Bobe Kryant - 1 year ago 124

Python Question

I generate a matrix that I want to get the covariance of:

`test=np.array([4,2,.6,4.2,2.1,.59,3.9,2,.58,4.3,2.1,.62,4.1,2.2,.63]).reshape(5,3)`

test

array([[ 4. , 2. , 0.6 ],

[ 4.2 , 2.1 , 0.59],

[ 3.9 , 2. , 0.58],

[ 4.3 , 2.1 , 0.62],

[ 4.1 , 2.2 , 0.63]])

I calculate the covariance with the numpy function:

`np.cov(test)`

array([[ 2.92 , 3.098 , 2.846 , 3.164 , 2.966 ],

[ 3.098 , 3.28703333, 3.0199 , 3.3566 , 3.1479 ],

[ 2.846 , 3.0199 , 2.7748 , 3.0832 , 2.8933 ],

[ 3.164 , 3.3566 , 3.0832 , 3.4288 , 3.2122 ],

[ 2.966 , 3.1479 , 2.8933 , 3.2122 , 3.0193 ]])

This however is different than following the covariance formula:

`mean=np.mean(test,0)`

np.dot(test-mean,(test-mean).T)/(5-1)

array([[ 0.004104, -0.002886, 0.006624, -0.005416, -0.002426],

[-0.002886, 0.002649, -0.005316, 0.005044, 0.000509],

[ 0.006624, -0.005316, 0.011744, -0.010496, -0.002556],

[-0.005416, 0.005044, -0.010496, 0.010164, 0.000704],

[-0.002426, 0.000509, -0.002556, 0.000704, 0.003769]])

This does not match the numpy calculations.

In fact, I take a peek at the source code and the equation is

`(x-m) * (x-m).T.conj() / (N - 1)`

Answer Source

The difference comes from the fact that the `np.cov`

calculates the covariance between row vectors, which is why the result is `5*5`

instead of `3*3`

, but `np.mean`

calculates the average of column vectors and when you do `test - mean`

the calculation is also broadcasted along column which differs from what `np.cov`

is doing, the fix would be a two-step:

Firstly, make sure the mean is calculated for each row, which can be done by simply transposing the `test`

matrix:

```
mean = np.mean(test.T, 0)
```

And then when calculate `x - x_bar`

, reshape the mean vector so that the minus is along the rows as well, and also since the vector under test is row vector the dimension is going to be `3`

instead of `5`

. After these fixing, it will give consistent results as `np.cov`

does:

```
np.dot(test-mean[:, None],(test-mean[:, None]).T)/(3-1)
# array([[ 2.92 , 3.098 , 2.846 , 3.164 , 2.966 ],
# [ 3.098 , 3.28703333, 3.0199 , 3.3566 , 3.1479 ],
# [ 2.846 , 3.0199 , 2.7748 , 3.0832 , 2.8933 ],
# [ 3.164 , 3.3566 , 3.0832 , 3.4288 , 3.2122 ],
# [ 2.966 , 3.1479 , 2.8933 , 3.2122 , 3.0193 ]])
```