Python frequency of records while combining combinations

I've started using python for a pet project as organizing this data was not possible in a program like excel. I hope you can give me some guidance on how to achieve the result I'm looking for. Please excuse my lack of python literacy.

I have the following code, which I have simplified by reducing the number of lists and the number of elements in those lists to make it easier to illustrate.

import itertools
from collections import Counter

a = list(itertools.permutations([32,36,41],3))
b = list(itertools.permutations([36,32,41],3))
c = list(itertools.permutations([19,31,7],3))

fulllist = a+b+c

print(Counter(map(tuple, fulllist)))


which gives the following result:

Counter((32, 36, 41): 2, (32, 41, 36): 2, (36, 32, 41): 2, (36, 41, 32): 2, (41, 32, 36): 2, (41, 36, 32): 2, (19, 31, 7): 1, (19, 7, 31): 1, (31, 19, 7): 1, (31, 7, 19): 1, (7, 19, 31): 1, (7, 31, 19): 1)


This is already quite good but not excatly what I need. Now that I have the first count of each list-combination generated by intertools, I don't care anymore about the order of each element inside of said list. So, the final result I would like to obtain is:

(32, 36, 41): 12
(19, 31, 7): 6


And sorted, just like I have written above.
I feel I might be going in circles and perhaps there's a simpler way to get the result I'm looking for. In reality, my lists have 15 elements in them and I have around 50 of these lists to process.

I hope you can help me with this. Thank you very much in advance.

counts =

key = tuple(sorted(inputlist))

count = math.factorial(len(inputlist))

counts[key] = counts.get(key, 0) + count

map(tuple, ...))

itertools.chain()

3 Answers
3

If you are not counting permutations (in that case use @Martin's answer), and that was just used to create an example list, just sort to make it clear order does not matter:

>>>print(Counter(tuple(sorted(x)) for x in fulllist))
Counter((32, 36, 41): 12, (7, 19, 31): 6)


If all you need is a count of possible permutations, then just calculate those numbers. That number is simply the factorial of the length of the input:

import math

permutationcount = math.factorial(len(inputlist))


If you are creating permutations that are shorter than len(inputlist) (say, 3 out of 20), then the formula is n! / (n - k)!:

len(inputlist)

n! / (n - k)!

n = len(inputlist)
k = 3
permutationcount = math.factorial(n) // math.factorial(n - k)


Of course, when k is equal to n, then you divide by 0!, which is 1.

k

n

You can sort the input list, and turn it into a tuple, to create a key into a mapping:

from collections import Counter

def count_permutations(lists, k=None):
counts = Counter()
if k is None:
k = len(lists[0])
for l in lists:
key = tuple(sorted(l))
permutationcount = math.factorial(len(l)) // math.factorial(len(l) - k)
counts[key] += permutationcount
return counts


Demo:

>>> count_permutations(([32,36,41], [36,32,41], [19,31,7]), k=3)
Counter((32, 36, 41): 12, (7, 19, 31): 6)


You don't really need to use a Counter here of course, but it may be convenient to use one anyway if you need the .most_common() method and don't want to look up how to sort a dictionary by value.

Counter

.most_common()

c = Counter(map(tuple, fulllist))
l = [ tuple(sorted(i[0])):i[1] for i in c.items()]

for e in l:
print(e)

(32, 36, 41): 2
(32, 36, 41): 2
(32, 36, 41): 2
(32, 36, 41): 2
(32, 36, 41): 2
(32, 36, 41): 2
(7, 19, 31): 1
(7, 19, 31): 1
(7, 19, 31): 1
(7, 19, 31): 1
(7, 19, 31): 1
(7, 19, 31): 1


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