CODEWITHSUNDEEP
pythonlist
3MIN3M
3MIN3M

Reputation: 67

python lists, appending something to the list changes the entire thing?

so im trying to implement selection sort in python.. and im appending the result of each iteration to a list to print at the end.. my code sorts the list of numbers properly but when i append it to the same list at the end it changes all other lists..

def s_sort(numbers):
  alist=[]
  #do actual sorting here and swap numbers/index if neccessary

       alist.append(numbers)
  return alist

def main():
    numbers=[5,7,3]
    print(s_sort(numbers))
main()

the alist returned is [[3,5,7],[3,5,7]] instead of [[3,7,5],[3,5,7]] !!!! somehow when i do the append to alist, the content of alist changes for both lists!

Upvotes: 0

Views: 145

Answers (3)

abarnert
abarnert

Reputation: 366213

Your code doesn't actually do what you say it does. In fact, it doesn't even run. But here's a simple example that does demonstrate the problem you're seeing:

def s_sort(numbers):
    alist=[]
    alist.append(numbers)
    numbers.sort()
    alist.append(numbers)
    return alist

The problem is that alist is not a list of two different lists, it's a list of the same list twice in a row. So, when you modify that one list, of course that one list is modified everywhere it appears—in numbers, and in alist[0], and in alist[1].

The solution is to not add the same list multiple times; instead, add a new one. For example:

def s_sort(numbers):
    alist=[]
    alist.append(numbers[:])
    alist.append(sorted(numbers))
    return alist

Now you've created two brand-new lists—one an exact copy of the original, one a sorted copy—and returned a list of them.

So, instead of returning [[3, 5, 7], [3, 5, 7]] (and also changing numbers to be [3, 5, 7]), it returns [[5, 7, 3], [3, 5, 7]] (and leaves numbers alone).

I have no idea why you expected [3, 7, 5] for the first element, but maybe you're doing some other work to the first element of alist which you didn't show us. In which case, as long as you do that work in a copying rather than mutating way (ala sorted(n) vs. n.sort()) or do it to a copy, everything will be fine.

Upvotes: 2

mhlester
mhlester

Reputation: 23251

I don't see the actual sort that you're doing, but in general:

Lists are mutable. Any change you make to it affects all links to that list. To make a copy of it and break its connection to other references, you need to return alist[:]

def s_sort(numbers):
  alist=[]
  #do actual sorting here and swap numbers/index if neccessary

       alist.append(numbers)
  return alist[:]  # this makes it a copy!

def main():
    numbers=[5,7,3]
    print(s_sort(numbers))
main()

Upvotes: 2

wim
wim

Reputation: 363616

Use a slice to make a copy

newlist = alist[:]

In your case, I guess it's:

alist.append(numbers[:])

Upvotes: 2

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