在Python中对切片进行高效迭代

本文介绍了在Python中对切片进行高效迭代的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 29岁程序员，3月因学历无情被辞！ Python中切片操作的迭代效率如何？如果使用切片不可避免地复制副本，还有替代方法吗？How efficient are iterations over slice operations in Python? And if a copy is inevitable with slices, is there an alternative?我知道列表上的切片操作是O（k），其中k是大小切片。I know that a slice operation over a list is O(k), where k is the size of the slice.x[5 : 5+k] # O(k) copy operation然而，当迭代列表的一部分时，我发现最干净（和大多数Pythonic？）的方式来做到这一点（没有要诉诸指数）：However, when iterating over a part of a list, I find that the cleanest (and most Pythonic?) way to do this (without having to resort to indices) is to do:for elem in x[5 : 5+k]: print elem然而我的直觉是，这仍然导致子列表的昂贵副本，而不是简单地迭代在现有列表上。However my intuition is that this still results in an expensive copy of the sublist, rather than simply iterating over the existing list.推荐答案您可以使用 itertools.islice 来从列表中获取切片迭代器：You can use itertools.islice to get a sliced iterator from the list:示例：>>> from itertools import islice>>> lis = range(20)>>> for x in islice(lis, 10, None, 1):... print x...10111213141516171819 更新： 正如@ user2357112所述， islice 的表现取决于切片的起点和可迭代的正常切片的大小几乎在所有情况下都会很快，应该是首选。以下是一些时间比较：Update:As noted by @user2357112 the performance of islice depends on the start point of slice and the size of the iterable, normal slice is going to be fast in almost all cases and should be preferred. Here are some more timing comparisons:对于巨大的列表 islice 略快或当切片的起始点小于列表大小的一半时，等于正常切片，对于更大的索引，正常切片是明显的赢家。For Huge lists islice is slightly faster or equal to normal slice when the slice's start point is less than half the size of list, for bigger indexes normal slice is the clear winner.>>> 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)对于小列表正常切片是几乎在所有情况下都快于 islice 。For Small lists normal slice is faster than islice for almost all cases.>>> 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： 寻找正常切片。 这篇关于在Python中对切片进行高效迭代的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！