Find permutations which also match other constraints-CodePudding

Given the following list:

dbset = [[{'id': '10556', 'nation': 'France', 'worth': '70'}], [{'id': '14808', 'nation': 'France', 'worth': '65'}], [{'id': '11446', 'nation': 'Ghana', 'worth': '69'}], [{'id': '11419', 'nation': 'France', 'worth': '69'}], [{'id': '11185', 'nation': 'Ghana', 'worth': '69'}], [{'id': '1527', 'nation': 'Ghana', 'worth': '64'}], [{'id': '12714', 'nation': 'Moldova', 'worth': '67'}], [{'id': '2855', 'nation': 'Moldova', 'worth': '63'}], [{'id': '9620', 'nation': 'Moldova', 'worth': '71'}]]

I know how to find all permutations of length 4 with:

from itertools import permutations
perms = permutations(dbset,4)

Now comes the part where I struggle; I want the maximum times a nation is in the permutation to be equal to 2 and I also would like the total worth to be above 300.

I haven't a clue where to start.

CodePudding user response：

What you want is as simple as:

perms = list(permutations(dbset,4))
out = [x in perms if CONDITION]

The second condition is as simple as:

out = [x for x in perms if sum([int(country[0]["worth"]) for country in x]) >= 300]

Note that this will be empty in your case, since the maximum worth of any nation is 71, and 300/4 = 75.

I'll let you figure out the way to implement the first condition, but it is very similar. Note: and statements are your friend!

CodePudding user response：

Write a function that tests your conditions. Then filter the permutations using it.

from collections import Counter

def valid_permutation(perm):
    if sum(int(p['worth']) for p in perm) <= 300:
        return False
    counts = Counter(p['nation'] for p in perm)
    return counts.most_common()[0][1] <= 2

perms = filter(valid_permutation, permutations([d[0] for d in dbset], 4))