[SOLVED] Outer product of average and average of outer product in numpy?

Issue

I have two 3d column vectors B_mu and B_nu that vary as a function of time:

import numpy as np
N = 5 # 5 time-steps

B_mu = np.array(
      [[5, 5, 8],
       [4, 8, 7],
       [2, 3, 1],
       [5, 7, 8],
       [6, 2, 7]]
)
B_nu = np.array(
      [[3, 2, 9],
       [9, 8, 8],
       [4, 9, 9],
       [4, 9, 6],
       [1, 9, 1]]
)

For every index i in the first vector, and every index j in the second vector, I want to compute the difference between the time-average of the product < B_mu[i] B_nu[j] > and the product of the time-averages <B_mu[i]> <B_nu[j]>.

In other words, I want to construct the matrix M such that:

M[i,j] = 1/N sum(B_mu[i] * B_nu[j]) - 1/N**2 * sum(B_mu[i]) * sum(B_nu[j])

where the sums are taken over the time parameter.

Here is the equation:

short equation

And an explicit, expanded version:

explicit equation

How can I express this equation in python?

Solution

Manipulation of matrices is relatively easy with module numpy. Here we’re looking for:

  • averaging (or summing) over one dimension (time);
  • taking the outer product of two column vectors.

We’re going to use:

These two functions combine straightforwardly to compute the product of averages. The average of the products, on the other hand, is straightforward to compute with python builtins sum and map; I don’t know how to do it in pure numpy.

import numpy as np

def diff_avgprod_prodavg(B_mu, B_nu):
    N = B_mu.shape[0]
    avg_of_prod = 1/N * sum(map(np.outer, B_mu, B_nu)) # not pure numpy
    prod_of_avg = 1/(N*N) * np.outer(B_mu.sum(axis=0), B_nu.sum(axis=0))
    return avg_of_prod - prod_of_avg

Testing:

B_mu = np.array(
      [[5, 5, 8],
       [4, 8, 7],
       [2, 3, 1],
       [5, 7, 8],
       [6, 2, 7]]
)
B_nu = np.array(
      [[3, 2, 9],
       [9, 8, 8],
       [4, 9, 9],
       [4, 9, 6],
       [1, 9, 1]]
)

print( diff_avgprod_prodavg(B_mu, B_nu) )
# [[-1.48 -0.76 -2.84]
#  [ 4.8  -0.6   3.  ]
#  [-0.04 -2.68 -2.52]]

print( diff_avgprod_prodavg(B_mu, B_mu) )
# [[1.84 0.   3.12]
#  [0.   5.2  2.8 ]
#  [3.12 2.8  6.96]]

print( diff_avgprod_prodavg(B_nu, B_nu) )
# [[ 6.96  0.72  4.28]
#  [ 0.72  7.44 -3.64]
#  [ 4.28 -3.64  9.04]]

Answered By – Stef

Answer Checked By – Robin (BugsFixing Admin)

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