计算numpy数组中列的出现次数

Question

Given a 2 xd dimensional numpy array M, I want to count the number of occurences of each column of M. That is, I'm looking for a general version of bincount .

What I tried so far: (1) Converted columns to tuples (2) Hashed tuples (via hash ) to natural numbers (3) used numpy.bincount .

This seems rather clumsy. Is anybody aware of a more elegant and efficient way?

Answer 1

You can use collections.Counter :

>>> import numpy as np
>>> a = np.array([[ 0,  1,  2,  4,  5,  1,  2,  3],
...               [ 4,  5,  6,  8,  9,  5,  6,  7],
...               [ 8,  9, 10, 12, 13,  9, 10, 11]])
>>> from collections import Counter
>>> Counter(map(tuple, a.T))
Counter({(2, 6, 10): 2, (1, 5, 9): 2, (4, 8, 12): 1, (5, 9, 13): 1, (3, 7, 11):
1, (0, 4, 8): 1})

Answer 2

Given:

a = np.array([[ 0,  1,  2,  4,  5,  1,  2,  3],
              [ 4,  5,  6,  8,  9,  5,  6,  7],
              [ 8,  9, 10, 12, 13,  9, 10, 11]])
b = np.transpose(a)

1. A more efficient solution than hashing (still requires manipulation):

   I create a view of the array with the flexible data type np.void (see here ) such that each row becomes a single element. Converting to this shape will allow np.unique to operate on it.

    %%timeit c = np.ascontiguousarray(b).view(np.dtype((np.void, b.dtype.itemsize*b.shape[1]))) _, index, counts = np.unique(c, return_index = True, return_counts = True) #counts are in the last column, remember original array is transposed >>>np.concatenate((b[idx], cnt[:, None]), axis = 1) array([[ 0, 4, 8, 1], [ 1, 5, 9, 2], [ 2, 6, 10, 2], [ 3, 7, 11, 1], [ 4, 8, 12, 1], [ 5, 9, 13, 1]]) 10000 loops, best of 3: 65.4 µs per loop 

   The counts appended to the unique columns of a .

2. Your hashing solution.

    %%timeit array_hash = [hash(tuple(row)) for row in b] uniq, index, counts = np.unique(array_hash, return_index= True, return_counts = True) np.concatenate((b[idx], cnt[:, None]), axis = 1) 10000 loops, best of 3: 89.5 µs per loop 

Update : Eph's solution is the most efficient and elegant.

%%timeit
Counter(map(tuple, a.T))
10000 loops, best of 3: 38.3 µs per loop