Changing subarrays in 3D arrays using values of a 2D array using numpy

Question

I'm working on a code and a question just pop up in my head. So basically I have a 3D numpy array with shape
(2, 5, 5) and also a 2D numpy array with shape (2, 4) (this is just an example, the arrays, can be much bigger). What I need is to replace the values of 1 in the subarrays of the 3D array (slice [:, 2:, 2:] ) by the values in my 2D array. I thought about getting the index from the values I want to change (the ones) in the 3D array and then use a for loop in the 2D array to iterate through the values but I'm not sure if it's efficient way and also I'm getting an error.

import numpy as np

arr = np.array([[[0.,  1., 43., 25., 21.],
                 [0.,  0.,  0.,  0.,  0.],
                 [0., 43.,  0.,  1.,  0.], 
                 [0., 43.,  1.,  0.,  1.],
                 [0., 45.,  0.,  1.,  0.]],

                [[0.,  1., 38., 29., 46.],
                 [0.,  0.,  0.,  0.,  0.],
                 [0., 32.,  0.,  0.,  1.],
                 [0., 26.,  0.,  0.,  1.],
                 [0., 30.,  1.,  1.,  0.]]])

values = [[2, 3, 1, 4],
          [4, 1, 5, 9]]

indexes = np.argwhere(newarr[:, 2:, 2:] == 1) + [0, 2, 2]

# indexes = [[0 2 3]
#            [0 3 2]
#            [0 3 4]
#            [0 4 3]
#            [1 2 4]
#            [1 3 4]
#            [1 4 2]
#            [1 4 3]]

for i in values:
    arr[indexes] == i

#Error
#index 2 is out of bounds for axis 0 with size 2

My desired output should be

newarr = [[[0.,  1., 43., 25., 21.],
           [0.,  0.,  0.,  0.,  0.],
           [0., 43.,  0.,  2.,  0.], 
           [0., 43.,  3.,  0.,  1.],
           [0., 45.,  0.,  4.,  0.]],

          [[0.,  1., 38., 29., 46.],            
           [0.,  0.,  0.,  0.,  0.],
           [0., 32.,  0.,  0.,  4.],
           [0., 26.,  0.,  0.,  1.],
           [0., 30.,  5.,  9.,  0.]]])

I think there should be a more efficient using only numpy, but I can't see how to do this, so any help will be appreciated, thank you!

yatu · Accepted Answer · 2020-07-09 15:39:59Z

2

You can slice arr and use the actual binary values cast to bool to mask the array and slice assign where there are 1s:

a_view = arr[:,2:,2:]
a_view[a_view.astype('bool')] = np.array(values).ravel()

print(arr)
array([[[ 0.,  1., 43., 25., 21.],
        [ 0.,  0.,  0.,  0.,  0.],
        [ 0., 43.,  0.,  2.,  0.],
        [ 0., 43.,  3.,  0.,  1.],
        [ 0., 45.,  0.,  4.,  0.]],

       [[ 0.,  1., 38., 29., 46.],
        [ 0.,  0.,  0.,  0.,  0.],
        [ 0., 32.,  0.,  0.,  4.],
        [ 0., 26.,  0.,  0.,  1.],
        [ 0., 30.,  5.,  9.,  0.]]])

answered Jul 9, 2020 at 15:39

yatu

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Comments

hpaulj · Accepted Answer · 2020-07-09 16:40:07Z

Two issues:

use np.where instead of argwhere; you won't need to iterate
the resulting indices apply to the slice, not the original array

With your array:

In [133]: arr = np.array([[[0.,  1., 43., 25., 21.], 
     ...:                  [0.,  0.,  0.,  0.,  0.], 
          ...
     ...:                  [0., 26.,  0.,  0.,  1.], 
     ...:                  [0., 30.,  1.,  1.,  0.]]]) 
     ...:

the slice:

In [136]: subarr = arr[:,2:,2:]

where the slice values are 1:

In [137]: np.nonzero(subarr==1)                                                                      
Out[137]: 
(array([0, 0, 0, 0, 1, 1, 1, 1]),
 array([0, 1, 1, 2, 0, 1, 2, 2]),
 array([1, 0, 2, 1, 2, 2, 0, 1]))

This like your argwhere except as a tuple of arrays, and not offset.

