When there is a zero mean applied to the numpy matrix, is there a difference expected between, the following 2 codes? I was learning andrew ng's ML course and he suggested to use X @ X^T to find the covariance matrix(considering the zero mean is applied). When I tried to visually examine the matrix, found it gives diff result with np.cov function.. Please help..
import numpy as np
X=np.random.randint(0,9,(3,3))
print(X)
[[2 1 5]
[7 4 8]
[4 7 6]]
X = (X - X.mean(axis=0)) # <- Zero Mean
print(X)
[[-2.33333333 -3. -1.33333333]
[ 2.66666667 0. 1.66666667]
[-0.33333333 3. -0.33333333]]
cov1 = (X @ X.T)/m # <- Find covariance manually as suggested in the course
print(cov1)
[[ 5.40740741 -2.81481481 -2.59259259]
[-2.81481481 3.2962963 -0.48148148]
[-2.59259259 -0.48148148 3.07407407]]
cov2 = np.cov(X,bias=True) # <- Find covariance with np.cov
print(cov2)
[[ 0.7037037 0.59259259 -1.2962963 ]
[ 0.59259259 1.81481481 -2.40740741]
[-1.2962963 -2.40740741 3.7037037 ]]
If your observations are in rows and variables are in columns (set rowvar to False), then it must be x.T @ x:
import numpy as np
x0 = np.array([[2, 1, 5], [7, 4, 8], [4, 7, 6]])
x = x0 - x0.mean(axis=0)
cov1 = x.T @ x / 3
cov2 = np.cov(x, rowvar=False, bias=True)
assert np.allclose(cov1, cov2)
x @ x.T is for the case when the observations are in columns and variables are in the rows:
x = x0 - x0.mean(axis=1)[:,None]
cov1 = x @ x.T / 3
cov2 = np.cov(x, bias=True) # rowvar=True by default
assert np.allclose(cov1, cov2)