N-D version of itertools.combinations in numpy

问题

I would like to implement itertools.combinations for numpy. Based on this discussion, I have a function that works for 1D input:

def combs(a, r):
    """
    Return successive r-length combinations of elements in the array a.
    Should produce the same output as array(list(combinations(a, r))), but 
    faster.
    """
    a = asarray(a)
    dt = dtype([('', a.dtype)]*r)
    b = fromiter(combinations(a, r), dt)
    return b.view(a.dtype).reshape(-1, r)

and the output makes sense:

In [1]: list(combinations([1,2,3], 2))
Out[1]: [(1, 2), (1, 3), (2, 3)]

In [2]: array(list(combinations([1,2,3], 2)))
Out[2]: 
array([[1, 2],
       [1, 3],
       [2, 3]])

In [3]: combs([1,2,3], 2)
Out[3]: 
array([[1, 2],
       [1, 3],
       [2, 3]])

However, it would be best if I could expand it to N-D inputs, where additional dimensions simply allow you to speedily do multiple calls at once. So, conceptually, if combs([1, 2, 3], 2) produces [1, 2], [1, 3], [2, 3], and combs([4, 5, 6], 2) produces [4, 5], [4, 6], [5, 6], then combs((1,2,3) and (4,5,6), 2) should produce [1, 2], [1, 3], [2, 3] and [4, 5], [4, 6], [5, 6] where "and" just represents parallel rows or columns (whichever makes sense). (and likewise for additional dimensions)

I'm not sure:

How to make the dimensions work in a logical way that's consistent with the way other functions work (like how some numpy functions have an axis= parameter, and a default of axis 0. So probably axis 0 should be the one I am combining along, and all other axes just represent parallel calculations?)
How to get the above code to work with ND (right now I get ValueError: setting an array element with a sequence.)
Is there a better way to do dt = dtype([('', a.dtype)]*r)?

回答1:

You can use itertools.combinations() to create the index array, and then use NumPy's fancy indexing:

import numpy as np
from itertools import combinations, chain
from scipy.special import comb

def comb_index(n, k):
    count = comb(n, k, exact=True)
    index = np.fromiter(chain.from_iterable(combinations(range(n), k)), 
                        int, count=count*k)
    return index.reshape(-1, k)

data = np.array([[1,2,3,4,5],[10,11,12,13,14]])

idx = comb_index(5, 3)
print(data[:, idx])

output:

[[[ 1  2  3]
  [ 1  2  4]
  [ 1  2  5]
  [ 1  3  4]
  [ 1  3  5]
  [ 1  4  5]
  [ 2  3  4]
  [ 2  3  5]
  [ 2  4  5]
  [ 3  4  5]]

 [[10 11 12]
  [10 11 13]
  [10 11 14]
  [10 12 13]
  [10 12 14]
  [10 13 14]
  [11 12 13]
  [11 12 14]
  [11 13 14]
  [12 13 14]]]

回答2:

Not sure how it will work out performance-wise, but you can do the combinations on an index array, then extract the actual array slices with np.take:

def combs_nd(a, r, axis=0):
    a = np.asarray(a)
    if axis < 0:
        axis += a.ndim
    indices = np.arange(a.shape[axis])
    dt = np.dtype([('', np.intp)]*r)
    indices = np.fromiter(combinations(indices, r), dt)
    indices = indices.view(np.intp).reshape(-1, r)
    return np.take(a, indices, axis=axis)

>>> combs_nd([1,2,3], 2)
array([[1, 2],
       [1, 3],
       [2, 3]])
>>> combs_nd([[1,2,3],[4,5,6]], 2, axis=1)
array([[[1, 2],
        [1, 3],
        [2, 3]],

       [[4, 5],
        [4, 6],
        [5, 6]]])

回答3:

When r = k = 2, you can also use numpy.triu_indices(n, 1) which indexes upper triangle of a matrix.

idx = comb_index(5, 2)

from HYRY's answer is equivalent to

idx = np.transpose(np.triu_indices(5, 1))

but built-in, and a few times faster for N above ~20:

timeit comb_index(1000, 2)
32.3 ms ± 443 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

timeit np.transpose(np.triu_indices(1000, 1))
10.2 ms ± 25.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

来源：https://stackoverflow.com/questions/16003217/n-d-version-of-itertools-combinations-in-numpy

标签

python

numpy

combinatorics

itertools