Slice numpy array wth list of wanted rows

问题

I have a numpy 2d array A, and a list of row numbers row_set. How can I get new array B such as if 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 don't have any idea, how can I get a row_num.

回答1:

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]])

回答2:

You can pass a list or an array as indexes 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]])

回答3:

Speed comparison: take() is 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

来源：https://stackoverflow.com/questions/13659266/slice-numpy-array-wth-list-of-wanted-rows

标签

python

arrays

numpy

slice