Reputation: 353
Suppose I have two vectors of length 25, and I want to compute their covariance matrix. I try doing this with numpy.cov, but always end up with a 2x2 matrix.
>>> import numpy as np
>>> x=np.random.normal(size=25)
>>> y=np.random.normal(size=25)
>>> np.cov(x,y)
array([[ 0.77568388, 0.15568432],
[ 0.15568432, 0.73839014]])
Using the rowvar flag doesn't help either - I get exactly the same result.
>>> np.cov(x,y,rowvar=0)
array([[ 0.77568388, 0.15568432],
[ 0.15568432, 0.73839014]])
How can I get the 25x25 covariance matrix?
Upvotes: 25
Views: 96728
Reputation: 5310
according the document, you should expect variable vector in column:
If we examine N-dimensional samples, X = [x1, x2, ..., xn]^T
though later it says each row is a variable
Each row of m represents a variable.
so you need input your matrix as transpose
x=np.random.normal(size=25)
y=np.random.normal(size=25)
X = np.array([x,y])
np.cov(X.T)
and according to wikipedia: https://en.wikipedia.org/wiki/Covariance_matrix
X is column vector variable
X = [X1,X2, ..., Xn]^T
COV = E[X * X^T] - μx * μx^T // μx = E[X]
you can implement it yourself:
# X each row is variable
X = X - X.mean(axis=0)
h,w = X.shape
COV = X.T @ X / (h-1)
Upvotes: 0
Reputation: 1039
To clarify the small confusion regarding what is a covariance matrix defined using two N-dimensional vectors, there are two possibilities.
The question you have to ask yourself is whether you consider:
[X1,X2,X3]
and [Y1,Y2,Y3]
, where you have 3 realizations for the variables X and Y respectively)[X1,Y1,Z1]
and [X2,Y2,Z2]
, where you have 1 realization for the variables X,Y and Z per vector)Since a covariance matrix is intuitively defined as a variance based on two different variables:
if you consider that you have 25 variables per vector (took 3 instead of 25 to simplify example code), so one realization for several variables in one vector, use rowvar=0
# [X1,Y1,Z1]
X_realization1 = [1,2,3]
# [X2,Y2,Z2]
X_realization2 = [2,1,8]
numpy.cov([X,Y],rowvar=0) # rowvar false, each column is a variable
Code returns, considering 3 variables:
array([[ 0.5, -0.5, 2.5],
[-0.5, 0.5, -2.5],
[ 2.5, -2.5, 12.5]])
otherwise, if you consider that one vector is 25 samples for one variable, use rowvar=1
(numpy's default parameter)
# [X1,X2,X3]
X = [1,2,3]
# [Y1,Y2,Y3]
Y = [2,1,8]
numpy.cov([X,Y],rowvar=1) # rowvar true (default), each row is a variable
Code returns, considering 2 variables:
array([[ 1. , 3. ],
[ 3. , 14.33333333]])
Upvotes: 5
Reputation: 22734
What you got (2 by 2) is more useful than 25*25. Covariance of X and Y is an off-diagonal entry in the symmetric cov_matrix.
If you insist on (25 by 25) which I think useless, then why don't you write out the definition?
x=np.random.normal(size=25).reshape(25,1) # to make it 2d array.
y=np.random.normal(size=25).reshape(25,1)
cov = np.matmul(x-np.mean(x), (y-np.mean(y)).T) / len(x)
Upvotes: 2
Reputation: 2901
You should change
np.cov(x,y, rowvar=0)
onto
np.cov((x,y), rowvar=0)
Upvotes: 2
Reputation: 7
i don't think you understand the definition of covariance matrix. If you need 25 x 25 covariance matrix, you need 25 vectors each with n data points.
Upvotes: -2
Reputation: 649
I suppose what youre looking for is actually a covariance function which is a timelag function. I'm doing autocovariance like that:
def autocovariance(Xi, N, k):
Xs=np.average(Xi)
aCov = 0.0
for i in np.arange(0, N-k):
aCov = (Xi[(i+k)]-Xs)*(Xi[i]-Xs)+aCov
return (1./(N))*aCov
autocov[i]=(autocovariance(My_wector, N, h))
Upvotes: 2
Reputation: 159
Try this:
import numpy as np
x=np.random.normal(size=25)
y=np.random.normal(size=25)
z = np.vstack((x, y))
c = np.cov(z.T)
Upvotes: 15
Reputation: 885
As pointed out above, you only have two vectors so you'll only get a 2x2 cov matrix.
IIRC the 2 main diagonal terms will be sum( (x-mean(x))**2) / (n-1) and similarly for y.
The 2 off-diagonal terms will be sum( (x-mean(x))(y-mean(y)) ) / (n-1). n=25 in this case.
Upvotes: 0
Reputation: 8568
You have two vectors, not 25. The computer I'm on doesn't have python so I can't test this, but try:
z = zip(x,y)
np.cov(z)
Of course.... really what you want is probably more like:
n=100 # number of points in each vector
num_vects=25
vals=[]
for _ in range(num_vects):
vals.append(np.random.normal(size=n))
np.cov(vals)
This takes the covariance (I think/hope) of num_vects
1xn
vectors
Upvotes: 13
Reputation: 548
Reading the documentation as,
>> np.cov.__doc__
or looking at Numpy Covariance, Numpy treats each row of array as a separate variable, so you have two variables and hence you get a 2 x 2 covariance matrix.
I think the previous post has right solution. I have the explanation :-)
Upvotes: 3