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並前往https://stackoverflow.com/questions/60916020/numpy-linalg-norm-vs-scipy-cdist-for-l2-norm

numpy.linalg.norm VS scipy cdist for L2 norm - Stack Overflow

You can use numpy broadcasting as follows: d = np.linalg.norm(a[:, None, :] - b[None, :, :], axis=2). Performace should be similar to ...

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