I have a vector [x,y,z,q] and I want to create a matrix:
[[x,y,z,q],
[x,y,z,q],
[x,y,z,q],
...
[x,y,z,q]]
with m rows. I think this could be done in some smart way, using broadcasting, but I can only think of doing it with a for loop.
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1 Answer
Certainly possible with broadcasting after adding with m zeros along the columns, like so -
np.zeros((m,1),dtype=vector.dtype) + vector
Now, NumPy already has an in-built function np.tile for exactly that same task -
np.tile(vector,(m,1))
Sample run -
In [496]: vector
Out[496]: array([4, 5, 8, 2])
In [497]: m = 5
In [498]: np.zeros((m,1),dtype=vector.dtype) + vector
Out[498]:
array([[4, 5, 8, 2],
[4, 5, 8, 2],
[4, 5, 8, 2],
[4, 5, 8, 2],
[4, 5, 8, 2]])
In [499]: np.tile(vector,(m,1))
Out[499]:
array([[4, 5, 8, 2],
[4, 5, 8, 2],
[4, 5, 8, 2],
[4, 5, 8, 2],
[4, 5, 8, 2]])
You can also use np.repeat after extending its dimension with np.newaxis/None for the same effect, like so -
In [510]: np.repeat(vector[None],m,axis=0)
Out[510]:
array([[4, 5, 8, 2],
[4, 5, 8, 2],
[4, 5, 8, 2],
[4, 5, 8, 2],
[4, 5, 8, 2]])
You can also use integer array indexing to get the replications, like so -
In [525]: vector[None][np.zeros(m,dtype=int)]
Out[525]:
array([[4, 5, 8, 2],
[4, 5, 8, 2],
[4, 5, 8, 2],
[4, 5, 8, 2],
[4, 5, 8, 2]])
And finally with np.broadcast_to, you can simply create a 2D view into the input vector and as such this would be virtually free and with no extra memory requirement. So, we would simply do -
In [22]: np.broadcast_to(vector,(m,len(vector)))
Out[22]:
array([[4, 5, 8, 2],
[4, 5, 8, 2],
[4, 5, 8, 2],
[4, 5, 8, 2],
[4, 5, 8, 2]])
Runtime test -
Here's a quick runtime test comparing the various approaches -
In [12]: vector = np.random.rand(10000)
In [13]: m = 10000
In [14]: %timeit np.broadcast_to(vector,(m,len(vector)))
100000 loops, best of 3: 3.4 µs per loop # virtually free!
In [15]: %timeit np.zeros((m,1),dtype=vector.dtype) + vector
10 loops, best of 3: 95.1 ms per loop
In [16]: %timeit np.tile(vector,(m,1))
10 loops, best of 3: 89.7 ms per loop
In [17]: %timeit np.repeat(vector[None],m,axis=0)
10 loops, best of 3: 86.2 ms per loop
In [18]: %timeit vector[None][np.zeros(m,dtype=int)]
10 loops, best of 3: 89.8 ms per loop
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