CODEWITHSUNDEEP
pythonarraysnumpymatrixcovariance
Ivan Castro
Ivan Castro

Reputation: 615

Implement this situation in numpy

Given these two arrays:

E = [[16.461, 17.015, 14.676],
 [15.775, 18.188, 14.459],
 [14.489, 18.449, 14.756],
 [14.171, 19.699, 14.406],
 [14.933, 20.644, 13.839],
 [16.233, 20.352, 13.555],
 [16.984, 21.297, 12.994],
 [16.683, 19.056, 13.875],
 [17.918, 18.439, 13.718],
 [17.734, 17.239, 14.207]]

S = [[0.213, 0.660, 1.287],
 [0.250, 2.016, 1.509],
 [0.016, 2.995, 0.619],
 [0.142, 4.189, 1.194],
 [0.451, 4.493, 2.459],
 [0.681, 3.485, 3.329],
 [0.990, 3.787, 4.592],
 [0.579, 2.170, 2.844],
 [0.747, 0.934, 3.454],
 [0.520, 0.074, 2.491]]

The problem states that I should get the 3x3 covariance matrix (C) between S and E using the following formula:

C = (1/(n-1))[S'E - (1/10)S'i i'E]

Here n is 10, and i is an n x 1 column vector consisting of only ones. S' and i' are the transpose of matrix S and column vector i, respectively.

So far, I can't get C because I don't understand the meaning of i (and i') and its implementation in the formula. Using numpy, so far I do:

import numpy as np

tS = numpy.array(S).T
C =  (1.0/9.0)*(np.dot(tS, E)-((1.0/10.0)*np.dot(tS, E))) #Here is where I lack the i and i' implementation.

I will really appreciate your help to understand and implement i and i' in the formula. The output should be:

C= [[0.2782, 0.2139, -0.1601],
    [-1.4028, 1.9619, -0.2744],
    [1.0443, 0.9712, -0.6610]]

Upvotes: 0

Views: 48

Answers (2)

Rouzbeh
Rouzbeh

Reputation: 2431

This is what you want I guess:

S = numpy.array(S)

E = numpy.array(E)

ones = np.ones((10,1))

C =  (1.0/9)*(np.dot(S.T, E)-((1.0/10)*    (np.dot(np.dot(np.dot(S.T,ones),ones.T),E))))

My output is :

array([[ 0.27842301,  0.21388842, -0.16011839],
   [-1.4017267 ,  1.96193373, -0.27441417],
   [ 1.04532836,  0.97120807, -0.66095656]])

Upvotes: 1

DSM
DSM

Reputation: 352969

It looks like the only part you're missing is making i:

>>> i = np.ones((N, 1))
>>> i
array([[ 1.],
       [ 1.],
       [ 1.],
       [ 1.],
       [ 1.],
       [ 1.],
       [ 1.],
       [ 1.],
       [ 1.],
       [ 1.]])

After that, we get

>>> C = (1.0/(N-1)) * (S.T.dot(E) - (1.0/N) * S.T.dot(i) * i.T.dot(E))
>>> C
array([[ 0.27842301,  0.21388842, -0.16011839],
       [-1.4017267 ,  1.96193373, -0.27441417],
       [ 1.04532836,  0.97120807, -0.66095656]])

Note that this doesn't quite produce the array you expected, which is more obvious if you round it, but maybe there are some minor typos in your data?

>>> C.round(4)
array([[ 0.2784,  0.2139, -0.1601],
       [-1.4017,  1.9619, -0.2744],
       [ 1.0453,  0.9712, -0.661 ]])

Upvotes: 2

Related Questions