您可以用来
itertools.islice从列表中获取切片式迭代器：
例：
>>> from itertools import islice>>> lis = range(20)>>> for x in islice(lis, 10, None, 1):... print x... 10111213141516171819
更新：
正如@
user2357112所指出的那样，的性能
islice取决于切片的起点和可迭代的大小，在几乎所有情况下，普通切片都将很快，因此应首选。以下是一些时序比较：
对于 巨大的名单
islice是稍快或等于正常片时片的起点是列表中只有不到一半的大小，更大的指标正常切片是明显的赢家。
>>> def func(lis, n): it = iter(lis) for x in islice(it, n, None, 1):pass... >>> def func1(lis, n): #it = iter(lis) for x in islice(lis, n, None, 1):pass... >>> def func2(lis, n): for x in lis[n:]:pass... >>> lis = range(10**6)>>> n = 100>>> %timeit func(lis, n)10 loops, best of 3: 62.1 ms per loop>>> %timeit func1(lis, n)1 loops, best of 3: 60.8 ms per loop>>> %timeit func2(lis, n)1 loops, best of 3: 82.8 ms per loop>>> n = 1000>>> %timeit func(lis, n)10 loops, best of 3: 64.4 ms per loop>>> %timeit func1(lis, n)1 loops, best of 3: 60.3 ms per loop>>> %timeit func2(lis, n)1 loops, best of 3: 85.8 ms per loop>>> n = 10**4>>> %timeit func(lis, n)10 loops, best of 3: 61.4 ms per loop>>> %timeit func1(lis, n)10 loops, best of 3: 61 ms per loop>>> %timeit func2(lis, n)1 loops, best of 3: 80.8 ms per loop>>> n = (10**6)/2>>> %timeit func(lis, n)10 loops, best of 3: 39.2 ms per loop>>> %timeit func1(lis, n)10 loops, best of 3: 39.6 ms per loop>>> %timeit func2(lis, n)10 loops, best of 3: 41.5 ms per loop>>> n = (10**6)-1000>>> %timeit func(lis, n)100 loops, best of 3: 18.9 ms per loop>>> %timeit func1(lis, n)100 loops, best of 3: 18.8 ms per loop>>> %timeit func2(lis, n)10000 loops, best of 3: 50.9 us per loop #clear winner for large index>>> %timeit func1(lis, n)
对于 小型列表， 普通切片比
islice几乎所有情况下都快。
>>> lis = range(1000)>>> n = 100>>> %timeit func(lis, n)10000 loops, best of 3: 60.7 us per loop>>> %timeit func1(lis, n)10000 loops, best of 3: 59.6 us per loop>>> %timeit func2(lis, n)10000 loops, best of 3: 59.9 us per loop>>> n = 500>>> %timeit func(lis, n)10000 loops, best of 3: 38.4 us per loop>>> %timeit func1(lis, n)10000 loops, best of 3: 33.9 us per loop>>> %timeit func2(lis, n)10000 loops, best of 3: 26.6 us per loop>>> n = 900>>> %timeit func(lis, n)10000 loops, best of 3: 20.1 us per loop>>> %timeit func1(lis, n)10000 loops, best of 3: 17.2 us per loop>>> %timeit func2(lis, n)10000 loops, best of 3: 11.3 us per loop
结论：
去正常切片。
