Repeat arrow on new axis along new axis
What's the best / most efficient way?
where x is an array and n is the number of times I want to repeat:
np.repeat(x[None,...], n, axis=0)
or
ones = [1] * len(x.shape) np.tile(x, tuple(n, *ones))
note that for a known size of the array, the latter becomes simple:
np.tile(x, (n, 1, 1)) # x is 2D
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Part of the code for np.tile
:
for i, nrep in enumerate(tup):
if nrep!=1:
c = c.reshape(-1, n).repeat(nrep, 0)
In other words, it repeat
has more than 1 repetition on each axis. It is, an effect, generalization repeat
to multiple axes.
Thus, I would expect the timings to be similar, although a simpler repeat
would have less Python overhead. repeat
compiled. (a few simple tests confirm this - repeat
2x faster for small arrays, slightly faster for large ones).
ps The step is x[None,...]
practically cost-free. And because of the broadcast, that might be all you need.
pss There is an even faster way to do this repetition using np.lib.index_tricks.as_strided
. For (20,50)
x
<,
as_strided(x,shape=(n,20,50),strides=(0,200,4))
np.broadcast_arrays
also uses as_strided
. So this gives the same thing:
np.broadcast_arrays(np.ones((n,1,1)),x)[1]
But to be honest, this is just a broadcast development, not a real replay. The data has not been replicated. The same values are used only n
once.
Broadcast can be used to populate a full array, but the timings are the same as for repeat
. This could be what it repeat
does under the hood.
z = np.empty((300,20,50),dtype=int)
z[:] = x[None,...]
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