按行对numpy数组排序，并根据原始数组对匹配值进行排序 [英] Sort numpy array by row and order matching values based on original array

问题描述

我有一个2D numpy数组，我想根据第一列的值对行进行排序.麻烦的是它的格式化方式:

我按以下顺序对列进行排序:0、1、2、3、4、0、1、2、3、4、0、1、2、3、4等->您可以看到它重复出现了

基本上，我想将1s，2s，3s，4s分组.匹配值的顺序很重要:我希望第1个"1"行成为未排序数组中的第一个行，然后是接下来显示的行，依此类推.我使用此命令: sortedData = myData [myData [:，0] .argsort()]

不幸的是，它似乎没有根据数组的原始顺序对匹配的列进行排序.我可以启用某些选项来启用此功能吗?

谢谢！

解决方案

您可以更改合并排序是一种稳定排序算法.)

I have a 2D numpy array and I would like to sort the rows based on first column values. The trouble is the way it is formatted:

Column I am sorting by: 0,1,2,3,4,0,1,2,3,4,0,1,2,3,4,etc ->you can see it repeats itself

Basically I want to group the 1s, then 2s, then 3s, then 4s. The ordering of the matching values matter: I want the 1st '1' row to be the first one that appeared in the unsorted array, followed by the one that shows up next, etc. I use this command: sortedData= myData[myData[:,0].argsort()]

Unfortunately, it doesn't not appear to order matching columns based on the original ordering of the array. Are there certain options I can turn on to enable this?

Thanks!

解决方案

You can change the sorting algorithm used by argsort with the kind argument.

Use

sortedData= myData[myData[:,0].argsort(kind='mergesort')]

to preserve the order of the equal items. (Merge sort is a stable sorting algorithm.)

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