwwws', ['ws', 'wws'])
[[0,3,7,12], [2,6,11]]
'''
How can I return a list of lists that have all the occurrences without import any modules.
You can use regular expressions
import re
def all_occurrences(a, b):
return [[occur.start() for occur in re.finditer(word, a)] for word in b]
Without imports it gets a little messy, but definitely still doable
def all_occurrences(a, b):
result = []
for word in b:
word_res = []
index = a.find(word)
while index != -1:
word_res.append(index)
index = a.find(word, index+1)
result.append(word_res)
return result
You can find all the occurrences by using the last found position as the start of the next search:
str.find(...)
S.find(sub [,start [,end]]) -> int
Return the lowest index in S where substring sub is found,
such that sub is contained within S[start:end]. Optional
arguments start and end are interpreted as in slice notation.
Return -1 on failure.
A loop that calls haystack.find(needle, last_pos + 1) repeatedly until it returns -1 should work.
you can also have simple list comprehensions to help with problems like these
[[i for i in range(len(a)-len(strng)+1) if strng == a[i:i+len(strng)]] for strng in b]
where
>>> a
'wswwswwwswwwws'
>>> b
['ws', 'wws']
Solution with a recursive procedure. I used a nested/inner function to maintain the OP's function signature:
def find_all_occurrences(a,b):
'''
>>>find_all_occurrences('wswwswwwswwwws', ['ws', 'wws'])
[[0,3,7,12], [2,6,11]]
'''
def r(a, b, count = 0, result = None):
if not a:
return result
if result is None:
# one sublist for each item in b
result = [[] for _ in b]
for i, thing in enumerate(b):
if a.startswith(thing):
result[i].append(count)
return r(a[1:], b, count = count + 1, result = result)
return r(a, b)
I have written this recursive code in Python:
def suma(i,l):
if i == 0:
return l[i]
else:
return suma(i-1,l)+l[i]
And whenever I call the function by suma(3,[7,2,3]) and run it, I get this error message:
List index out of range on line return suma(i-1,l)+l[i]
Ok, I'm going to assume that the intent here is to recursively add the first i elements of l and return the result. If so, here's a concise way to do so:
def suma(i,l):
return 0 if i == 0 else suma(i-1,l)+l[i-1]
This is equivalent to:
def suma(i,l):
if i == 0:
return 0
else
return suma(i-1,l)+l[i-1]
It's unorthodox, but you could just call your suma() function with the first argument reduced by 1:
>>> l = [7, 2, 3]
>>> suma(len(l)-1, l)
12
But it could be better written like this:
def suma(l):
if len(l) > 1:
return suma(l[:-1]) + l[-1]
return l[0]
>>> suma([7,2,3])
12
>>> suma(range(1,11))
55
This has the advantage of not requiring the length of the list to be passed to the sum function at all - the length is always available using len().
I know there are easier ways to create a function which gives you the largest number in a list of numbers but I wanted to use recursion. When I call the function greatest, i get none. For example greatest([1,3,2]) gives me none. If there are only two elements in the list, I get the right answer so I know the problem must be with the function calling itself. Not sure why though.
def compare(a,b):
if a==b:
return a
if a > b:
return a
if a < b:
return b
def greatest(x):
if len(x)==0:
return 0
i=0
new_list=[]
while i< len(x):
if len(x)-i>1:
c=compare(x[i],x[i+1])
else:
c=x[i]
new_list.append(c)
i=i+2
if len(new_list)>1:
greatest(new_list)
else:
return new_list[0]
print greatest([1,3,2])
This line:
if len(new_list)>1:
greatest(new_list) # <- this one here
calls greatest but doesn't do anything with the value it returns. You want
return greatest(new_list)
After fixing that, your function seems to behave (although I didn't look too closely):
>>> import itertools
>>> for i in range(1, 6):
... print i, all(max(g) == greatest(g) for g in itertools.product(range(-5, 5), repeat=i))
...
1 True
2 True
3 True
4 True
5 True
A simple recursion can be like this :
from random import *
def greatest(x,maxx=float("-inf")):
if len(x)>0:
if x[0] > maxx:
maxx=x[0]
return greatest(x[1:],maxx)
else:
return maxx
lis=range(10,50)
shuffle(lis)
print greatest(lis) #prints 49
This question already has answers here:
Removing elements that have consecutive duplicates
(9 answers)
Closed 3 years ago.
Google Python Class | List Exercise -
Given a list of numbers, return a list where
all adjacent == elements have been reduced to a single element,
so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or
modify the passed in list.
My solution using a new list is -
def remove_adjacent(nums):
a = []
for item in nums:
if len(a):
if a[-1] != item:
a.append(item)
else: a.append(item)
return a
The question even suggests that it could be done by modifying the passed in list. However, the python documentation warned against modifying elements while iterating a list using the for loop.
