Get count of tuples in list regardless of elements orders

问题

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 number combinations regardless of the order. For example, if it were to be simply printed, I'd like the output to be:

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

Basically, I have a list of connections but the direction doesn't matter, so 1 to 12 is the same as 12 to 1.

I have been working on this for a while and can't come up with a clean solution. Everything I've come up with would have to check for both directions, but the list of tuples is random, so covering every possibility would be ridiculous. I could easily count the unique tuples with set or something, but again, every solution for the "bi-directional" count I've thought about is sloppy.

I feel like this shouldn't be very hard, but I am thoroughly at brain fry from working on this so long. Any help would be greatly appreciated!

回答1:

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

回答2:

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

回答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}

回答4:

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}

来源：https://stackoverflow.com/questions/49724651/get-count-of-tuples-in-list-regardless-of-elements-orders

标签

python

list

count

tuples