Get count of tuples in list regardless of elements orders

Question

I have a list of tuples that looks something like this:

my_list = [(1,12),(12,1),(12,1),(20,15),(7,8),(15,20)]

I want to get a count of the num

Answer 1

You can use collections.Counter, normalizing your elements with sorted first.

from collections import Counter

lst = [(1,12),(12,1),(12,1),(20,15),(7,8),(15,20)]

count = Counter(tuple(sorted(t)) for t in lst)
# output: Counter({(1, 12): 3, (15, 20): 2, (7, 8): 1})

You can then print like so.

for value, amount in count.items():
    print(value, '=', amount)

# Prints: 
# (1, 12) = 3
# (15, 20) = 2
# (7, 8) = 1

Answer 2

You can try simple approach without importing anything like this:

my_list = [(1,12),(12,1),(12,1),(20,15),(7,8),(15,20)]

count={}
for i in my_list:
    if tuple(sorted(i)) not in count:
        count[tuple(sorted(i))]=1
    else:
        count[tuple(sorted(i))]+=1

print(count)

output:

{(15, 20): 2, (7, 8): 1, (1, 12): 3}

Answer 3

First you can sort your list

lst2 = [tuple(sorted(i)) for i in lst]

Then you can use set() to find the unique values in this list and count their occurence.

count = {}
for i in set(lst2):
    num = len([1 for j in lst2 if j == i])
    count.update({i:num})
print(count)

{(1, 12): 3, (7, 8): 1, (15, 20): 2}

Answer 4

You can swap all positions by size and then use collections.counter:

from collections import Counter
l = [(1,12),(12,1),(12,1),(20,15),(7,8),(15,20)]
new_l = Counter([[(b, a), (a, b)][a < b] for a, b in l])
for a, b in new_l.items():
  print('{},{}:{}'.format(*(list(a)+[b])))

Output:

15,20:2
7,8:1
1,12:3