I am wondering what else can I try apart from iterating over the list, to get this done. I am not looking for the solution, but maybe a hint that can take me into a right direction.
UPDATE
-updated the above code with suggested improvements.
-tried the following with a while loop using suggested hints -
def remove_adjacent(nums):
i = 1
while i < len(nums):
if nums[i] == nums[i-1]:
nums.pop(i)
i -= 1
i += 1
return nums
Here's the traditional way, deleting adjacent duplicates in situ, while traversing the list backwards:
Python 1.5.2 (#0, Apr 13 1999, 10:51:12) [MSC 32 bit (Intel)] on win32
Copyright 1991-1995 Stichting Mathematisch Centrum, Amsterdam
>>> def dedupe_adjacent(alist):
... for i in xrange(len(alist) - 1, 0, -1):
... if alist[i] == alist[i-1]:
... del alist[i]
...
>>> data = [1,2,2,3,2,2,4]; dedupe_adjacent(data); print data
[1, 2, 3, 2, 4]
>>> data = []; dedupe_adjacent(data); print data
[]
>>> data = [2]; dedupe_adjacent(data); print data
[2]
>>> data = [2,2]; dedupe_adjacent(data); print data
[2]
>>> data = [2,3]; dedupe_adjacent(data); print data
[2, 3]
>>> data = [2,2,2,2,2]; dedupe_adjacent(data); print data
[2]
>>>
Update: If you want a generator but (don't have itertools.groupby or (you can type faster than you can read its docs and understand its default behaviour)), here's a six-liner that does the job:
Python 2.3.5 (#62, Feb 8 2005, 16:23:02) [MSC v.1200 32 bit (Intel)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> def dedupe_adjacent(iterable):
... prev = object()
... for item in iterable:
... if item != prev:
... prev = item
... yield item
...
>>> data = [1,2,2,3,2,2,4]; print list(dedupe_adjacent(data))
[1, 2, 3, 2, 4]
>>>
Update 2: Concerning the baroque itertools.groupby() and the minimalist object() ...
To get the dedupe_adjacent effect out of itertools.groupby(), you need to wrap a list comprehension around it to throw away the unwanted groupers:
>>> [k for k, g in itertools.groupby([1,2,2,3,2,2,4])]
[1, 2, 3, 2, 4]
>>>
... or muck about with itertools.imap and/or operators.itemgetter, as seen in another answer.
Expected behaviour with object instances is that none of them compares equal to any other instance of any class, including object itself. Consequently they are extremely useful as sentinels.
>>> object() == object()
False
It's worth noting that the Python reference code for itertools.groupby uses object() as a sentinel:
self.tgtkey = self.currkey = self.currvalue = object()
and that code does the right thing when you run it:
>>> data = [object(), object()]
>>> data
[<object object at 0x00BBF098>, <object object at 0x00BBF050>]
>>> [k for k, g in groupby(data)]
[<object object at 0x00BBF098>, <object object at 0x00BBF050>]
Update 3: Remarks on forward-index in-situ operation
The OP's revised code:
def remove_adjacent(nums):
i = 1
while i < len(nums):
if nums[i] == nums[i-1]:
nums.pop(i)
i -= 1
i += 1
return nums
is better written as:
def remove_adjacent(seq): # works on any sequence, not just on numbers
i = 1
n = len(seq)
while i < n: # avoid calling len(seq) each time around
if seq[i] == seq[i-1]:
del seq[i]
# value returned by seq.pop(i) is ignored; slower than del seq[i]
n -= 1
else:
i += 1
#### return seq #### don't do this
# function acts in situ; should follow convention and return None
Use a generator to iterate over the elements of the list, and yield a new one only when it has changed.
itertools.groupby does exactly this.
You can modify the passed-in list if you iterate over a copy:
for elt in theList[ : ]:
...
Just to show one more way here is another single liner version without indexes:
def remove_adjacent(nums):
return [a for a,b in zip(nums, nums[1:]+[not nums[-1]]) if a != b]
The not part puts the last value to result as only a ends up to result.
As usual, I am just here to advertise the impressive recipes in the Python itertools documentation.
What you are looking for is the function unique_justseen:
from itertools import imap, groupby
from operator import itemgetter
def unique_justseen(iterable, key=None):
"List unique elements, preserving order. Remember only the element just seen."
# unique_justseen('AAAABBBCCDAABBB') --> A B C D A B
# unique_justseen('ABBCcAD', str.lower) --> A B C A D
return imap(next, imap(itemgetter(1), groupby(iterable, key)))
list(unique_justseen([1,2,2,3])) # [1, 2, 3]
Well, katrielalex is right about itertools, but the OP seems to be rather more interested (or should be!) in learning to manipulate the basics of the built-in data structures. As for manipulating a list in place, it does need thought, but my recommendation would be to read through this section of the documentation and try a few list methods (hint: list.pop(), list.remove(), and learn everything about slices.)
