Combinatoric / cartesian product of Numpy arrays without iterators and/or loop(s) [duplicate]

Question

The following code

import numpy as np
import itertools

a_p1 = np.arange(0, 4, 1)
a_p2 = np.arange(20, 25, 1)

params = itertools.product(a_p1, a_p2)
for (p1, p2) in params:
    print(p1, p2)

outputs

(0, 20) (0, 21) (0, 22) (0, 23) (0, 24) (1, 20) (1, 21) (1, 22) (1, 23) (1, 24) (2, 20) (2, 21) (2, 22) (2, 23) (2, 24) (3, 20) (3, 21) (3, 22) (3, 23) (3, 24)

2 nested for loops can also outputs same results

for i, p1 in enumerate(a_p1):
    for j, p2 in enumerate(a_p2):
        print(p1, p2)

I'm looking for a solution to directly output a Numpy array with such combination (a Numpy array of tuples).

Is there a way to generate such a Numpy array without iterators and/or for loop(s) ?

I'm aware that such a solution will be more memory consuming than using iterators.

see related: stackoverflow.com/questions/11144513/…

EdChum
– EdChum

2015-04-18 08:30:06 +00:00
Commented Apr 18, 2015 at 8:30 — EdChum
– EdChum, Commented Apr 18, 2015 at 8:30

Community · Accepted Answer · 2017-05-23 12:13:43Z

14

Install Scikit-Learn http://scikit-learn.org/

from sklearn.utils.extmath import cartesian
print cartesian([a_p1, a_p2])

It should output

[[ 0 20]
 [ 0 21]
 [ 0 22]
 [ 0 23]
 [ 0 24]
 [ 1 20]
 [ 1 21]
 [ 1 22]
 [ 1 23]
 [ 1 24]
 [ 2 20]
 [ 2 21]
 [ 2 22]
 [ 2 23]
 [ 2 24]
 [ 3 20]
 [ 3 21]
 [ 3 22]
 [ 3 23]
 [ 3 24]]

This solution was taken from a similar question: Using numpy to build an array of all combinations of two arrays

edited May 23, 2017 at 12:13

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answered Apr 18, 2015 at 8:37

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