Difference between scipy pairwise distance and X.X+Y.Y - X.Y^t

Question

Let\'s imagine we have data as

d1 = np.random.uniform(low=0, high=2, size=(3,2))
d2 = np.random.uniform(low=3, high=5, size=(3,2))
X = np.vstack((d1,d2))

X
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Answer 1

pdist(..., metric='seuclidean') computes the standardized Euclidean distance, not the squared Euclidean distance (which is what cal_pdist returns).

From the docs:

Y = pdist(X, 'seuclidean', V=None)

Computes the standardized Euclidean distance. The standardized Euclidean distance between two n-vectors u and v is

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  √∑(ui−vi)^2 / V[xi]

V is the variance vector; V[i] is the variance computed over all the i’th components of the points. If not passed, it is automatically computed.

Try passing metric='sqeuclidean', and you will see that both functions return the same result to within rounding error.