Concatenate two numpy arrays in the 4th dimension
Question:
I have two numpy arrays with three dimensions (3 x 4 x 5) and I want to concatenate them so the result has four dimensions (3 x 4 x 5 x 2). In Matlab, this can be done with cat(4, a, b), but not in Numpy.
For example:
a = ones((3,4,5))
b = ones((3,4,5))
c = concatenate((a,b), axis=3) # error!
To clarify, I wish c[:,:,:,0] and c[:,:,:,1] to correspond to the original two arrays.
Answers:
This works for me:
c = numpy.array([a,b])
Though it would be nice if it worked your way, too.
How about the following:
c = concatenate((a[:,:,:,None],b[:,:,:,None]), axis=3)
This gives a (3 x 4 x 5 x 2) array, which I believe is laid out in the manner you require.
Here, None is synonymous to np.newaxis: Numpy: Should I use newaxis or None?
edit As suggested by @Joe Kington, the code could be cleaned up a little bit by using an ellipsis:
c = concatenate((a[...,None],b[...,None]), axis=3)
Here you go:
import numpy as np
a = np.ones((3,4,5))
b = np.ones((3,4,5))
c = np.concatenate((a[...,np.newaxis],b[...,np.newaxis]),axis=3)
It’s not necessarily the most elegant, but I’ve used variations of
c = rollaxis(array([a,b]), 0, 4)
in the past.
The accepted answer above is great. But I’ll add the following because I’m a math dork and it’s a nice use of the fact that a.shape is a.T.shape[::-1]…i.e. taking a transpose reverses the order of the indices of a numpy array. So if you have your building blocks in an array called blocks, then the solution above is:
new = np.concatenate([block[..., np.newaxis] for block in blocks],
axis=len(blocks[0].shape))
but you could also do
new2 = np.array([block.T for block in blocks]).T
which I think reads more cleanly. It’s worth noting that the already-accepted answer runs more quickly:
%%timeit
new = np.concatenate([block[..., np.newaxis] for block in blocks],
axis=len(blocks[0].shape))
1000 loops, best of 3: 321 µs per loop
while
%%timeit
new2 = np.array([block.T for block in blocks]).T
1000 loops, best of 3: 407 µs per loop
What about
c = np.stack((a,b), axis=3)
I have two numpy arrays with three dimensions (3 x 4 x 5) and I want to concatenate them so the result has four dimensions (3 x 4 x 5 x 2). In Matlab, this can be done with cat(4, a, b), but not in Numpy.
For example:
a = ones((3,4,5))
b = ones((3,4,5))
c = concatenate((a,b), axis=3) # error!
To clarify, I wish c[:,:,:,0] and c[:,:,:,1] to correspond to the original two arrays.
This works for me:
c = numpy.array([a,b])
Though it would be nice if it worked your way, too.
How about the following:
c = concatenate((a[:,:,:,None],b[:,:,:,None]), axis=3)
This gives a (3 x 4 x 5 x 2) array, which I believe is laid out in the manner you require.
Here, None is synonymous to np.newaxis: Numpy: Should I use newaxis or None?
edit As suggested by @Joe Kington, the code could be cleaned up a little bit by using an ellipsis:
c = concatenate((a[...,None],b[...,None]), axis=3)
Here you go:
import numpy as np
a = np.ones((3,4,5))
b = np.ones((3,4,5))
c = np.concatenate((a[...,np.newaxis],b[...,np.newaxis]),axis=3)
It’s not necessarily the most elegant, but I’ve used variations of
c = rollaxis(array([a,b]), 0, 4)
in the past.
The accepted answer above is great. But I’ll add the following because I’m a math dork and it’s a nice use of the fact that a.shape is a.T.shape[::-1]…i.e. taking a transpose reverses the order of the indices of a numpy array. So if you have your building blocks in an array called blocks, then the solution above is:
new = np.concatenate([block[..., np.newaxis] for block in blocks],
axis=len(blocks[0].shape))
but you could also do
new2 = np.array([block.T for block in blocks]).T
which I think reads more cleanly. It’s worth noting that the already-accepted answer runs more quickly:
%%timeit
new = np.concatenate([block[..., np.newaxis] for block in blocks],
axis=len(blocks[0].shape))
1000 loops, best of 3: 321 µs per loop
while
%%timeit
new2 = np.array([block.T for block in blocks]).T
1000 loops, best of 3: 407 µs per loop
What about
c = np.stack((a,b), axis=3)