Python:将列表切成n个几乎相等长度的分区

我正在寻找一个快速,干净,pythonic的方式将列表分成几乎几乎相等的分区。

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)

在这里有几个答案Iteration over list slices运行非常接近我想要的,除了他们专注于列表的大小,我关心的列表数量(其中一些也用无填充)。这些都是简单的转换,显然,但我在寻找一个最佳实践。

同样,人们在这里指出了一个很好的解决方案How do you split a list into evenly sized chunks?一个非常类似的问题,但我更感兴趣的分区数量比具体的大小,只要它在1之内。再次,这是trivially可转换,但我寻找最佳实践。

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:

def partition(lst, n):
    division = len(lst) / n
    return [lst[round(division * i):round(division * (i + 1))] for i in range(n)]
http://stackoverflow.com/questions/2659900/python-slicing-a-list-into-n-nearly-equal-length-partitions

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