Suppose I have a Nx3 array A, and another empty MxNx3 array B. I want to copy the values from A to B such that those sets of values appear M times in B. How to do this efficiently other than using a loop?
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3 Answers
You can write b[:] = a and let broadcasting take over. For example:
>>> a = np.arange(6).reshape(2, 3)
>>> b = np.zeros((3, 2, 3))
>>> a
array([[0, 1, 2],
[3, 4, 5]])
>>> b
array([[[ 0., 0., 0.],
[ 0., 0., 0.]],
[[ 0., 0., 0.],
[ 0., 0., 0.]],
[[ 0., 0., 0.],
[ 0., 0., 0.]]])
Then to copy a into b:
>>> b[:] = a
>>> b
array([[[ 0., 1., 2.],
[ 3., 4., 5.]],
[[ 0., 1., 2.],
[ 3., 4., 5.]],
[[ 0., 1., 2.],
[ 3., 4., 5.]]])
Note that b has to be able to hold the datatype of a. If a was an array of complex numbers, the imaginary part would be lost when copying to b (because it can only hold float values).
1 Comment
Eric
TIL broadcasting works for slice assignment instead of just on mathematical operators
You also could write: suppose your array a is numpy array
ans = numpy.array([a.tolist()*M])
Comments
It seems that for large arrays numpy.repeat(a, repeats, axis) is the fastest option, despite allocating b's memory on the fly.
import numpy as np
a = np.arange(20 * 30 * 40).reshape(20, 30, 40)
b = np.empty((16, 20, 30, 40))
%timeit b[:] = a
# 272 µs ± 22.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit np.empty((16, 20, 30, 40))[:] = a
# 277 µs ± 19 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit np.repeat(a[np.newaxis, ...], repeats=16, axis=0)
# 140 µs ± 483 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
