从numpy数组中获取最小的N值，忽略inf和nan

I need a good, quick method for finding the 10 smallest real values from a numpy array that could have arbitrarily many nan and/or inf values.

我需要一个好的，快速的方法来从numpy数组中找到10个最小的实数值，这个数组可能有任意多个nan和/或inf值。

I need to identify the indices of these smallest real values, not the values themselves.

我需要确定这些最小实际值的索引，而不是值本身。

I have found the argmin and nanargmin functions from numpy. They aren't really getting the job done because I also want to specify more than 1 value, like I want the smallest 100 values, for example. Also they both return -inf values as being the smallest value when it is present in the array.

我从numpy找到了argmin和nanargmin函数。他们并没有真正完成工作，因为我还想指定多于1个值，例如我想要最小的100个值。此外，它们都将-inf值返回为数组中存在的最小值。

heapq.nsmallest kind of works, but it also returns nan and -inf values as smallest values. Also it doesn't give me the indices that I am looking for.

heapq.nsmallest类的工作，但它也返回nan和-inf值作为最小值。它也没有给我我正在寻找的指数。

Any help here would be greatly appreciated.

这里的任何帮助将不胜感激。

3 个解决方案

#1

The only values that should be throwing this out are the negative infinite ones. So try:

应该抛出的唯一值是负无限值。所以尝试：

import numpy as np
a = np.random.rand(20)
a[4] = -np.inf
k = 10
a[np.isneginf(a)] = inf
result = a[np.argsort(a)[:k]]

#2

It seems to me like you could just take the first n finite values from your sorted array, instead of trying to modify the original array, which could be dangerous.

在我看来，你可以从排序数组中取出前n个有限值，而不是试图修改原始数组，这可能很危险。

n = 10
b = np.sort(a)
smalls = b[np.isfinite(b)][n:]

#3

you can find the index of inf and Nan like this:

你可以找到inf和Nan的索引，如下所示：

a=np.array([[12,12,111],[np.inf,np.inf,1,2,3],[np.nan,7,8]])

the you can loop through a and check it with:

你可以循环并检查它：

for item in a:    
    idxInf=(np.isnan(a[item])).nonzero()
    idxNan=(np.isnan(a[item])).nonzero()

i.e:

即：

In [17]: (np.isnan(a[2]))
Out[17]: array([ True, False, False], dtype=bool)

In [18]: (np.isnan(a[2])).nonzero()
Out[18]: (array([0]),)

#1