Find frequency of pair of numbers in 2D array in python

Question

I want to find frequency of pairs in a 2D array. Sample inputs is as follows:

list_of_items = [[12,14,18],[12,19,54,89,105],[ 14, 19],[54, 88 ,105,178]]

Expected Output is as following:

(12,14):1
(12,18):1
(12,19):1
(12,54):1
(12,88):0
.
.
.
(54,105):2
.
.

I have tried following code but I think it is not optimal solution:

number_set = [ 12, 14, 18,19,54,88,89 , 105, 178]

def get_frequency_of_pairs(list_of_items, number_set):
    x=1
    combination_list = []
    result = {}
    for i in number_set:
       for j in range(x,len(number_set)):
           combination_list = combination_list +[(i,number_set[j])]
       x = x+1
    for t in combination_list:
        result[t]=0
    for t in combination_list:
       for items in list_of_items:
           if( set(t).issubset(items) ):
              result[t]=result[t]+1
    return result

Is `number_set` always the set of numbers present in `list_of_items`? The best solution to the problem depends on the answer to this question. — Eric O. Lebigot, Aug 25 '15 at 12:04
Yes @EOL number_set is union of all the numbers present in list_of_items — Amit Jaiswal, Aug 25 '15 at 12:07
1. Should the pairs be in present in same array? For example, is 19 from one sub list and 54 from another sublist a valid pair? 2. Also, in number set, what is the way pairs can be formed? Is (14, 12) a valid pair? — Priyesh, Aug 25 '15 at 12:09
@EOL I have corrected it. Now it will give expected results. Thanks for pointing out the problem — Amit Jaiswal, Aug 25 '15 at 12:25
Your code also only produces counts for pairs *ordered* like in the original sublists (for example, (105, 54) does not appear there)): I guess that this is what you want? — Eric O. Lebigot, Aug 26 '15 at 00:34

score 5 · Answer 1 · edited Aug 26 '15 at 00:25

You can use combinations from itertools and use a Counter from collections as follows:

counts = collections.Counter()
list_of_items = [[12,14,18], [12,19,54,89,105], [14,19], [54,88,105,178]]
for sublist in list_of_items:
    counts.update(itertools.combinations(sublist, 2))

print counts
Counter({(54, 105): 2, (88, 105): 1, (54, 89): 1, (19, 105): 1, (12, 14): 1, (14, 19): 1, (14, 18): 1, (12, 89): 1, (12, 19): 1, (89, 105): 1, (12, 18): 1, (19, 89): 1, (19, 54): 1, (105, 178): 1, (88, 178): 1, (54, 178): 1, (12, 105): 1, (12, 54): 1, (54, 88): 1})

Each pair has to be enumerated to be counted, and this method enables you to enumerate each pair only once. Should be best possible time complexity.

Find frequency of pair of numbers in 2D array in python

1 Answers1

Linked