在迭代过程中读取python列表中的其他项[重复]

你可以使用

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

尽管在您的案例中(因为您只是在寻找下一个值),您也可以简单地使用

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:])]

>>> 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)]

在……的帮助下呢? pairwise recipe from itertools document

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〕

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

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

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