Pythonic way to split a list into first and rest in Python 2.x?
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.
Answers:
first, rest = l[0], l[1:]
Basically the same, except that it’s a oneliner. Tuple assigment rocks.
This is a bit longer and less obvious, but generalized for all iterables (instead of being restricted to sliceables):
i = iter(l)
first = next(i) # i.next() in older versions
rest = list(i)
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.
If l is string typeI would suggest:
first, remainder = l.split(None, maxsplit=1)
Yet another one, working with python 2.7. Just use an intermediate function. Logical as the new behavior mimics what happened for functions parameters passing.
li = [1, 2, 3]
first, rest = (lambda x, *y: (x, y))(*li)
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.
first, rest = l[0], l[1:]
Basically the same, except that it’s a oneliner. Tuple assigment rocks.
This is a bit longer and less obvious, but generalized for all iterables (instead of being restricted to sliceables):
i = iter(l)
first = next(i) # i.next() in older versions
rest = list(i)
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.
If l is string typeI would suggest:
first, remainder = l.split(None, maxsplit=1)
Yet another one, working with python 2.7. Just use an intermediate function. Logical as the new behavior mimics what happened for functions parameters passing.
li = [1, 2, 3]
first, rest = (lambda x, *y: (x, y))(*li)