Converting Specified Elements of a NumPy Array by a New Value

Question

I wanted to convert the specified elements of the NumPy array A: 1, 5, and 8 into 0.

So I did the following:

import numpy as np

A = np.array([[1,2,3,4,5],[6,7,8,9,10]])

bad_values = (A==1)|(A==5)|(A==8)

A[bad_values] = 0

print A

Yes, I got the expected result, i.e., new array.

However, in my real world problem, the given array (A) is very large and is also 2-dimensional, and the number of bad_values to be converted into 0 are also too many. So, I tried the following way of doing that:

bads = [1,5,8] # Suppose they are the values to be converted into 0

bad_values = A == x for x in bads  # HERE is the problem I am facing

How can I do this?

Then, of course the remaining is the same as before.

A[bad_values] = 0

print A

Alex Riley · Accepted Answer · 2014-09-28 12:20:57Z

4

If you want to get the index of where a bad value occurs in your array A, you could use in1d to find out which values are in bads:

>>> np.in1d(A, bads)
array([ True, False, False, False,  True, False, False,  True, False, False], dtype=bool)

So you can just write A[np.in1d(A, bads)] = 0 to set the bad values of A to 0.

EDIT: If your array is 2D, one way would be to use the in1d method and then reshape:

>>> B = np.arange(9).reshape(3, 3)
>>> B
array([[0, 1, 2],
       [3, 4, 5],
       [6, 7, 8]])

>>> np.in1d(B, bads).reshape(3, 3)
array([[False,  True, False],
       [False, False,  True],
       [False, False,  True]], dtype=bool)

So you could do the following:

>>> B[np.in1d(B, bads).reshape(3, 3)] = 0
>>> B
array([[0, 0, 2],
       [3, 4, 0],
       [6, 7, 0]])

edited Sep 28, 2014 at 12:20

answered Sep 28, 2014 at 12:07

Alex Riley

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3 Comments

Roman Over a year ago

Thanks. Sorry I forget to mention that my REAL array A is two dimensional. Is your solution still okay?

Roman Over a year ago

Thank you so much. Accepted. Would up vote if I had >15 points.

Alex Riley Over a year ago

No problem at all! Glad you found the answer helpful.

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