大多数pythonic（和有效）的方式将列表成对嵌套

Question

my list is:

mylist=[1,2,3,4,5,6]

I would like to convert mylist into a list of pairs:

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

Is there a pythonic way of doing so? List comprehension? Itertools?

Answer 1

Yeppers, list comprehension is my usual way of doing it:

>>> groupsize = 2
>>> [mylist[x:x+groupsize] for x in range(0,len(mylist),groupsize)]
[[1,2],[3,4],[5,6]]
>>> groupsize = 3
>>> [mylist[x:x+groupsize] for x in range(0,len(mylist),groupsize)]
[[1,2,3],[4,5,6]]

I use range for portability, if you are using python 2 (you probably are) change the range to xrange to save memory.

Answer 2

My preferred technique:

>>> mylist = [1, 2, 3, 4, 5, 6]
>>> mylist = iter(mylist)
>>> zip(mylist, mylist)
[(1, 2), (3, 4), (5, 6)]

I usually use generators instead of lists anyway, so line 2 usually isn't required.

Answer 3

An alternate way:

zip( mylist[:-1:2], mylist[1::2] )

Which produces a list of tuples:

>>> zip(mylist[:-1:2],mylist[1::2])
[(1, 2), (3, 4), (5, 6)]

If you really want a list of lists:

map(list, zip(mylist[:-1:2],mylist[1::2]))

Answer 4

Check out the "grouper" recipe from the itertools documentation :

def grouper(n, iterable, fillvalue=None):
    "Collect data into fixed-length chunks or blocks"
    # grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

Answer 5

[mylist[2*n:2*n+2] for n in xrange(len(mylist)/2)]

This solution combines the use of list comprehensions and slicing to extract pairs in sequence from the original list, and build a list of the slices.

Alternatively, [mylist[n:n+2] for n in xrange(0, len(mylist), 2)] which is the same except xrange counts by twos instead of the slices. Thanks to Steven Rumbalski for the suggestion.

And now for something completely different: here is a solution (ab)using zip and an ephemeral function instead of intermediate assignment:

>>> (lambda i: zip(i, i))(iter(mylist))
[(1, 2), (3, 4), (5, 6)]