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avoid double loop of xi-xj in calculation of s #8

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avigdev opened this issue Aug 21, 2017 · 0 comments

avigdev commented Aug 21, 2017

could be implemented with a numpy trick:

s = np.sum(np.sign(np.tril(np.subtract.outer(a,a)))) # sum of sign(xi-xj) where i>j

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