Assume a 1D array A is given. Is there an easy way to construct a 3D array B, such that B[i,j,k] = A[k] for all i,j,k? You can assume that the shape of B is prescribed, and that B.shape[2] = A.shape[0].
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1Do you mean you want to tile the array? docs.scipy.org/doc/numpy/reference/generated/numpy.tile.htmlYXD– YXD2011-04-05 23:49:34 +00:00Commented Apr 5, 2011 at 23:49
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3 Answers
>>> k = 4
>>> a = np.arange(k)
>>> j = 3
>>> i = 2
>>> np.tile(a,j*i).reshape((i,j,k))
array([[[0, 1, 2, 3],
[0, 1, 2, 3],
[0, 1, 2, 3]],
[[0, 1, 2, 3],
[0, 1, 2, 3],
[0, 1, 2, 3]]]
3 Comments
D R
You can replace
np.tile(np.tile(a,j),i).reshape((i,j,k)) with np.tile(a, i*j).reshape((i,j,k))
marshall.ward
I removed my
np.tile(A,(i,j,1)) suggestion because it had terrible performance for large arrays. Do you know why np.tile(A,i*j).reshape(i,j,k) scales so much better?
Paul
@MLW: I typed
numpy.source(numpy.tile) and saw that tile calls reshape and repeat for each element in the reps tuple. reshape returns a copy which means re-allocating all that memory each time. Seems poorly implemented to me.Another easy way to do this is simple assignment -- broadcasting will automatically do the right thing:
i = 2
j = 3
k = 4
a = numpy.arange(k)
b = numpy.empty((i, j, k))
b[:] = a
print b
prints
[[[ 0. 1. 2. 3.]
[ 0. 1. 2. 3.]
[ 0. 1. 2. 3.]]
[[ 0. 1. 2. 3.]
[ 0. 1. 2. 3.]
[ 0. 1. 2. 3.]]]
Comments
for k,v in enumerate(A): B[:,:,k] = v
2 Comments
D R
Isn't this very inefficient since it involves a slow for loop?
Apalala
@celli It is O(n) on the size of
A, and no algorithm can do better than that because each element of A must be examined to perform the computation. Other answers have suggested that using numppy.tile() is a better option than a direct assignment. The solution I propose is clear and pythonic, and I expect NumPy to handle the assignments efficiently (perhaps with tile()), but one would have to measure (time it) to know.