Replace subarrays in numpy

Question

Given an array,

>>> n = 2
>>> a = numpy.array([[[1,1,1],[1,2,3],[1,3,4]]]*n)
>>> a
array([[[1, 1, 1],
        [1, 2, 3],
        [1, 3, 4]],

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

I know that it's possible to replace values in it succinctly like so,

>>> a[a==2] = 0
>>> a
array([[[1, 1, 1],
        [1, 0, 3],
        [1, 3, 4]],

       [[1, 1, 1],
        [1, 0, 3],
        [1, 3, 4]]])

Is it possible to do the same for an entire row (last axis) in the array? I know that a[a==[1,2,3]] = 11 will work and replace all the elements of the matching subarrays with 11, but I'd like to substitute a different subarray. My intuition tells me to write the following, but an error results,

>>> a[a==[1,2,3]] = [11,22,33]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: array is not broadcastable to correct shape

In summary, what I'd like to get is:

array([[[1, 1, 1],
        [11, 22, 33],
        [1, 3, 4]],

       [[1, 1, 1],
        [11, 22, 33],
        [1, 3, 4]]])

... and n of course is, in general, a lot larger than 2, and the other axes are also larger than 3, so I don't want to loop over them if I don't need to.

Update: The [1,2,3] (or whatever else I'm looking for) is not always at index 1. An example:

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

Saullo G. P. Castro · Accepted Answer · 2014-03-23 12:05:51Z

3

You can achieve this with a much higher performance using np.all to check if all the columns have a True value for your comparison, then using the created mask to replace the values:

mask = np.all(a==[1,2,3], axis=2)
a[mask] = [11, 22, 23]

print(a)
#array([[[ 1,  1,  1],
#        [11, 22, 33],
#        [ 1,  3,  4]],
# 
#       [[ 1,  1,  1],
#        [11, 22, 33],
#        [ 1,  3,  4]]])

answered Mar 23, 2014 at 12:05

Saullo G. P. Castro

59.4k28 gold badges191 silver badges244 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Dunes · Accepted Answer · 2012-06-05 09:26:59Z

2

You have to do something a little more complicated to acheive what you want.

You can't select slices of arrays as such, but you can select all the specific indexes you want.

So first you need to construct an array that represents the rows you wish to select. ie.

data = numpy.array([[1,2,3],[55,56,57],[1,2,3]])

to_select = numpy.array([1,2,3]*3).reshape(3,3) # three rows of [1,2,3]

selected_indices = data == to_select
# array([[ True,  True,  True],
#        [False, False, False],
#        [ True,  True,  True]], dtype=bool)

data = numpy.where(selected_indices, [4,5,6], data)
# array([[4, 5, 6],
#        [55, 56, 57],
#        [4, 5, 6]])

# done in one step, but perhaps not very clear as to its intent
data = numpy.where(data == numpy.array([1,2,3]*3).reshape(3,3), [4,5,6], data)

numpy.where works by selecting from the second argument if true and the third argument if false.

You can use where to select from 3 different types of data. The first is an array that has the same shape as selected_indices, the second is just a value on its own (like 2 or 7). The first is most complicated as can be of shape that can be broadcast into the same shape as selected_indices. In this case we provided [1,2,3] which can be stacked together to get an array with shape 3x3.

answered Jun 5, 2012 at 9:26

Dunes

42.1k7 gold badges86 silver badges107 bronze badges

1 Comment

Karol Over a year ago

Brilliant! I didn't think about where.

BrenBarn · Accepted Answer · 2012-06-05 08:39:28Z

0

Note sure if this is what you want, your code example does not create the array you say it does. But:

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

       [[1, 1, 1],
        [1, 2, 3],
        [1, 3, 4]]])
>>> a[:,1,:] = [[8, 8, 8], [8,8,8]]
>>> a
array([[[1, 1, 1],
        [8, 8, 8],
        [1, 3, 4]],

       [[1, 1, 1],
        [8, 8, 8],
        [1, 3, 4]]])
>>> a[:,1,:] = [88, 88, 88]
>>> a
array([[[ 1,  1,  1],
        [88, 88, 88],
        [ 1,  3,  4]],

       [[ 1,  1,  1],
        [88, 88, 88],
        [ 1,  3,  4]]])

answered Jun 5, 2012 at 8:39

BrenBarn

253k39 gold badges421 silver badges392 bronze badges

1 Comment

Karol Over a year ago

You're right, there was a mistake in my question... fixed it now. Your solution works for my specific example, but [1,2,3] will not always be at index 1, so a[:,1,:] is not something that will work in general. Take, for example this array: a = numpy.array([[[1,1,1],[1,2,3],[1,3,4]], [[1,2,3],[1,1,1],[1,3,4]]]). I need to find the places where [1,2,3] occurs.

Collectives™ on Stack Overflow

Replace subarrays in numpy

3 Answers 3

Comments

1 Comment

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

1 Comment

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related