Putting a list in the same order as another list

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暖寄归人
暖寄归人 2021-01-19 23:40

There\'s a bunch of questions that are phrased similarly, but I was unable to find one that actually mapped to my intended semantics.

There are two lists, A

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  •  有刺的猬
    2021-01-20 00:09

    You can easily do this with numpy by sorting both lists (to get a mapping between the two lists) and by inverting one of the sorting permutations:

    import numpy as np
    
    a = [7, 14, 0, 9, 19, 9]
    b = [45, 42, 0, 1, -1, 0]
    
    a = np.array(a)
    b = np.array(b)
    
    ai = np.argsort(a)
    bi = np.argsort(b)
    aiinv = np.empty(ai.shape,dtype=int)
    aiinv[ai] = np.arange(a.size)  # inverse of ai permutation
    
    b_new = b[bi[aiinv]]
    # array([ 0, 42, -1,  0, 45,  1])
    

    numpy.argsort gives the indices (permutation) that will sort your array. This needs to be inverted to be used inside b, which can be done by the inverted assignment

    aiinv[ai] = np.arange(a.size)
    

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