Numpy – create matrix with rows of vector

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.

Answers:

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Method 1

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

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0