How to make this function return pair of combination from array rows in Python?

Question

i have the following function

import numpy as np

a = [['a','b','c'],['d','e','f'],['g','h','i'],['j','k','l']]
b = np.array(a)

def func01(matrix):
    m,n = np.shape(matrix)
    for jump in range (m-1):
        for row in range (jump):
            for col in range (n):
                print (matrix[row][col],matrix[row+jump][col])

func01(b)

this results in:

('a', 'd') ('b', 'e') ('c', 'f') ('a', 'g') ('b', 'h') ('c', 'i') ('d', 'j') ('e', 'k') ('f', 'l')

however i want my result to look like this:

('a', 'd') ('b', 'e') ('c', 'f') ('a', 'g') ('b', 'h') ('c', 'i') ('a', 'j') ('b', 'k') ('c', 'l') ('d', 'g') ('e', 'h') ('f', 'i') ('d', 'j') ('e', 'k') ('f', 'l') ('g', 'j') ('h', 'k') ('i', 'l')

What have I done wrong? Sorry for my bad English

Make it do like you want? What's the algorithm that you want. Seems like there is a missing 3rd loop somewhere? — Matthieu Brucher
– Matthieu Brucher, Commented Nov 27, 2018 at 17:14
I want the first to the last value of the row 0 to be paired with first to the last value of the row 1 respectively, then row 0 with row 2, row 0 with row 3, after that row 1 with row 2, and so on. Without repeating any pair, (a,d) is the same with (d,a) — Surakri Karto
– Surakri Karto, Commented Nov 27, 2018 at 17:22
@SurakriKarto See my answer or Christian Sloper's answer below. Should be what you want =) — Stephen C
– Stephen C, Commented Nov 27, 2018 at 17:28

Stephen C · Accepted Answer · 2018-11-27 17:26:45Z

1

How about this:

a = [['a','b','c'],['d','e','f'],['g','h','i'],['j','k','l']]

new_list = []   
for i in range(len(a)): # Loop over the first subgroup
    for j in range(i+1, len(a)): # Loop over the second subgroup
        for k in range(len(a[0])): # Loop through the elements of each subgroup
            new_list.append((a[i][k], a[j][k]))


new_list
[('a', 'd'), ('b', 'e'), ('c', 'f'),
 ('a', 'g'), ('b', 'h'), ('c', 'i'),
 ('a', 'j'), ('b', 'k'), ('c', 'l'),
 ('d', 'g'), ('e', 'h'), ('f', 'i'),
 ('d', 'j'), ('e', 'k'), ('f', 'l'),
 ('g', 'j'), ('h', 'k'), ('i', 'l')]

This could be more concise with a list-comprehension:

new_list = [(a[i][k], a[j][k]) for i in range(len(a)) 
            for j in range(i+1, len(a)) 
            for k in range(len(a[0])]

edited Nov 27, 2018 at 17:26

answered Nov 27, 2018 at 17:17

Stephen C

2,0461 gold badge18 silver badges30 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Christian Sloper · Accepted Answer · 2018-11-27 17:31:25Z

1

Using combinations from itertools and some numpy slicing:

import numpy as np
import itertools


a = [['a','b','c'],['d','e','f'],['g','h','i'],['j','k','l']]
b = np.array(a)
m,n = b.shape

res = sorted([k for i in range(n) for k in itertools.combinations(b[:,i],2) ])

yields:

[('a', 'd'), ('a', 'g'), ('a', 'j'), ('b', 'e'), ('b', 'h'), ('b', 'k'), ('c', 'f'), ('c', 'i'), ('c', 'l'), ('d', 'g'), ('d', 'j'), ('e', 'h'), ('e', 'k'), ('f', 'i'), ('f', 'l'), ('g', 'j'), ('h', 'k'), ('i', 'l')]

edited Nov 27, 2018 at 17:31

answered Nov 27, 2018 at 17:19

Christian Sloper

7,5303 gold badges17 silver badges30 bronze badges

5 Comments

Stephen C Over a year ago

Dunno if OP cares: Although these are the same combinations, the ordering is different than OP expected output... Though it's admittedly much cleaner than my posted answer :P

Christian Sloper Over a year ago

@StephenCowley ok, good catch. I'll see if i can add some sorting.

Surakri Karto Over a year ago

ah, indeed this is clean, only the matter of order is a bit off. thank you!

Stephen C Over a year ago

@ChristianSloper Still not quite right. OP's goes [a b c a b c... I've been playing with your solution a little as well and haven't figured out how to get that order either... shrug

Christian Sloper Over a year ago

@StephenCowley yeah, i guess it is not obvious how to get the correct order with this. a pity :-)

Rohan Saxena · Accepted Answer · 2018-11-27 17:37:44Z

0

This is exactly the use case that zip was made for:

from itertools import combinations

a = [['a','b','c'], ['d','e','f'], ['g','h','i'], ['j','k','l']]

result = [list(zip(*x)) for x in combinations(a, 2)]

# If you want to flatten the result:
flat_result = [item for row in result for item in row]

print(flat_result)

result:

[[('a', 'd'), ('b', 'e'), ('c', 'f')],
 [('a', 'g'), ('b', 'h'), ('c', 'i')],
 [('a', 'j'), ('b', 'k'), ('c', 'l')],
 [('d', 'g'), ('e', 'h'), ('f', 'i')],
 [('d', 'j'), ('e', 'k'), ('f', 'l')],
 [('g', 'j'), ('h', 'k'), ('i', 'l')]]

flat_result:

[('a', 'd'), ('b', 'e'), ('c', 'f'), ('a', 'g'), ('b', 'h'), ('c', 'i'), ('a', 'j'), ('b', 'k'), ('c', 'l'), ('d', 'g'), ('e', 'h'), ('f', 'i'), ('d', 'j'), ('e', 'k'), ('f', 'l'), ('g', 'j'), ('h', 'k'), ('i', 'l')]

Or if you directly want to compute the flat version:

result = []
for x in combinations(a, 2):
    result += list(zip(*x))

answered Nov 27, 2018 at 17:37

Rohan Saxena

3,3882 gold badges18 silver badges34 bronze badges

2 Comments

Surakri Karto Over a year ago

woah, thank you! I'm still not familiar with all the Python libraries and stuff, hehe

Rohan Saxena Over a year ago

Yeah, I guessed. That's why I linked the doc to zip :-). Python has a rich library which will usually be more efficient and easier to use than anything that you could write (at least as a beginner). So before coding it yourself, browse the docs to see if you can find a built-in functionality which does what you wanna do.

Collectives™ on Stack Overflow

How to make this function return pair of combination from array rows in Python?

3 Answers 3

Comments

5 Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

5 Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related