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直接计数,而不是增量计数解决方案

counter combinations dictionary python

vash_the_stampede · 技术社区 · 6 年前

这就产生了一个计数3个元组之间的频率列表,它可以通过直接计数而不是增量计数(不 += )?

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})

1 回复 | 直到 6 年前

jpp 6 年前

存储 collections.Counter 使用生成器表达式以避免显式 for 回路:

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)))

结果:

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})

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