Quickly Compute Eigenvectors For Each Element Of An Array In Python

I want to compute eigenvectors for an array of data (in my actual case, i cloud of polygons) To do so i wrote this function: import numpy as np def eigen(data): eigenvectors =

Solution 1:

Hack It!

Well I hacked into covariance func definition and put in the stated input states : ddof=0, rowvar=False and as it turns out, everything reduces to just three lines -

nC = m.shape[1]  # m is the 2D input arrayX = m - m.mean(0)
out = np.dot(X.T, X)/nC

To extend it to our 3D array case, I wrote down the loopy version with these three lines being iterated for the 2D arrays sections from the 3D input array, like so -

for i,d in enumerate(m):

    # Using np.cov :
    org_cov = np.cov(d, ddof=0, rowvar=False)

    # Using earlier 2D array hacked version :
    nC = m[i].shape[0]
    X = m[i] - m[i].mean(0,keepdims=True)
    hacked_cov = np.dot(X.T, X)/nC

Boost-it-up

We are needed to speedup the last three lines there. Computation of X across all iterations could be done with broadcasting -

diffs = data - data.mean(1,keepdims=True)

Next up, the dot-product calculation for all iterations could be done with transpose and np.dot, but that transpose could be a costly affair for such a multi-dimensional array. A better alternative exists in np.einsum, like so -

cov3D = np.einsum('ijk,ijl->ikl',diffs,diffs)/data.shape[1]

Use it!

To sum up :

for d in data:
    # compute covariance for each triangle
    cov = np.cov(d, ddof=0, rowvar=False)

Could be pre-computed like so :

diffs = data - data.mean(1,keepdims=True)
cov3D = np.einsum('ijk,ijl->ikl',diffs,diffs)/data.shape[1]

These pre-computed values could be used across iterations to compute eigen vectors like so -

for i,d inenumerate(data):
    # Directly use pre-computed covariances for each triangle
    vals, vecs = np.linalg.eig(cov3D[i])

Test It!

Here are some runtime tests to assess the effect of pre-computing covariance results -

In [148]: def original_app(data):
     ...:     cov = np.empty(data.shape)
     ...:     for i,d in enumerate(data):    
     ...:         # compute covariance for each triangle
     ...:         cov[i] = np.cov(d, ddof=0, rowvar=False)
     ...:     return cov
     ...: 
     ...: def vectorized_app(data):            
     ...:     diffs = data - data.mean(1,keepdims=True)
     ...:     return np.einsum('ijk,ijl->ikl',diffs,diffs)/data.shape[1]
     ...: 

In [149]: data = np.random.randint(0,10,(1000,3,3))

In [150]: np.allclose(original_app(data),vectorized_app(data))
Out[150]: True

In [151]: %timeit original_app(data)
10 loops, best of 3: 64.4 ms per loop

In [152]: %timeit vectorized_app(data)
1000 loops, best of 3: 1.14 ms per loop

In [153]: data = np.random.randint(0,10,(5000,3,3))

In [154]: np.allclose(original_app(data),vectorized_app(data))
Out[154]: True

In [155]: %timeit original_app(data)
1 loops, best of 3: 324 ms per loop

In [156]: %timeit vectorized_app(data)
100 loops, best of 3: 5.67 ms per loop

Solution 2:

I don't know how much of a speed up you can actually achieve.

Here is a slight modification that can help a little:

%timeit -n 10values, vectors = \
    eigen(triangles)
10 loops, best of3: 745 ms per loop

%timeit values, vectors = \
    zip(*(np.linalg.eig(np.cov(d, ddof=0, rowvar=False)) for d in triangles))
10 loops, best of3: 705 ms per loop