Changing specified values on numpy array in a certain dimension

Question

Wondering if there is an easy way to do this:

Say I have an numpy array with the shape (2,3,2), for example:

x = 
[[[ 0, 1],
  [ 2, 3],
  [ 4, 5]],

 [[ 6, 7],
  [ 8, 9],
  [10,11]]]

If I wanted to replace all the entries that corresponded to axis=1 and position=0, with zero, I could do this easily:

x[:,0,:] = 0
x = 
[[[ 0  0]
  [ 2  3]
  [ 4  5]]

 [[ 0  0]
  [ 8  9]
  [10 11]]]

However, what If I had a list of axes that I wanted to perform these operations on. Is there a built-in numpy function for this? Ideally it'd look something like this:

array_replace(array=x,axis=1,pos=0,replace_val=0)

Which would give the same array as above.

I can think of a way to do this by flattening matrices and calculating where the positions of each variable would be based on the dimension of each array, but I'm wondering if there is already something built into numpy.

I'm not sure what you want, but this isn't the best way to do it, I can guarantee it. Numpy is powerful enough that such operations can be vectorised easily. — cs95
– cs95, Commented Aug 5, 2017 at 22:36

Jonas Adler · Accepted Answer · 2017-08-05 22:39:04Z

2

You can construct a tuple of slices and then use the specific axis for the position. So you can define such function as:

def array_replace(array, axis, pos, replace_val):
    array[(slice(None),) * axis + (pos,)] = replace_val

So what we do is constructing a 1-tuple with a slice object (slice(None),). A slice object is what Python generates behind the curtains for a colon :. So x[:,0,:] is a simple representation for x[(slice(None),0,slice(None))].

Next we repeat this slice axis times (the number of times before the specified axis), followed by the position we want. The remaining slices are optional, so we do not specify these here. We then use numpy's broadcasting to assign the replace_val to all these indices.

This then generates:

>>> x
array([[[ 0,  1],
        [ 2,  3],
        [ 4,  5]],

       [[ 6,  7],
        [ 8,  9],
        [10, 11]]])
>>> array_replace(array=x, axis=1, pos=0, replace_val=0)
>>> x
array([[[ 0,  0],
        [ 2,  3],
        [ 4,  5]],

       [[ 0,  0],
        [ 8,  9],
        [10, 11]]])

edited Aug 5, 2017 at 22:39

Jonas Adler

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answered Aug 5, 2017 at 22:28

willeM_ Van Onsem

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2 Comments

lstbl Over a year ago

Thanks! Do you know of a reference for what the slice function is doing "behind the curtains"? I haven't come across this usage before but it seems very useful.

willeM_ Van Onsem Over a year ago

@lstbl: slice is constructor. It construct a slice object. It does not do much, it is simply there as a representation of :. Suc that an object with a __getitem__ function, can handle slices. It has a start, step and stop attribute.

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