CODEWITHSUNDEEP
pythondictionarycombinationscounter
vash_the_stampede
vash_the_stampede

Reputation: 4606

Direct count, not incremental count solution

This produces a count of frequency of tuples between 3 list, can it be done via direct count not incremental count (not +=)?

For clarification, I would like to work around having to increase the value of the key for every iteration using += and directly apply the count to corresponding pair

from collections import defaultdict
from itertools import combinations

dd = defaultdict(int)

L1 = ["cat", "toe", "man"]
L2 = ["cat", "toe", "ice"]
L3 = ["cat", "hat", "bed"]

for L in [L1, L2, L3]:
    for pair in map(frozenset, (combinations(L, 2))):
        dd[pair] += 1

defaultdict(int,
            {frozenset({'cat', 'toe'}): 2,
             frozenset({'cat', 'man'}): 1,
             frozenset({'man', 'toe'}): 1,
             frozenset({'cat', 'ice'}): 1,
             frozenset({'ice', 'toe'}): 1,
             frozenset({'cat', 'hat'}): 1,
             frozenset({'bed', 'cat'}): 1,
             frozenset({'bed', 'hat'}): 1})

Upvotes: 0

Views: 84

Answers (1)

jpp
jpp

Reputation: 164753

The counts are not stored anywhere, so iteration is unavoidable. But you can use collections.Counter with a generator expression to avoid an explicit for loop:

from collections import Counter
from itertools import chain, combinations

L1 = ["cat", "toe", "man"]
L2 = ["cat", "toe", "ice"]
L3 = ["cat", "hat", "bed"]

L_comb = [L1, L2, L3]

c = Counter(map(frozenset, chain.from_iterable(combinations(L, 2) for L in L_comb)))

Result:

Counter({frozenset({'cat', 'toe'}): 2,
         frozenset({'cat', 'man'}): 1,
         frozenset({'man', 'toe'}): 1,
         frozenset({'cat', 'ice'}): 1,
         frozenset({'ice', 'toe'}): 1,
         frozenset({'cat', 'hat'}): 1,
         frozenset({'bed', 'cat'}): 1,
         frozenset({'bed', 'hat'}): 1})

Upvotes: 4

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