问题描述
给定一个数组"a",我想按列对数组进行排序.sort(a, axis=0)
对数组做一些处理,然后撤消排序.我所说的不是重新排序,而是基本上颠倒了每个元素的移动方式.我认为argsort()
是我所需要的,但是我不清楚如何用argsort()
的结果对数组进行排序,或更重要的是应用argsort()
Given an array 'a' I would like to sort the array by columns sort(a, axis=0)
do some stuff to the array and then undo the sort. By that I don't mean re sort but basically reversing how each element was moved. I assume argsort()
is what I need but it is not clear to me how to sort an array with the results of argsort()
or more importantly apply the reverse/inverse of argsort()
这里有更多细节
我有一个数组a
,shape(a) = rXc
我需要对每一列进行排序
I have an array a
, shape(a) = rXc
I need to sort each column
aargsort = a.argsort(axis=0) # May use this later
aSort = a.sort(axis=0)
现在平均每一行
aSortRM = asort.mean(axis=1)
现在用行均值替换一行中的每个col.有没有比这更好的方法
now replace each col in a row with the row mean.is there a better way than this
aWithMeans = ones_like(a)
for ind in range(r) # r = number of rows
aWithMeans[ind]* aSortRM[ind]
现在,我需要撤消在第一步中所做的排序. ????
Now I need to undo the sort I did in the first step. ????
推荐答案
我不确定如何在numpy
中做到最好,但是,在纯Python中,推理将是:
I'm not sure how best to do it in numpy
, but, in pure Python, the reasoning would be:
aargsort
保留了range(len(a))
的排列,告诉您aSort
的项目来自何处-就像在纯Python中一样:
aargsort
is holding a permutation of range(len(a))
telling you where the items of aSort
came from -- much like, in pure Python:
>>> x = list('ciaobelu')
>>> r = range(len(x))
>>> r.sort(key=x.__getitem__)
>>> r
[2, 4, 0, 5, 1, 6, 3, 7]
>>>
即sorted(x)
的第一个参数是x[2]
,第二个参数x[4]
,依此类推.
i.e., the first argument of sorted(x)
will be x[2]
, the second one x[4]
, and so forth.
因此,给定排序的版本,您可以通过将项目放回它们的来源"来重建原始版本:
So given the sorted version, you can reconstruct the original by "putting items back where they came from":
>>> s = sorted(x)
>>> s
['a', 'b', 'c', 'e', 'i', 'l', 'o', 'u']
>>> original = [None] * len(s)
>>> for i, c in zip(r, s): original[i] = c
...
>>> original
['c', 'i', 'a', 'o', 'b', 'e', 'l', 'u']
>>>
当然,会有更严格,更快速的方法来用numpy
表示(不幸的是,我对Python的了解不如我对Python本身的了解;-),但是我希望这可以通过显示您需要执行的将东西放回原位"操作的基本逻辑.
Of course there are going to be tighter and faster ways to express this in numpy
(which unfortunately I don't know inside-out as much as I know Python itself;-), but I hope this helps by showing the underlying logic of the "putting things back in place" operation you need to perform.
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