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pythonnumpyscipy
tsawallis
tsawallis

Reputation: 1035

Calculate all possible columnwise differences in a matrix

I would like to compute all possible pairwise differences (without repetition) between the columns of a matrix. What's an efficient / pythonic way to do this? 

mat = np.random.normal(size=(10, 3))
mat 

array([[ 1.57921282,  0.76743473, -0.46947439],
       [ 0.54256004, -0.46341769, -0.46572975],
       [ 0.24196227, -1.91328024, -1.72491783],
       [-0.56228753, -1.01283112,  0.31424733],
       [-0.90802408, -1.4123037 ,  1.46564877],
       [-0.2257763 ,  0.0675282 , -1.42474819],
       [-0.54438272,  0.11092259, -1.15099358],
       [ 0.37569802, -0.60063869, -0.29169375],
       [-0.60170661,  1.85227818, -0.01349722],
       [-1.05771093,  0.82254491, -1.22084365]])

In this matrix there are 3 pairwise differences (N choose k unique combinations, where order doesn't matter).

pair_a = mat[:, 0] - mat[:, 1]
pair_b = mat[:, 0] - mat[:, 2]
pair_c = mat[:, 1] - mat[:, 2]

is one (ugly) way. You can easily imagine using nested for loops for larger matrices, but I am hoping there's a nicer way.

I would like the result to be another matrix, with scipy.misc.comb(mat.shape[1], 2) columns and mat.shape[0] rows.

Upvotes: 0

Views: 121

Answers (3)

tsawallis
tsawallis

Reputation: 1035

Incidentally, here is the solution I came up with. Much less elegant than moarningsun's.

def pair_diffs(mat):
    n_pairs = int(sp.misc.comb(mat.shape[1], 2))
    pairs = np.empty([mat.shape[0], n_pairs])

    this_pair = 0

    # compute all differences:
    for i in np.arange(mat.shape[1]-1):
        for j in np.arange(i+1, mat.shape[1]):
            pairs[:, this_pair] = mat[:, i] - mat[:, j]
            this_pair += 1
    return pairs

Upvotes: 0

user2379410
user2379410

Reputation:

Combinations of length 2 can be found using the following trick:

N = mat.shape[1]
I, J = np.triu_indices(N, 1)
result = mat[:,I] - mat[:,J]

Upvotes: 5

ev-br
ev-br

Reputation: 26040

In [7]: arr = np.arange(m*n).reshape((m, n))

In [8]: arr
Out[8]: 
array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11],
       [12, 13, 14, 15],
       [16, 17, 18, 19]])

In [9]: from itertools import combinations

In [10]: def diffs(arr):
   ....:     arr = np.asarray(arr)
   ....:     n = arr.shape[1]
   ....:     for i, j in combinations(range(n), 2):
   ....:         yield arr[:, i] - arr[:, j]
   ....:         

In [11]: for x in diffs(arr): print x
[-1 -1 -1 -1 -1]
[-2 -2 -2 -2 -2]
[-3 -3 -3 -3 -3]
[-1 -1 -1 -1 -1]
[-2 -2 -2 -2 -2]
[-1 -1 -1 -1 -1]

If you need them in an array, then just preallocate the array and assign the rows (or columns, as desired).

Upvotes: 1

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