Python: why list[1:].append(list[0]) is None?

Asked May 20 '16 at 09:09

Active Sep 14 '22 at 16:02

Viewed 46 times

Strange thing:

>>> a = [1, 2, 3, 4, 5, 6, 7]

>>> b = a[1:]

>>> b

[2, 3, 4, 5, 6, 7]

>>> b.append(a[0])

>>> b

[2, 3, 4, 5, 6, 7, 1]

It works fine - b is the circular shift of elements from a array.

But when I'm trying to do the same in one line:

>>> b = a[1:].append(a[0])

>>> b

>>> # b is none

Why?!?!?!

a[1:] - is a new list and it has the append method - everything is fine but why the results of two line and one line realizations are different?

edited Jun 20 '20 at 09:12

Community

asked May 20 '16 at 09:09

Dmitriy Kulagin

1

`append()` doesn't return the list you apply it to. It returns `None`. – Frédéric Hamidi May 20 '16 at 09:11
@FrédéricHamidi that should be an answer, and it's exactly right. – John Szakmeister May 20 '16 at 09:12
1

This behaviour is consistent across most of Python: a method that mutates an object almost always returns None (the exception being things like list `list.pop()`). – Duncan May 20 '16 at 09:12
Oh! Right! Thanks! a[1:]+[a[0]] - returns the list I need. – Dmitriy Kulagin May 20 '16 at 09:15

0 Answers0