How to use np.where on 3D array with all()

Question

I've been stuck for quite some time on this problem.

I have an array of shape (4,3,3). In my real case, the shape is much bigger.

a = np.array([
    [[1,2,3], [3,4,2], [1,3,4]],
    [[1,2,3], [3,6,2], [1,4,4]],
    [[1,2,3], [3,6,2], [1,4,4]],
    [[1,2,3], [3,6,2], [1,2,4]]
])

I want to use the np.where function to replace some of the list inside the array. For example, each time I have the list [1,2,3], I want to replace it by the list [99,99,99].

Without using np.where, to check if the list is completely equal to [1,2,3], I would use some function like all().

Using np.where:

np.where(a == [1,2,3], 99, a)

Will result in:

array([[[99, 99, 99],
    [ 3,  4,  2],
    [99,  3,  4]],

   [[99, 99, 99],
    [ 3,  6,  2],
    [99,  4,  4]],

   [[99, 99, 99],
    [ 3,  6,  2],
    [99,  4,  4]],

   [[99, 99, 99],
    [ 3,  6,  2],
    [99, 99,  4]]])

As you can see on the 3rd row, a part of the list has been replaced by 99.

I only want to replace the full list (never a part of the list) when it's exactly [1,2,3] and in the right order.

But I cannot solve this problem. I've been looking a lot online, with no luck.

a[(a == [1,2,3]).all(-1)] = 99, IIUC. Please include your desired result. — Michael Szczesny
– Michael Szczesny, Commented Feb 28, 2022 at 14:30
Is this for an assignment? The "Without using np.where, ..., i would use some function like all(). is very peculiar phrasing ^^ — mozway
– mozway, Commented Feb 28, 2022 at 14:34
@michael Is there an easy solution without using all() ? I'm interested in. — Marie
– Marie, Commented Feb 28, 2022 at 14:53
@Marie np.where works at the element level. I think you have to use .all() at some point to tell np to look at a whole row, now just a single element. @Michael's solution is very elegant imho. It doesn't get much more concise than that! — joanis
– joanis, Commented Feb 28, 2022 at 15:11
Does this answer your question? Identify and replace rows in a ndarray — Michael Szczesny
– Michael Szczesny, Commented Feb 28, 2022 at 16:20

hpaulj · Accepted Answer · 2022-02-28 16:21:38Z

The cond array is central to the workings of where:

In [167]: a == [1, 2, 3]
Out[167]: 
array([[[ True,  True,  True],
        [False, False, False],
        [ True, False, False]],

       [[ True,  True,  True],
        [False, False, False],
        [ True, False, False]],

       [[ True,  True,  True],
        [False, False, False],
        [ True, False, False]],

       [[ True,  True,  True],
        [False, False, False],
        [ True,  True, False]]])

You get 99 everywhere there's a True. We can use all to identify rows:

In [168]: (a == [1, 2, 3]).all(axis=-1)
Out[168]: 
array([[ True, False, False],
       [ True, False, False],
       [ True, False, False],
       [ True, False, False]])

But we can't use that in the where because it is 2d, and a is 3d:

In [169]: np.where(_, 99, a)
Traceback (most recent call last):
  Input In [169] in <module>
    np.where(_, 99, a)
  File <__array_function__ internals>:180 in where
ValueError: operands could not be broadcast together with shapes (4,3) () (4,3,3)

But we can adjust the shape of the cond so it works:

In [171]: np.where(Out[168][...,None], 99, a)
Out[171]: 
array([[[99, 99, 99],
        [ 3,  4,  2],
        [ 1,  3,  4]],

       [[99, 99, 99],
        [ 3,  6,  2],
        [ 1,  4,  4]],

       [[99, 99, 99],
        [ 3,  6,  2],
        [ 1,  4,  4]],

       [[99, 99, 99],
        [ 3,  6,  2],
        [ 1,  2,  4]]])

Michael Szczesny's suggestion also works, because the 2d mask selects the desired rows from a:

In [172]: a[Out[168]]
Out[172]: 
array([[1, 2, 3],
       [1, 2, 3],
       [1, 2, 3],
       [1, 2, 3]])

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