How to get matrix of pairwise distance with given ndarray of coordinates using solely numpy-CodePudding

I have point_coords ndarray:

point_coord_x = np.array([np.random.randint(low=-100, high=100) for i in range(40)])
point_coord_y = np.array([np.random.randint(low=-100, high=100) for i in range(40)])
point_coords = np.array([point_coord_x, point_coord_y]).transpose()

It looks like:

point_coords
array([[  62,  -31],
       [  49,   33],
       [  -2,  -86],
       [ -29,   49],
        ...

I want to get a square matrix with distance between points. How am I supposed to do it?

CodePudding user response：

>>> from scipy.spatial import distance_matrix
>>> distance_matrix(point_coords, point_coords)
array([[  0.        , 149.21461054,  88.64536085, ...,  44.94441011,  24.73863375,  60.5309838 ],
       [149.21461054,   0.        , 122.64175472, ..., 136.47344064, 163.60012225, 201.07958623],
       [ 88.64536085, 122.64175472,   0.        , ...,  45.01110974, 113.35784049, 147.2752525 ],
       ...,
       [ 44.94441011, 136.47344064,  45.01110974, ...,   0.        ,  69.57010852, 102.3132445 ],
       [ 24.73863375, 163.60012225, 113.35784049, ...,  69.57010852,   0.        ,  38.62641583],
       [ 60.5309838 , 201.07958623, 147.2752525 , ..., 102.3132445 ,  38.62641583,   0.        ]])

If only numpy is to be used:

np.linalg.norm(point_coords[:, None, :] - point_coords[None, :, :], axis=-1)

CodePudding user response：

You can do it using solely Numpy:

Compute deltas by each coordinate (square matrices):

dx = point_coord_x[:, np.newaxis] - point_coord_x[np.newaxis, :]
dy = point_coord_y[:, np.newaxis] - point_coord_y[np.newaxis, :]

Then compute the distance array from these deltas:

result = np.sqrt(dx ** 2   dy ** 2)