Question

This is a very simple question, but I haven’t seem to be able to find a satisfactory answer for it.

What is the best way, in Python, make the last item of a list become the first one “pushing” the rest of the list.

Something that does:

>>> a=[1,2,3,4]
>>> a[?????]
[4, 1, 2, 3]

I know I can always play with len, list concatenation…

>>> a=[1,2,3,4]
>>> [a[len(a)-1]] + a[0:len(a)-1]
[4, 1, 2, 3]

But that doesn’t look right… “Pythonic“, if you may

Thank you in advance.

Answer 1

Slicing is a little smarter than that; you can use negative indices to count from the end:

a[-1:] + a[:-1]

Demo:

>>> a=[1,2,3,4]
>>> a[-1:] + a[:-1]
[4, 1, 2, 3]

This works for an arbitrary number of elements to be moved to the front:

>>> a[-2:] + a[:-2]
[3, 4, 1, 2]

Using slicing like this is comparable in speed to using .insert() + .pop() (on a short list):

>>> timeit.timeit('a[-1:] + a[:-1]', 'a=[1,2,3,4]')
0.59950494766235352
>>> timeit.timeit('a.insert(0,a.pop(-1))', 'a=[1,2,3,4]')
0.52790379524230957

but wins hands down if you need to shift more than one element:

>>> timeit.timeit('a[-2:] + a[:-2]', 'a=[1,2,3,4]')
0.58687901496887207
>>> timeit.timeit('a.insert(0,a.pop(-1));a.insert(0,a.pop(-1))', 'a=[1,2,3,4]')
1.0615170001983643