BUG: Different behaviour of scipy.spatial.distance.cdist depending on parameters

Describe your issue.

There is a different behavior of the function scipy.distance.spatial.cdist depending on whether the parameter w is specified or not: if it is explicitly passed then the C implementation of the distance measure is used otherwise the python implementation is used.

Reproducing Code Example

import numpy, scipy.spatial
v1 = numpy.ndarray((1, 4))
v2 = numpy.ndarray((1, 4))
v1[0] = [0, 1, 0, 0]
v2[0] = [0, 1, 0, 0]

# Cosine similarity measures the cosine of the angle between the two vectors

print(scipy.spatial.distance.cdist(v1, v2, 'cosine')) # no parameter `w` so the C implementation of `cosine` is used
# array([[0.]])    -> 0 means that we have a 0 degree angle, so a perfect match

v2[0] = [0, 0, 0, 0] # Change the vector

print(scipy.spatial.distance.cdist(v1, v2, 'cosine')) # no parameter `w` so the C implementation of `cosine` is used
# array([[nan]])  # We get nan because the result of the dot product is divided by the product of the norm of the vector, so in this case 0

print(scipy.spatial.distance.cdist(v1, v2, 'cosine', w=[1,1,1,1])) # parameter `w` has been specified so the Python implementation of `cosine` is used
# array([[0.]])  # Different behavior

Error message

None

SciPy/NumPy/Python version information

1.8.0 1.22.3 sys.version_info(major=3, minor=10, micro=4, releaselevel=‘final’, serial=0)

[czgdp1807]

I think #14611 fixes the issue. Output I got on my PR,

[[0.]]
[[nan]]
[[nan]]

[czgdp1807]

Update - Went out of my mind. Will do the checking tomorrow for sure. Apologies.