通过遍历矩阵的每一行来查找索引 [英] Finding index by iterating over each row of matrix

问题描述

我有一个大小为5000x10的numpy数组'A'.我还有另一个号码'Num'.我想将以下内容应用于A的每一行:

I have an numpy array 'A' of size 5000x10. I also have another number 'Num'. I want to apply the following to each row of A:

import numpy as np
np.max(np.where(Num > A[0,:]))

是否有比上面编写for循环更好的pythonic方法.

Is there a pythonic way than writing a for loop for above.

推荐答案

您可以使用argmax-

A.shape[1] - 1 - (Num > A)[:,::-1].argmax(1)

或者使用cumsum和argmax-

(Num > A).cumsum(1).argmax(1)

说明:基本上，我们希望使用np.max(np.where(..)来获取比较中每一行的最后匹配项.

Explanation : With np.max(np.where(..), we are basically looking to get the last occurrence of matches along each row on the comparison.

同样，我们可以使用argmax.但是，布尔数组上的argmax给我们第一个出现而不是最后一个出现.因此，一种技巧是执行比较并使用[:,::-1]翻转列，然后使用argmax.然后，将列索引减去数组中cols的数量，以使其追溯到原始顺序.

For the same, we can use argmax. But, argmax on a boolean array gives us the first occurrence and not the last one. So, one trick is to perform the comparison and flip the columns with [:,::-1] and then use argmax. The column indices are then subtracted by the number of cols in the array to make it trace back to the original order.

在第二种方法上，它与 related post 非常相似，因此引用了它:

On the second approach, it's very similar to a related post and therefore quoting from it :

argmax的用途之一是获取max元素在数组中沿轴的第一次出现的ID.因此，我们沿行获得了总和，并获得了第一个最大ID，该ID代表了最后一个非零的elem.这是因为剩余元素的总和不会增加最后一个非零元素之后的总和.

One of the uses of argmax is to get ID of the first occurence of the max element along an axis in an array . So, we get the cumsum along the rows and get the first max ID, which represents the last non-zero elem. This is because cumsum on the leftover elements won't increase the sum value after that last non-zero element.

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