In [138]: values = [[2, 3, 1, 4], 
     ...:           [4, 1, 5, 9]] 
     ...:

That tuple can be used to index the slice, both for fetching and setting:

In [139]: subarr[np.nonzero(subarr==1)]                                                              
Out[139]: array([1., 1., 1., 1., 1., 1., 1., 1.])
In [140]: subarr[np.nonzero(subarr==1)]=np.ravel(values)

Since subarr is a view, setting values in it also sets values in arr. No need to convert the indices to the arr framework.

In [141]: arr                                                                                        
Out[141]: 
array([[[ 0.,  1., 43., 25., 21.],
        [ 0.,  0.,  0.,  0.,  0.],
        [ 0., 43.,  0.,  2.,  0.],
        [ 0., 43.,  3.,  0.,  1.],
        [ 0., 45.,  0.,  4.,  0.]],

       [[ 0.,  1., 38., 29., 46.],
        [ 0.,  0.,  0.,  0.,  0.],
        [ 0., 32.,  0.,  0.,  4.],
        [ 0., 26.,  0.,  0.,  1.],
        [ 0., 30.,  5.,  9.,  0.]]])

boolean indexing

As the other answer notes, we can select elements with the boolean mask, without the nonzero step (under the covers the indexing is similar if not identical).

In [144]: arr = np.array([[[0.,  1., 43., 25., 21.], 
     ...:                  [0.,  0.,  0.,  0.,  0.], 
    ...
     ...:                  [0., 30.,  1.,  1.,  0.]]]) 
     ...:                                                                                            
In [145]: subarr = arr[:,2:,2:]                                                                      
In [146]: subarr[subarr==1]                                                                          
Out[146]: array([1., 1., 1., 1., 1., 1., 1., 1.])

In [148]: subarr[subarr==1] = np.ravel(values)                                                       
In [149]: arr                                                                                        
Out[149]: 
array([[[ 0.,  1., 43., 25., 21.],
        [ 0.,  0.,  0.,  0.,  0.],
        ....
        [ 0., 30.,  5.,  9.,  0.]]])

closer to your attempt

Actually your iteration could have worked if you'd iterated on indices:

for i,v in zip(idx,np.ravel(values)):
     arr[tuple(i)] == v

Starting with my where tuple:

In [159]: Out[137]                                                                                   
Out[159]: 
(array([0, 0, 0, 0, 1, 1, 1, 1]),
 array([0, 1, 1, 2, 0, 1, 2, 2]),
 array([1, 0, 2, 1, 2, 2, 0, 1]))
In [160]:

Your offset argwhere:

In [160]: idx = np.transpose(Out[137])+[0,2,2]                                                       
In [161]: idx                                                                                        
Out[161]: 
array([[0, 2, 3],
       [0, 3, 2],
       [0, 3, 4],
       [0, 4, 3],
       [1, 2, 4],
       [1, 3, 4],
       [1, 4, 2],
       [1, 4, 3]])

Using that iteratively to index arr (note the use of tuple):

In [162]: [arr[tuple(i)] for i in idx]                                                               
Out[162]: [2.0, 3.0, 1.0, 4.0, 4.0, 1.0, 5.0, 9.0]

Iterating directly on values doesn't work, since that returns 2 lists. It needs to be flattened/raveled.

In [163]: for v in values: print(v)                                                                  
[2, 3, 1, 4]
[4, 1, 5, 9]

Thank you for the long answer, I can see a lot of things I was doing wrong, thanks!

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Changing subarrays in 3D arrays using values of a 2D array using numpy

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boolean indexing

closer to your attempt

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boolean indexing

closer to your attempt

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