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感谢 Philip Cloud对上一个问题的出色回答，我去挖掘了pairwise_distancesin scikit 的源代码。

相关部分是：

def pairwise_distances(X, Y=None, metric="euclidean", n_jobs=1, **kwds):
    if metric == "precomputed":
        return X
    elif metric in PAIRWISE_DISTANCE_FUNCTIONS:
        func = PAIRWISE_DISTANCE_FUNCTIONS[metric]
        if n_jobs == 1:
            return func(X, Y, **kwds)
        else:
            return _parallel_pairwise(X, Y, func, n_jobs, **kwds)
    elif callable(metric):
        # Check matrices first (this is usually done by the metric).
        X, Y = check_pairwise_arrays(X, Y)
        n_x, n_y = X.shape[0], Y.shape[0]
        # Calculate distance for each element in X and Y.
        # FIXME: can use n_jobs here too
        D = np.zeros((n_x, n_y), dtype='float')
        for i in range(n_x):
            start = 0
            if X is Y:
                start = i
            for j in range(start, n_y):
                # distance assumed to be symmetric.
                D[i][j] = metric(X[i], Y[j], **kwds)
                if X is Y:
                    D[j][i] = D[i][j]
        return D

从中理解是否正确，如果我要计算成对距离矩阵，例如：

matrix = pairwise_distances(foo, metric=lambda u,v: haversine(u,v), n_jobs= -1)

哪里haversine(u,v)有一个计算两点之间Haversine距离的函数，而这个函数不在，PAIRWISE_DISTANCE_FUNCTIONS即使计算也不会并行化n_jobs= -1？

我意识到#FIXME评论似乎暗示了这一点，但我想确保我没有发疯，因为似乎有点奇怪，不会抛出信息警报，说明当你通过时计算实际上不会并行化n_jobs= -1a不在的可调用函数PAIRWISE_DISTANCE_FUNCTIONS。

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1 回答 1

3

确认的。metric如果可调用n_jobs= -1对象不在PAIRWISE_DISTANCE_FUNCTIONS.

于 2013-09-03T16:26:20.833 回答