Appending a list to itself in Python

Question

I want to attach a list to itself and I thought this would work:

x = [1,2]
y = x.extend(x)
print y

I wanted to get back [1,2,1,2] but all I get back is the builtin None. What am I doing wrong? I'm using Python v2.6

score 11 · Accepted Answer · answered Jul 22 '11 at 17:07

11

x.extend(x) does not return a new copy, it modifies the list itself.

Just print x instead.

You can also go with x + x

answered Jul 22 '11 at 17:07

Stefan Kanev

score 2 · Answer 2 · answered Jul 22 '11 at 17:08

2

x.extend(x) modifies x in-place.

If you want a new, different list, use y = x + x.

answered Jul 22 '11 at 17:08

Amber

score 1 · Answer 3 · answered Jul 22 '11 at 17:10

1

or just:

x = [1,2]
y = x * 2
print y

answered Jul 22 '11 at 17:10

fransua

You are creating `[x, x]`, if one element of `x` changes, it will change in the other index of `y`, too: http://stackoverflow.com/questions/240178/python-list-of-lists-changes-reflected-across-sublists-unexpectedly – Schorsch Nov 08 '16 at 17:58
1

@Schorsch That is true for `[x] * 2` which is equivalent to `[x] + [x]`. However, `x * 2` is equivalent to `x + x` which creates a "shallow" list of values in this case. – Shmuel Valariola Jun 11 '20 at 00:47

score 0 · Answer 4 · answered Jul 22 '11 at 17:08

0

x.extend(x) will extend x inplace.

>>> print x

[1, 2, 1, 2]

answered Jul 22 '11 at 17:08

Dogbert

score 0 · Answer 5 · answered Jul 22 '11 at 17:09

0

If you want a new copy of the list try:

x = [1,2]
y = x + x
print y # prints [1,2,1,2]

The difference is that extend modifies the list "in place", meaning it will always return None even though the list is modified.

answered Jul 22 '11 at 17:09

Kenan Banks

5 Answers5