CODEWITHSUNDEEP
pythonarraysnumpy
Jason
Jason

Reputation: 3266

Replace values in an nd numpy array at given axis index

I want to replace the values in a given numpy array (A) at a given index (e.g. 0), along a given axis (e.g. -2) with a given value (e.g. 0), equivalently:

A[:,:,0,:]=0

Problem is the input array A may come in as 3D or 4D or other shapes, so for 3D data I would need

A[:,0,:]=0

If it's 5D: A[:,:,:,0,:]=0

Currently I'm using an exec() to get this done:

slicestr=[':']*numpy.ndim(var)
slicestr[-2]=str(0)
slicestr=','.join(slicestr)
cmd='A[%s]=0' %slicestr
exec(cmd)
return A

I'm a bit concerned that the usage of exec() might not be quite a nice approach. I know that numpy.take() can give me the column at specific index along specific axis, but to replace values I still need to construct the slicing/indexing string, which is dynamic. So I wonder is there any native numpy way of achieving this?

Thanks.

Upvotes: 3

Views: 6406

Answers (1)

akuiper
akuiper

Reputation: 214927

You can use Ellipsis (see numpy indexing for more) to skip the first few dimensions:

# assert A.ndim >= 2
A[...,0,:]=0

A2d = np.arange(12).reshape(2,6)
A3d = np.arange(12).reshape(2,3,2)
A4d = np.arange(12).reshape(2,1,3,2)

A2d[...,0,:] = 0

A2d
#array([[ 0,  0,  0,  0,  0,  0],
#       [ 6,  7,  8,  9, 10, 11]])

A3d[...,0,:] = 0

A3d
#array([[[ 0,  0],
#        [ 2,  3],
#        [ 4,  5]],

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

A4d[...,0,:] = 0

A4d
#array([[[[ 0,  0],
#         [ 2,  3],
#         [ 4,  5]]],  

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

Upvotes: 7

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