如何在python中按行对2d数组排序? [英] How to sort 2d array by row in python?

问题描述

我有2d数组，尺寸为3x10，我想按第二行中的值(从最低到最高)进行排序.

I have 2d array, dimension 3x10, and I want to sort by values in 2nd row, from lowest to highest value.

推荐答案

Python本身没有"2d数组"-它具有(1d)内置列表以及标准库模块中的(1d)数组数组.有一些第三方库，例如numpy，它们确实提供了Python可用的多维数组，但是如果您正在使用其中的某些库，那么您当然会提到此类第三方库，而不仅仅是说在Python中" ，正确?-)

Python, per se, has no "2d array" -- it has (1d) lists as built-ins, and (1d) arrays in standard library module array. There are third-party libraries such as numpy which do provide Python-usable multi-dimensional arrays, but of course you'd be mentioning such third party libraries if you were using some of them, rather than just saying "in Python", right?-)

因此，我假设"2d数组"是指列表列表，例如:

So I'll assume that by "2d array" you mean a list of lists, such as:

lol = [ range(10), range(2, 12), range(5, 15) ]

或类似内容-即包含3个项目的列表，每个项目是包含10个项目的列表，而第二行"将是子列表项目lol[1].是的，有很多假设，但是您的问题是如此的令人困惑，以至于无法避免做出假设-如果您不喜欢试图读懂想法的人(可能是失败的话)，请编辑您的Q以更精确地阐明问题，并举一个例子！因为您目前无法避免.

or the like -- i.e. a list with 3 items, each item being a list with 10 items, and the "second row" would be the sublist item lol[1]. Yeah, lots of assumptions, but your question is so maddeningly vague that there's no way to avoid making assumptions - edit your Q to clarify with more precision, and an example!, if you dislike people trying to read your mind (and probably failing) as you currently make it impossible to avoid.

因此，在这些假设下，您可以按照对第二个子列表进行排序所需的顺序对3个子列表中的每个列表进行排序，例如:

So under these assumptions you can sort each of the 3 sublists in the order required to sort the second one, for example:

indices = range(10)
indices.sort(key = lol[1].__getitem__)
for i, sublist in enumerate(lol):
  lol[i] = [sublist[j] for j in indices]

这里的一般方法是对索引范围进行排序，然后使用适当排序的范围对正在播放的所有子列表进行重新排序.

The general approach here is to sort the range of indices, then just use that appropriately sorted range to reorder all the sublists in play.

如果您实际上有其他问题，当然会有不同的解决方案；-).

If you actually have a different problem, there will of course be different solutions;-).

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