Efficient way of making a list of pairs from an array in Numpy

Question

I have a numpy array x (with (n,4) shape) of integers like:

[[0 1 2 3],
[1 2 7 9],
[2 1 5 2],
...]

I want to transform the array into an array of pairs:

[0,1]
[0,2]
[0,3]
[1,2]
...

so first element makes a pair with other elements in the same sub-array. I have already a for-loop solution:

y=np.array([[x[j,0],x[j,i]] for i in range(1,4) for j in range(0,n)],dtype=int)

but since looping over numpy array is not efficient, I tried slicing as the solution. I can do the slicing for every column as:

y[1]=np.array([x[:,0],x[:,1]]).T
# [[0,1],[1,2],[2,1],...]

I can repeat this for all columns. My questions are:

How can I append y[2] to y[1],... such that the shape is (N,2)?
If number of columns is not small (in this example 4), how can I find y[i] elegantly?
What are the alternative ways to achieve the final array?

@BlacklightShining: Not really. The pattern needed isn't actually combinations, though it might look that way at first due to poor choice of example data, and for NumPy, you want to avoid itertools about as much as you want to avoid loops and comprehensions. — user2357112
– user2357112, Commented Apr 12, 2015 at 3:20

Jaime · Accepted Answer · 2015-04-12 03:19:28Z

The cleanest way of doing this I can think of would be:

>>> x = np.arange(12).reshape(3, 4)
>>> x
array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11]])
>>> n = x.shape[1] - 1
>>> y = np.repeat(x, (n,)+(1,)*n, axis=1)
>>> y
array([[ 0,  0,  0,  1,  2,  3],
       [ 4,  4,  4,  5,  6,  7],
       [ 8,  8,  8,  9, 10, 11]])
>>> y.reshape(-1, 2, n).transpose(0, 2, 1).reshape(-1, 2)
array([[ 0,  1],
       [ 0,  2],
       [ 0,  3],
       [ 4,  5],
       [ 4,  6],
       [ 4,  7],
       [ 8,  9],
       [ 8, 10],
       [ 8, 11]])

This will make two copies of the data, so it will not be the most efficient method. That would probably be something like:

>>> y = np.empty((x.shape[0], n, 2), dtype=x.dtype)
>>> y[..., 0] = x[:, 0, None]
>>> y[..., 1] = x[:, 1:]
>>> y.shape = (-1, 2)
>>> y
array([[ 0,  1],
       [ 0,  2],
       [ 0,  3],
       [ 4,  5],
       [ 4,  6],
       [ 4,  7],
       [ 8,  9],
       [ 8, 10],
       [ 8, 11]])

Mahdi · Accepted Answer · 2015-04-12 06:47:38Z

2

Like Jaimie, I first tried a repeat of the 1st column followed by reshaping, but then decided it was simpler to make 2 intermediary arrays, and hstack them:

x=np.array([[0,1,2,3],[1,2,7,9],[2,1,5,2]])
m,n=x.shape
x1=x[:,0].repeat(n-1)[:,None]
x2=x[:,1:].reshape(-1,1)
np.hstack([x1,x2])

producing

array([[0, 1],
       [0, 2],
       [0, 3],
       [1, 2],
       [1, 7],
       [1, 9],
       [2, 1],
       [2, 5],
       [2, 2]])

There probably are other ways of doing this sort of rearrangement. The result will copy the original data in one way or other. My guess is that as long as you are using compiled functions like reshape and repeat, the time differences won't be significant.

edited Apr 12, 2015 at 6:47

Mahdi

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answered Apr 12, 2015 at 4:54

hpaulj

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Comments

Abinash Panda · Accepted Answer · 2015-04-12 03:16:14Z

1

Suppose the numpy array is

arr = np.array([[0, 1, 2, 3],
                [1, 2, 7, 9],
                [2, 1, 5, 2]])

You can get the array of pairs as

import itertools
m, n = arr.shape
new_arr = np.array([x for i in range(m) 
                    for x in itertools.product(a[i, 0 : 1], a[i, 1 : n])])

The output would be

array([[0, 1],
       [0, 2],
       [0, 3],
       [1, 2],
       [1, 7],
       [1, 9],
       [2, 1],
       [2, 5],
       [2, 2]])

answered Apr 12, 2015 at 3:16

Abinash Panda

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