Updating the same list with square values

Question

I want to declare a list in python3 and then update it with square values. e.g.

list1 = [1,2,3,4]

and Output should be the same list with square values:

list1 = [1,4,9,16]

I don't want to use another list. Please help?

@meowgoesthedog That replaces the list, not the values in the list. — Bill the Lizard, Jan 28 '19 at 13:42
@RahulAgarwal None of the answers in the "dupe" does it in-place. — tobias_k, Jan 28 '19 at 13:42
You _could_ use a list comprehension and then do slice-assignment: `list[:] = [see comment above]`, but this will still create a new list temporarily. — tobias_k, Jan 28 '19 at 13:43
@BilltheLizard OP should specify whether he meant a separate variable or a new list object. — willywonkadailyblah, Jan 28 '19 at 13:43
@tobias_k: The below posted answer is also the 2nd answer of the dupe!! — Rahul Agarwal, Jan 28 '19 at 13:43
@RahulAgarwal Does not mean that the dupe is a dupe, just that the answer does not do what OP asks for. (Well, in fact, it does now) — tobias_k, Jan 28 '19 at 13:44
This looked like a funny way to do this: https://stackoverflow.com/a/3000471/4121573 — Adonis, Jan 28 '19 at 13:47
@tobias_k there is a target of modifying a list in place, that is considered a dup to a target that is not modifying a list in place... It seems like the other dup target was a poor choice of dup target, but this is an exact duplicate of the target that is pointing to a duplicate that is not an adequate target... — d_kennetz, Jan 28 '19 at 15:50
See [this post](https://stackoverflow.com/questions/24201926/in-place-replacement-of-all-occurrences-of-an-element-in-a-list-in-python) as a duplicate target that is adequate. — d_kennetz, Jan 28 '19 at 15:51

score 4 · Answer 1 · answered Jan 28 '19 at 13:47

You could use a slice-assignment with a generator expression:

>>> list1 = [1,2,3,4]
>>> list2 = list1
>>> list1[:] = (x**2 for x in list1)
>>> list2
[1, 4, 9, 16]

With the [:], it changes the list in-place, and by using a generator (...) instead of a list comprehension [...] it does not create a temporary list (in case that's the problem). (But note that if you have a reference to an element of that list, that reference will not be updated.)

Roel van Endhoven · Answer 2 · 2019-01-28T13:45:11.603

2

You could use list1 = list(map(lambda x: x**2, list1)) but this doesn't work in place. It replaces the list. For doing it truly in place you loop over every item:

for i, x in enumerate(list1):
    list1[i] = x ** 2

edited Jan 28 '19 at 13:45

answered Jan 28 '19 at 13:40

Roel van Endhoven

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1
9

score 0 · Answer 3 · answered Jan 28 '19 at 13:46

0

Try this:

list1 = list(map(lambda x:x*x, list1))

or:

list1 = [i*i for i in list1]

answered Jan 28 '19 at 13:46

Mehrdad Pedramfar

9,989
4
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Updating the same list with square values

3 Answers3