将列表切成n个几乎相等长度的分区 [英] Slicing a list into n nearly-equal-length partitions

问题描述

我正在寻找一种快速，干净，pythonic的方式将列表分为正好n个几乎相等的分区.

I'm looking for a fast, clean, pythonic way to divide a list into exactly n nearly-equal partitions.

partition([1,2,3,4,5],5)->[[1],[2],[3],[4],[5]]
partition([1,2,3,4,5],2)->[[1,2],[3,4,5]] (or [[1,2,3],[4,5]])
partition([1,2,3,4,5],3)->[[1,2],[3,4],[5]] (there are other ways to slice this one too)

在这里在列表切片上进行迭代的答案很接近想要的，除了它们专注于列表的 size ，而且我关心列表的 number (其中有些还填充了None).显然，这些都是微不足道的转换，但是我正在寻找最佳实践.

There are several answers in here Iteration over list slices that run very close to what I want, except they are focused on the size of the list, and I care about the number of the lists (some of them also pad with None). These are trivially converted, obviously, but I'm looking for a best practice.

类似地，人们在这里指出了很好的解决方案

Similarly, people have pointed out great solutions here How do you split a list into evenly sized chunks? for a very similar problem, but I'm more interested in the number of partitions than the specific size, as long as it's within 1. Again, this is trivially convertible, but I'm looking for a best practice.

推荐答案

def partition(lst, n):
    division = len(lst) / float(n)
    return [ lst[int(round(division * i)): int(round(division * (i + 1)))] for i in xrange(n) ]

>>> partition([1,2,3,4,5],5)
[[1], [2], [3], [4], [5]]
>>> partition([1,2,3,4,5],2)
[[1, 2, 3], [4, 5]]
>>> partition([1,2,3,4,5],3)
[[1, 2], [3, 4], [5]]
>>> partition(range(105), 10)
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [11, 12, 13, 14, 15, 16, 17, 18, 19, 20], [21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31], [32, 33, 34, 35, 36, 37, 38, 39, 40, 41], [42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52], [53, 54, 55, 56, 57, 58, 59, 60, 61, 62], [63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73], [74, 75, 76, 77, 78, 79, 80, 81, 82, 83], [84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94], [95, 96, 97, 98, 99, 100, 101, 102, 103, 104]]

Python 3版本:

Python 3 version:

def partition(lst, n):
    division = len(lst) / n
    return [lst[round(division * i):round(division * (i + 1))] for i in range(n)]

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