Pythonic way to split a list into first and rest?

Question

I think in Python 3 I'll be able to do:

first, *rest = l

which is exactly what I want, but I'm using 2.6. For now I'm doing:

first = l[0]
rest = l[1:]

This is fine, but I was just wondering if there's something more elegant.

Solution 1

You can do

first = l.pop(0)

and then l will be the rest. It modifies your original list, though, so maybe it’s not what you want.

Solution 2

If l is string typeI would suggest:

first, remainder = l.split(None, maxsplit=1)

Comments

[Uyghur Lives Matter]

list does not have a .split() method.

[fghoussen]

ooops ! my mistake, though it was for a string

[avyfain]

This won't work for empty lists: x = []; x.pop(0) Traceback (most recent call last): File "<stdin>", line 1, in <module> IndexError: pop from empty list

[kriss]

@avyfain: neither would work the Python3 example of the OP. first, *rest = [] raise a ValueError.

[Gnudiff]

Still +1 for while it is wrong in this case, it was a good thing, because I wanted to split string and assign head + rest of result to different variables.

[Peter Gibson]

Note this throws an error if the split only returns a list of 1 item