Replacing fill values in 2D array with values from a 1D array

Question

My goal is to fill a 2D array with values from a 1D array that exactly matches the pattern of values in the 2D array. For example:

array_a = 
([[nan,nan,0],
 [0,nan,0],
 [nan,0,0], 
 [0,0,nan]])

array_b = 
([0.324,0.254,0.204,
0.469,0.381,0.292,
0.550])

And I want to get this:

array_c = 
([[nan,nan,0.324],
 [0.254,nan,0.204],
 [nan,0.469,0.381], 
 [0.292,0.550,nan]])

The number of values that need to be filled in array_a will exactly match the number of values in array_b. The main issue is that I want to have the nan values in the appropiate order throughout the array and I'm not sure how best to do that.

flatten your array_a, get the indices of non nan elements, replace the elements at these indices with array_b and reshape back — Sheldore
– Sheldore, Commented Aug 29, 2018 at 20:50
@Bazingaa, no need to flatten or reshape. Boolean indexing flattens automatically. — hpaulj
– hpaulj, Commented Aug 29, 2018 at 21:04

hpaulj · Accepted Answer · 2018-08-29 21:01:48Z

3

boolean indexing does the job nicely:

Locate the nan:

In [229]: mask = np.isnan(array_a)
In [230]: mask
Out[230]: 
array([[ True,  True, False],
       [False,  True, False],
       [ True, False, False],
       [False, False,  True]])

boolean mask applied to the array produces a 1d array:

In [231]: array_a[~mask]
Out[231]: array([0., 0., 0., 0., 0., 0., 0.])

Use that same array in a set context:

In [232]: array_a[~mask]=array_b
In [233]: array_a[~mask]
Out[233]: array([0.324, 0.254, 0.204, 0.469, 0.381, 0.292, 0.55 ])
In [234]: array_a
Out[234]: 
array([[  nan,   nan, 0.324],
       [0.254,   nan, 0.204],
       [  nan, 0.469, 0.381],
       [0.292, 0.55 ,   nan]])

answered Aug 29, 2018 at 21:01

hpaulj

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Onyambu · Accepted Answer · 2018-08-29 21:52:09Z

2

You can also do:

np.place(array_a, array_a == 0, array_b)

array_a

array([[  nan,   nan, 0.324],
       [0.254,   nan, 0.204],
       [  nan, 0.469, 0.381],
       [0.292, 0.55 ,   nan]])

answered Aug 29, 2018 at 21:52

Onyambu

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J. Blackadar · Accepted Answer · 2018-08-29 20:56:08Z

1

This should do the trick, although there might be a pre-written solution or a list comprehension to do the same.

import numpy as np
b_index = 0
array_c = np.zeros(np.array(array_a).shape)
for row_index, row in enumerate(array_a):
    for col_index, col in enumerate(row):
        if not np.isnan(col):
            array_c[row_index, col_index] = array_b[b_index]
            b_index += 1
        else:
            array_c[row_index, col_index] = np.nan

>>> print(array_c)
[[  nan   nan 0.324]
 [0.254   nan 0.204]
 [  nan 0.469 0.381]
 [0.292 0.55    nan]]

answered Aug 29, 2018 at 20:56

J. Blackadar

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