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pythonnumpymatrix
Jakub Arnold
Jakub Arnold

Reputation: 87210

Numpy: create a matrix from a cartesian product of two vectors (or one with itself) while applying a function to all pairs

To give a bit of explanation, I want to create a covariance matrix where each element is defined by a kernel function k(x, y), and I want to do this for a single vector. It should be something like:

# This is given
x = [x1, x2, x3, x4, ...]

# This is what I want to compute
result = [[k(x1, x1), k(x1, x2), k(x1, x3), ...],
          [k(x2, x1), k(x2, x2), ...],
          [k(x3, x1), k(x3, x2), ...],
          ...]

but of course this should be done in numpy arrays, ideally without doing Python interations, because of performance. If I didn't care about performance, I'd probably just write:

result = np.zeros((len(x), len(x)))

for i in range(len(x)):
  for j in range(len(x)):
     result[i, j] = k(x[i], x[j])

But I feel like there must be a more idiomatic way to write this pattern.

Upvotes: 4

Views: 1387

Answers (2)

anishtain4
anishtain4

Reputation: 2402

Another solution is to let numpy do the broadcasting itself:

import numpy as np

def k(x,y):
    return x**2+y

def meshgrid1D_view(x):
    shp = (len(x),len(x))
    mesh1 = np.broadcast_to(x,shp)
    mesh2 = np.broadcast_to(x[:,None],shp)    
    return mesh1, mesh2

x = np.random.randint(0,10,(10000))
b=k(a[:,None],a[None,:])

def sol0(x):
    k(x[:,None],x[None,:])

def sol1(x):
    x,y=np.meshgrid(x,x)
    k(x,y)

def sol2(x):
    x,y=meshgrid1D_view(x)
    k(x,y)

%timeit sol0(x)
165 ms ± 1.67 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

%timeit sol1(x)
655 ms ± 6.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit sol2(x)
341 ms ± 2.91 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

You see this is more efficient and there's less code.

Upvotes: 2

Divakar
Divakar

Reputation: 221524

If k operates on 2D arrays, you could use np.meshgrid. But, this would have extra memory overhead. One alternative would be to create 2D mesh views same as with np.meshgrid, like so -

def meshgrid1D_view(x):
    shp = (len(x),len(x))
    mesh1 = np.broadcast_to(x,shp)
    mesh2 = np.broadcast_to(x[:,None],shp)    
    return mesh1, mesh2

Sample run -

In [140]: x
Out[140]: array([3, 5, 6, 8])

In [141]: np.meshgrid(x,x)
Out[141]: 
[array([[3, 5, 6, 8],
        [3, 5, 6, 8],
        [3, 5, 6, 8],
        [3, 5, 6, 8]]), array([[3, 3, 3, 3],
        [5, 5, 5, 5],
        [6, 6, 6, 6],
        [8, 8, 8, 8]])]
In [142]: meshgrid1D(x)
Out[142]: 
(array([[3, 5, 6, 8],
        [3, 5, 6, 8],
        [3, 5, 6, 8],
        [3, 5, 6, 8]]), array([[3, 3, 3, 3],
        [5, 5, 5, 5],
        [6, 6, 6, 6],
        [8, 8, 8, 8]]))

How does this help?

It helps with memory efficiency and hence performance. Let's test out on large arrays to see the difference -

In [143]: x = np.random.randint(0,10,(10000))

In [144]: %timeit np.meshgrid(x,x)
10 loops, best of 3: 171 ms per loop

In [145]: %timeit meshgrid1D(x)
100000 loops, best of 3: 6.91 µs per loop

Upvotes: 4

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