Numpy replacing elements based on logic and value in an identically shaped array

Question

I have 2 numpy arrays. One is filled with boolean values and the other numerical values.

How would I perform logic on the numerical array based on also the current value in the boolean array.

e.g. if true and > 5 then make the value false

matrix1
matrix2

newMatrix = matrix1 > 5 where matrix2 value is false

Please note that these arrays have the same shape e.g.

[[0, 1, 1],  
[1, 0, 0]]

and

[[3, 1, 0]  
[6, 2, 6]]

And the result I would like would be a new boolean matrix that is true if its value is true in the boolean array and the equivalent value in the numerical array is more than 5 e.g.

[[0, 0, 0]  
[1, 0, 0]]

Please, provide an example of the data you work with and the result you want to achieve. — sentence, Jul 03 '19 at 16:34
https://stackoverflow.com/questions/16343752/numpy-where-function-multiple-conditions — Joe, Jul 03 '19 at 18:16
https://stackoverflow.com/questions/20528328/numpy-logical-or-for-more-than-two-arguments — Joe, Jul 03 '19 at 18:17
https://stackoverflow.com/questions/19766757/replacing-numpy-elements-if-condition-is-met — Joe, Jul 03 '19 at 18:18

score 1 · Accepted Answer · answered Jul 03 '19 at 16:38

newMatrix = np.logical_and(matrix2 == 0, matrix1 > 5 )

This will iterate over all elements, and make an 'and' between pairs of booleans from matrix == 0 and matrix1 > 5. Note that matrix1 > 5 type of expression generates a matrix of boolean values.

If you want 0,1 instead of False,True, you can add +0 to the result:

newMatrix = np.logical_and(matrix2 == 0, matrix1 > 5 ) + 0

sentence · Answer 2 · 2019-07-03T17:32:09.623

The clearest way:

import numpy as np

matrix1 = np.array([[3, 1, 0],
                    [6, 2, 6]])

matrix2 = np.array([[0, 1, 1],
                    [1, 0, 0]])

r,c = matrix1.shape

res = np.zeros((r,c))

for i in range(r):
    for j in range(c):
        if matrix1[i,j]>5 and matrix2[i,j]==1:
            res[i,j]=1

result

array([[0., 0., 0.],
       [1., 0., 0.]])

A fancier way, using numpy.where():

import numpy as np

matrix1 = np.array([[3, 1, 0],
                    [6, 2, 6]])

matrix2 = np.array([[0, 1, 1],
                    [1, 0, 0]])

r,c = matrix1.shape

res = np.zeros((r,c))

res[np.where((matrix1>5) & (matrix2==1))]=1

result

array([[0., 0., 0.],
       [1., 0., 0.]])

Numpy replacing elements based on logic and value in an identically shaped array

2 Answers2