The posted code could be simplified, by the way (you should however add handling of error conditions):
def remove_adjacent(nums):
a = nums[:1]
for item in nums[1:]:
if item != a[-1]:
a.append(item)
return a
Extremely elegant solution from Google (source is here: https://developers.google.com/edu/python/exercises/basic):
def remove_adjacent(nums):
result = []
for num in nums:
if len(result) == 0 or num != result[-1]:
result.append(num)
return result
You can use list comprehension. For example something like this should do the job:
def remove_adjacent(L):
return [elem for i, elem in enumerate(L) if i == 0 or L[i-1] != elem]
or:
def remove_adjacent(L):
return [L[i] for i in xrange(len(L)) if i == 0 or L[i-1] != L[i]]
Try this:
def remove_adjacent(nums):
result = []
if len(nums) > 0:
result = [nums[0]]
for i in range(len(nums)-1):
if nums[i] != nums[i+1]:
result.append(nums[i+1])
return result
itertools.groupby is superior, but there is also
reduce(lambda x, y: x + [y] if x[-1] != y else x, seq[1:], seq[0:1])
e.g.
>>> seq = [[1,1], [2,2], [3,3], [3,3], [2,2], [2,2], [1,1]]
>>> print reduce(lambda x, y: x + [y] if x[-1] != y else x, seq[1:], seq[0:1])
[[1, 1], [2, 2], [3, 3], [2, 2], [1, 1]]
When coming from functional languages where this sort of thing is done with a fold, then using reduce often feels natural.
You can modify a list you're iterating over if you use indices explicitly:
def remove_adjacent(l):
if len(l)<2:
return l
prev,i = l[0],1
while i < len(l):
if l[i] == prev:
del l[i]
else:
prev = l[i]
i += 1
It doesn't work with iterators because iterators don't "know" how to modify the index when you remove arbitrary elements, so it's easier to just forbid it. Some languages have iterators with functions to remove the "current item".
#katrielalex's solution is more pythonic, but if you did need to modify the list in-place without making a copy, you could use a while loop and break when you catch an IndexError.
e.g.
nums = [1,1,1,2,2,3,3,3,5,5,1,1,1]
def remove_adjacent(nums):
"""Removes adjacent items by modifying "nums" in-place. Returns None!"""
i = 0
while True:
try:
if nums[i] == nums[i+1]:
# Letting you figure this part out,
# as it's a homework question
except IndexError:
break
print nums
remove_adjacent(nums)
print nums
Edit: pastebin of one way to do it here, in case you get stuck and want to know..
def remove_adjacent(nums):
newList=[]
for num in nums:
if num not in newList:
newList.append(num)
newList.sort()
return newList
Another approach. Comments welcome.
def remove_adjacent(nums):
'''modifies the list passed in'''
l, r = 0, 1
while r < len(nums):
if nums[l] == nums[r]:
r += 1
else:
l += 1
nums[l] = nums[r]
r += 1
del nums[l+1:]
Seeing the code written by Google is a humbling lol. This is what I came up with:
def remove_adjacent(nums):
rmvelement = []
checkedIndex = []
for num in nums:
if nums.index(num) not in checkedIndex:
index = nums.index(num)
checkedIndex.append(index)
skip = False
else:
skip = True
if skip == False:
for x in nums[index+1:]:
if x == num:
rmvelement.append(x)
else:
break
[nums.remove(_) for _ in rmvelement]
return nums
This should work for a transparent (albeit roundabout) solution:
def remove_adjacent(nums):
numstail = [i for i in range(0,len(nums))]
nums = nums + numstail
for i in nums:
if nums[i] == nums[i-1]:
del nums[i]
return nums[:-len(numstail)]
The logic is as follows:
Create a tail-list equal to the length of the original list of numbers and append this to the end of the original list.
Run a 'for-loop' that checks if a given element of nums is the same as the previous element. If so, delete it.
Return the new nums list, with the necessary deletions, up to len(numtails) index positions from the end of the list.
(numstail is defined to avoid indices being out of range for any length list)
def removeDupAdj2(a):
b=[]
for i in reversed(range(1,len(a))):
if(a[i-1] == a[i]):
del(a[i])
#print(a)
return a
a = [int(i) for i in '1 2 3 3 4 4 3 5 4 4 6 6 6 7 8 8 8 9 1 1 0 0'.split(' ')]
a
res = removeDupAdj2(a)
res
Since you are in a Python class, I'm going to guess that you are new to the language. Thus, for you and any other beginners out there, I wrote a simple version of the code to help others get through the logic.
original= [1, 2, 2, 3]
newlist=[]
for item in original:
if item in newlist:
print "You don't need to add "+str(item)+" again."
else:
newlist.append(item)
print "Added "+str(item)
print newlist