Question

可能重复：
Python: Looping through all but the last item of a list

当您还需要列表中的下一个项目（或任何其他任意项目）时，是否有更好的方法来遍历列表？我用这个，但也许有人可以做得更好......

values = [1, 3, 6, 7 ,9]
diffs = []
for i in range(len(values)):
    try: diffs.append(values[i+1] - values[i])
    except: pass

print diffs

给出：

[2, 3, 1, 2]

Answer 1

>>> values = range(10)
>>> values
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> zip(values[0:],values[1:])
[(0, 1), (1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8), (8, 9)]

Answer 2

借助pairwise recipe from itertools document

来做什么

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)
def main():
    values = [1, 3, 6, 7 ,9]
    diffs = [post - prior for prior, post in pairwise(values)]
    print diffs


if __name__ == "__main__":
    main()

输出
[2,3,1,2]

Answer 3

您可以使用

for pos, item in enumerate(values):
    try:
        diffs.append(values[pos+1] - item)
    except IndexError:
        pass

虽然在你的情况下（因为你只是在寻找下一个值），你也可以简单地使用

for item,nextitem in zip(values, values[1:]):
    diffs.append(nextitem-item)

也可以表示为列表理解：

diffs = [nextitem-item for item,nextitem in zip(values, values[1:])]

Answer 4

for i, j in zip(values, values[1:]):
     j - i

Answer 5

diff = [values[i+1] - values[i] for i in range(len(values)-1)]

Answer 6

[y - x for x,y in zip(L,L[1:])]

在迭代它的同时读取python列表中的其他项

6 个答案: