Let A and B be two numpy arrays.
I want to append n copies of B at the end of A :
C = [A, B, B, B, ... (n times) ... B, B]
How to do this simply/efficiently with numpy ?
Something like
numpy.append(A, [B * n]) # B * n is not n copies of B,
# but rather B multiplied by constant n ?
or with numpy.concatenate ?
It appears you want to use tile()
C = np.concatenate((A, np.tile(B,n)))
First lets check out concatenate routines:
A = np.arange(1E4)
#Baseline
%timeit np.concatenate((A,A))
100000 loops, best of 3: 11.1 µs per loop
%timeit np.hstack((A,A))
10000 loops, best of 3: 20.9 µs per loop
%timeit np.append(A,A)
100000 loops, best of 3: 19 µs per loop
Note that this is only the case for small arrays, append, hstack, and concatenate should be asymptotically convergent as all of these functions call concatenate, the main difference is python overhead. Now the only question is how to create array B:
#Using python
%timeit np.concatenate((A,[5]*10000))
1000 loops, best of 3: 1.1 ms per loop
#Tile small array
%timeit C = np.concatenate((A, np.tile(np.array(5),1E4)))
10000 loops, best of 3: 92.1 µs per loop
#Create an array of ones and multiply by a number
%timeit np.concatenate((A,np.ones(1E4)*5))
10000 loops, best of 3: 39.5 µs per loop
#Create empty array and fill from A and then set
%timeit C = np.empty(2E4); C[:10000]=A; C[10000:]=5
100000 loops, best of 3: 16.8 µs per loop
Looks like our winner is: create an empty array and then set the elements. This is assuming a certain array size, but most of these should scale similarly.
B = A[-1000:]) but I don't think this will be important here, it should work in a more general case for which B is any array.