Efficiently generating multiple instances of numpy.random.choice without replacement

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一个人的身影
一个人的身影 2020-12-10 18:48

I\'m new to Python. While reading, please mention any other suggestions regarding ways to improve my Python code.

Question: How do I generate a 8x

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  •  没有蜡笔的小新
    2020-12-10 19:02

    Create a random array of specified shape and then sort along the axis where you want to keep the limits, thus giving us a vectorized and very efficient solution. This would be based on this smart answer to MATLAB randomly permuting columns differently. Here's the implementation -

    Sample run -

    In [122]: N = 10
    
    In [123]: np.argsort(np.random.rand(8,N),axis=0)+1
    Out[123]: 
    array([[7, 3, 5, 1, 1, 5, 2, 4, 1, 4],
           [8, 4, 3, 2, 2, 8, 5, 5, 6, 2],
           [1, 2, 4, 6, 5, 4, 4, 3, 4, 7],
           [5, 6, 2, 5, 8, 2, 7, 8, 5, 8],
           [2, 8, 6, 3, 4, 7, 1, 1, 2, 6],
           [6, 7, 7, 8, 6, 6, 3, 2, 7, 3],
           [4, 1, 1, 4, 3, 3, 8, 6, 8, 1],
           [3, 5, 8, 7, 7, 1, 6, 7, 3, 5]], dtype=int64)
    

    Runtime tests -

    In [124]: def sortbased_rand8(N):
         ...:     return np.argsort(np.random.rand(8,N),axis=0)+1
         ...: 
         ...: def rand_M(N):
         ...:     M = np.zeros(shape = (8, N))
         ...:     for i in range (0, N):
         ...:         M[:, i] = np.random.choice(8, size = 8, replace = False) + 1 
         ...:     return M
         ...: 
    
    In [125]: N = 5000
    
    In [126]: %timeit sortbased_rand8(N)
    100 loops, best of 3: 1.95 ms per loop
    
    In [127]: %timeit rand_M(N)
    1 loops, best of 3: 233 ms per loop
    

    Thus, awaits a 120x speedup!

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