Head and tail in one line

问题:

Is there a pythonic way to unpack a list in the first element and the "tail" in a single command?

For example:

>> head, tail = **some_magic applied to** [1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
>> head
1
>>> tail
[1, 2, 3, 5, 8, 13, 21, 34, 55]

回答1:

Under Python 3.x, you can do this nicely:

>>> head, *tail = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
>>> head
1
>>> tail
[1, 2, 3, 5, 8, 13, 21, 34, 55]

A new feature in 3.x is to use the * operator in unpacking, to mean any extra values. It is described in PEP 3132 - Extended Iterable Unpacking. This also has the advantage of working on any iterable, not just sequences.

It's also really readable.

As described in the PEP, if you want to do the equivalent under 2.x (without potentially making a temporary list), you have to do this:

it = iter(iterable)
head, tail = next(it), list(it)

As noted in the comments, this also provides an opportunity to get a default value for head rather than throwing an exception. If you want this behaviour, next() takes an optional second argument with a default value, so next(it, None) would give you None if there was no head element.

Naturally, if you are working on a list, the easiest way without the 3.x syntax is:

head, tail = seq[0], seq[1:]

回答2:

>>> mylist = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
>>> head, tail = mylist[0], mylist[1:]
>>> head
1
>>> tail
[1, 2, 3, 5, 8, 13, 21, 34, 55]

回答3:

For O(1) complexity of head,tail operation you should use deque however.

Following way:

from collections import deque
l = deque([1,2,3,4,5,6,7,8,9])
head, tail = l.popleft(), l

It's useful when you must iterate through all elements of the list. For example in naive merging 2 partitions in merge sort.

回答4:

Python 2, using lambda

>>> head, tail = (lambda lst: (lst[0], lst[1:]))([1, 1, 2, 3, 5, 8, 13, 21, 34, 55])
>>> head
1
>>> tail
[1, 2, 3, 5, 8, 13, 21, 34, 55]

回答5:

Building on the Python 2 solution from @GarethLatty, the following is a way to get a single line equivalent without intermediate variables in Python 2.

t=iter([1, 1, 2, 3, 5, 8, 13, 21, 34, 55]);h,t = [(h,list(t)) for h in t][0]

If you need it to be exception-proof (i.e. supporting empty list), then add:

t=iter([]);h,t = ([(h,list(t)) for h in t]+[(None,[])])[0]

If you want to do it without the semicolon, use:

h,t = ([(h,list(t)) for t in [iter([1,2,3,4])] for h in t]+[(None,[])])[0]