等价于基本python中的Numpy.argsort（）？[重复]

等价于基本python中的Numpy.argsort（）？[重复]

我在上面建议了时间，这是我的结果。

首先，功能：

def f(seq):    # http://stackoverflow.com/questions/3382352/equivalent-of-numpy-argsort-in-basic-python/3383106#3383106    #non-lambda version by Tony Veijalainen    return [i for (v, i) in sorted((v, i) for (i, v) in enumerate(seq))]def g(seq):    # http://stackoverflow.com/questions/3382352/equivalent-of-numpy-argsort-in-basic-python/3383106#3383106    #lambda version by Tony Veijalainen    return [x for x,y in sorted(enumerate(seq), key = lambda x: x[1])]def h(seq):    #http://stackoverflow.com/questions/3382352/equivalent-of-numpy-argsort-in-basic-python/3382369#3382369    #by unutbu    return sorted(range(len(seq)), key=seq.__getitem__)

现在，IPython会话：

In [16]: seq = rand(10000).tolist()In [17]: %timeit f(seq)100 loops, best of 3: 10.5 ms per loopIn [18]: %timeit g(seq)100 loops, best of 3: 8.83 ms per loopIn [19]: %timeit h(seq)100 loops, best of 3: 6.44 ms per loop

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