I would like to obtain a new matrix which calculates the sum of inverse of non-zero elements of each row. The desired output is attached.
import numpy as np
A=np.array([[1,2,0],[4,0,6],[0,8,9]])
Desired output:
array([[(1/1) + (1/2),
(1/4)+(1/6),
(1/8)+(1/9)]])
Run:
result = np.divide(1, A, where=A != 0).sum(1)
The result is:
array([1.5 , 0.41666667, 0.23611111])
Details:
np.divide(1, A, where=A != 0)
computes "conditional" inverse of each element.
Actually 1 / A
(for each element) is computed only for elements.= 0.
And the last step is to sum this intermediate result by each row.
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.