replace values in an array

Question

as a replacement value for another within an operation with arrays, or how to search within an array and replace a value by another

for example:

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

where it can replace NaN by 0. thanks for any response

Answer 1

You could do this:

import numpy as np
x=np.array([[np.NaN, 1., 1., 1., 1., 1., 1.],[1., np.NaN, 1., 1., 1., 1., 1.],[1., 1., np.NaN, 1., 1., 1., 1.], [1., 1., 1., np.NaN, 1., 1., 1.], [1., 1., 1., 1., np.NaN, 1., 1.],[1., 1., 1., 1., 1., np.NaN, 1.], [1., 1., 1., 1., 1., 1., np.NaN]])
x[np.isnan(x)]=0

np.isnan(x) returns a boolean array which is True wherever x is NaN. x[ boolean_array ] = 0 employs fancy indexing to assign the value 0 wherever the boolean array is True.

For a great introduction to fancy indexing and much more, see also the numpybook.

Answer 2

these days there is the special function:

a = numpy.nan_to_num(a)

Answer 3

Here is the example array in the question:

import numpy as np
a = np.where(np.eye(7), np.nan, 1)

You can either use numpy.where and numpy.isnan functions to create a new array b:

b = np.where(np.isnan(a), 0, a)

Or use an in-place function to directly modify the a array:

np.place(a, np.isnan(a), 0)  # returns None