How to change numpy array based on mask array?

Question

I have an array data_set, size:(172800,3) and mask array, size (172800) consists of 1's and 0's. I would like to replace value form data_set array based on values (0 or 1) in mask array by the value defined by me: ex : [0,0,0] or [128,16,128].

I have tried, "np.placed" function but here the problem is the incorrect size of mask array.

I have also checked the more pythonic way: data_set[mask]= [0,0,0] it worked fine but for some raison only for 2 first elements.

data_set[mask]= [0,0,0]

data_set = np.place(data_set, mask, [0,0,0])

My expected output is to change the value of element in data_set matrix to [0,0,0] if the mask value is 1.

ex.

data_set = [[134,123,90] , [234,45,65] , [32,233,45]]
mask = [ 1, 0, 1]

output = [[0,0,0] , [234, 45,65] , [0,0,0]]

Or data_set[np.where(mask)] = 0., data_set[mask == 1] = 0. — wwii
– wwii, Commented Nov 3, 2019 at 15:49
thank you. it works with [0,0,0] but what if i need to replace it with other values like ex. [128,16,128]? — donchuan
– donchuan, Commented Nov 3, 2019 at 15:50

Aleon · Accepted Answer · 2019-11-03 15:59:40Z

2

When you try to index your data with mask numpy assumes you are giving it a list of indices. Use boolean arrays, or convert your mask to a list of indices:

import numpy as np

data_set = np.array([[134,123,90] , [234,45,65] , [32,233,45]])
mask = np.array([1, 0, 1])
val = np.zeros(data_set.shape[1])

data_set[mask.astype(bool),:] = val
# or
data_set[np.where(mask),:] = val

The first one converts your array of ints to an array of bools, while the second one creates a list of indexes where the mask is not zero.

You can set val to whatever value you need as long as it matches the remaining dimension of the dataset (in this case, 3).

edited Nov 3, 2019 at 15:59

answered Nov 3, 2019 at 15:54

Aleon

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