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Scaling an array of slices wth list of required rows

I have a numd 2d A array and a list of row_set row row_set . How can I get a new array B , for example row_set = [0, 2, 5] , then B = [A_row[0], A_row[2], A_row[5]] ?

I thought of something like this:

 def slice_matrix(A, row_set): slice = array([row for row in A if row_num in row_set])

but i have no idea how i can get row_num.

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Use take() :

 In [87]: m = np.random.random((6, 2)) In [88]: m Out[88]: array([[ 0.6641412 , 0.31556053], [ 0.11480163, 0.00143887], [ 0.4677745 , 0.43055324], [ 0.49749099, 0.15678506], [ 0.48024596, 0.65701218], [ 0.48952677, 0.97089177]]) In [89]: m.take([0, 2, 5], axis=0) Out[89]: array([[ 0.6641412 , 0.31556053], [ 0.4677745 , 0.43055324], [ 0.48952677, 0.97089177]])
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You can pass a list or array as indices to any np array.

 >>> r = np.random.randint(0,10,(5,5)) >>> r array([[3, 8, 9, 8, 4], [4, 1, 5, 9, 1], [3, 6, 8, 8, 0], [5, 1, 7, 6, 1], [6, 1, 7, 7, 7]]) >>> idx = [0,3,1] >>> r[idx] array([[3, 8, 9, 8, 4], [5, 1, 7, 6, 1], [4, 1, 5, 9, 1]])
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Speed ​​comparison: take() faster.

 In [1]: m = np.random.random((1000, 2)) i = np.random.randint(1000, size=500) %timeit m[i] Out[1]: 10000 loops, best of 3: 27.2 µs per loop In [2]: %timeit m.take(i, axis=0) Out[2]: 100000 loops, best of 3: 7.24 µs per loop

This remains true for very large m and i

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