python list item combinations

Question

I want to do a comparison between every pair of elements in the list without the following: DIFFERENCE=[1,2,3,4,5,6]

no self comparisons
no reverse comparisons

So the answer will be [1,2],[1,3],[1,4],[1,5],[1,6],[2,3],[4,5],[5,6],[3,4],[2,4],[2,5],[3,6],[2,6],[3,5],[4,6] I have written this so far but I was looking for a faster way.

for i in DIFFERENCE:
       for j in DIFFERENCE:
           if(some condition and i!=j and i+'_'+j not in COMPARISON and j+'_'+i not in COMPARISON):
               COMPARISON.append(i+'_'+j);
               COMPARISON.append(j+'_'+i);
               ANS_COUNT=ANS_COUNT+1;

nneonneo · Accepted Answer · 2014-04-09 19:31:20Z

6

You should just use itertools.combinations:

>>> import itertools
>>> list(itertools.combinations([1,2,3,4,5,6], 2))
[(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6)]

answered Apr 9, 2014 at 19:31

nneonneo

181k37 gold badges331 silver badges412 bronze badges

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

3 Comments

Process1 Over a year ago

Hi I was wondering if you can give me some insights into why I am getting the following error when I run : TMP_DIFF=[list(i) for i in list(itertools.combinations(DIFFERENCE2, 2))]; MemoryError. Atfirst, I thought may be len(DIFFERENCE2) =0,1,2 was a problem so I wrote,if(len(DIFFERENCE2)>2):TMP_DIFF=[list(i) for i in list(itertools.combinations(DIFFERENCE2, 2))];but the error this persists. Please comment.

nneonneo Over a year ago

@mparida: How big is DIFFERENCE2? If it's more than a few tens of thousands of entries long, you'll run out of memory constructing every possible combination. In that case, consider iterating over the itertools.combinations output directly, which will "lazily" generate the combinations for you.

Process1 Over a year ago

Thanks for quick reply nneonneo I am not sure how big is DIFFERENCE2 but like you said I am assuming it is pretty big to flag such an error. I am not sure if I understand the last line In that case, consider iterating over the itertools.combinations output directly, which will "lazily" generate the combinations for you..please explain. Appreciate it.

Madhusudan · Accepted Answer · 2014-04-09 19:49:31Z

0

You can also do this using for loop:

l = list()
for i in range (1,7):
    for j in range (2,7):
        if(i == j):
            j = j + 1
        elif ((i,j) in l or (j,i) in l):
            continue
        else:
            l.append((i,j))

print l

output:

[(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6)]

edited Apr 9, 2014 at 19:49

answered Apr 9, 2014 at 19:43

Madhusudan

4652 gold badges11 silver badges26 bronze badges

Collectives™ on Stack Overflow

python list item combinations

2 Answers 2

3 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related