迭代Python列表中的项目对

Question

可能的重复:
在Python中迭代列表(当前,下一个)
迭代列表中的每两个元素

是否可以在Python中以下列方式迭代列表(将此代码视为伪代码)？

a = [5, 7, 11, 4, 5]
for v, w in a:
    print [v, w]

它应该产生

[5, 7]
[7, 11]
[11, 4]
[4, 5]

Answer 1

你可以使用这个zip功能:

a = [5, 7, 11, 4, 5]
for v, w in zip(a[:-1], a[1:]):
    print [v, w]

Answer 2

来自itertools receipes:

from itertools import tee, izip
def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = tee(iterable)
    next(b, None)
    return izip(a, b)

for v, w in pairwise(a):
    ...

Answer 3

要做到这一点,你应该这样做:

a =  [5, 7, 11, 4, 5]
for i in range(len(a)-1):
    print [a[i], a[i+1]]

Answer 4

几乎逐字逐句迭代Python中的列表(循环方式)中的对:

def pairs(seq):
    i = iter(seq)
    prev = next(i)
    for item in i:
        yield prev, item
        prev = item

Answer 5

另一种方法是itertools.pairwise（Python 3.10 或更高版本）：

from itertools import pairwise

temp = [-39.5, -27.5, -15.5, -3.5, 8.5, 20.5, 32.5, 44.5, 56.5, 68.5, 80.5, 92.5,104.5]
res = list(pairwise(temp))

[(-39.5, -27.5), (-27.5, -15.5), (-15.5, -3.5), (-3.5, 8.5), (8.5, 20.5), (20.5, 32.5), (32.5, 44.5), (44.5, 56.5), (56.5, 68.5), (68.5, 80.5), (80.5, 92.5), (92.5, 104.5)]

Answer 6

>>> a = [5, 7, 11, 4, 5]
>>> for n,k in enumerate(a[:-1]):
...     print a[n],a[n+1]
...
5 7
7 11
11 4
4 5