Suppose I have an 2D numpy array a=[[1,-2,1,0], [1,0,0,-1]], but I want to convert it to an 3D numpy array by element-wise multiply a vector t=[[x0,x0,x0,x0],[x1,x1,x1,x1]] where xi is a 1D numpy array with 3072 size. So the result would be a*t=[[x0,-2x0,x0,0],[x1,0,0,-x1]] with the size (2,4,3072). So how should I do that in Python numpy?
3 Answers
Code:
import numpy as np
# Example data taken from bendl's answer !!!
a = np.array([[1,-2,1,0], [1,0,0,-1]])
xi = np.array([1, 2, 3])
b = np.outer(a, xi).reshape(a.shape[0], -1, len(xi))
print('a:')
print(a)
print('b:')
print(b)
Output:
a:
[[ 1 -2 1 0]
[ 1 0 0 -1]]
b:
[[[ 1 2 3]
[-2 -4 -6]
[ 1 2 3]
[ 0 0 0]]
[[ 1 2 3]
[ 0 0 0]
[ 0 0 0]
[-1 -2 -3]]]
As i said: it looks like an outer-product and splitting/reshaping this one dimension is easy.
7 Comments
David Hx
Thanks, that's exactly what I need.
Nils Werner
Isn't
xi supposed to be 2D, like xi = np.array([[1, 2, 3], [4, 5, 6]])?
sascha
Not according to his words (where xi is a 1D numpy array with 3072 size; at least that's my interpretation).
sascha
@DavidHx I edited the reshape-operation (more generic now). You will need it with different number of rows!
Nils Werner
No, but
t is supposed to be 2D: (2, 3072). And the product we are looking for is between a and t, not between a and xi (which is only one row of t). |
You can use numpy broadcasting for this:
a = numpy.array([[1, -2, 1, 0], [1, 0, 0, -1]])
t = numpy.arange(3072 * 2).reshape(2, 3072)
# array([[ 0, 1, 2, ..., 3069, 3070, 3071], # = x0
# [3072, 3073, 3074, ..., 6141, 6142, 6143]]) # = x1
a.shape
# (2, 4)
t.shape
# (2, 3072)
c = (a.T[None, :, :] * t.T[:, None, :]).T
# array([[[ 0, 1, 2, ..., 3069, 3070, 3071], # = 1 * x0
# [ 0, -2, -4, ..., -6138, -6140, -6142], # = -2 * x0
# [ 0, 1, 2, ..., 3069, 3070, 3071], # = 1 * x0
# [ 0, 0, 0, ..., 0, 0, 0]], # = 0 * x0
#
# [[ 3072, 3073, 3074, ..., 6141, 6142, 6143], # = 1 * x1
# [ 0, 0, 0, ..., 0, 0, 0], # = 0 * x1
# [ 0, 0, 0, ..., 0, 0, 0], # = 0 * x1
# [-3072, -3073, -3074, ..., -6141, -6142, -6143]]]) # = -1 * x1
c.shape
# (2, 4, 3072)
Comments
Does this do what you need?
import numpy as np
a = np.array([[1,-2,1,0], [1,0,0,-1]])
xi = np.array([1, 2, 3])
a = np.dstack([a * i for i in xi])
The docs for this are here: https://docs.scipy.org/doc/numpy-1.13.0/reference/generated/numpy.dstack.html
1 Comment
David Hx
The result is what I need, but can you do it without any for-loop? Because it may affect performance.
np.outer(np.array([[1, -2, 1, 0], [1, 0, 0, -1]]), np.random.random(size=3072))returns a shape of(8, 3072), so can it be that you are looking for this kind of operation?