>>> l = [1,2,3,4,5,6]

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

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

>>> [a+b for a,b in zip(l,l[1:])[::2]]
[3, 7, 11]

>>> ["%d + %d = %d" % (a,b,a+b) for a,b in zip(l,l[1:])[::2]]
['1 + 2 = 3', '3 + 4 = 7', '5 + 6 = 11']

简单的方法:

[(a[i],a[i+1]) for i in range(0,len(a),2)]

如果你的数组是a，并且你想成对迭代它，这是很有用的。 要迭代三胞胎或更多，只需更改"range" step命令，例如:

[(a[i],a[i+1],a[i+2]) for i in range(0,len(a),3)]

(如果你的数组长度和步长不匹配，你必须处理多余的值)

>>> l = [1,2,3,4,5,6]

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

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

>>> [a+b for a,b in zip(l,l[1:])[::2]]
[3, 7, 11]

>>> ["%d + %d = %d" % (a,b,a+b) for a,b in zip(l,l[1:])[::2]]
['1 + 2 = 3', '3 + 4 = 7', '5 + 6 = 11']

我需要把一个列表除以一个数字，然后像这样固定。

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

def divideByN(data, n):
        return [data[i*n : (i+1)*n] for i in range(len(data)//n)]  

>>> print(divideByN(l,2))
[[1, 2], [3, 4], [5, 6]]

>>> print(divideByN(l,3))
[[1, 2, 3], [4, 5, 6]]

一个简单的解决方案。

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

for i in range(0, len(l), 2):
    print str(l[i]), '+', str(l[i + 1]), '=', str(l[i] + l[i + 1])

对于可能有所帮助的人，这里有一个类似问题的解决方案，但有重叠的对(而不是互斥的对)。

来自Python itertools文档:

from itertools import izip

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

或者，更一般地说:

from itertools import izip

def groupwise(iterable, n=2):
    "s -> (s0,s1,...,sn-1), (s1,s2,...,sn), (s2,s3,...,sn+1), ..."
    t = tee(iterable, n)
    for i in range(1, n):
        for j in range(0, i):
            next(t[i], None)
    return izip(*t)