cod*_*nce 5 python collections counter dot-product python-collections
Python集合计数器好奇,如果有更好的方法来做到这一点.重写Counter类方法?内置乘法产生两个计数器的点积
from collections import Counter
a = Counter({'b': 4, 'c': 2, 'a': 1})
b = Counter({'b': 8, 'c': 4, 'a': 2})
newcounter = Counter()
for x in a.elements():
for y in b.elements():
if x == y:
newcounter[x] = a[x]*b[y]
$ newcounter
Counter({'b': 32, 'c': 8, 'a': 2})
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假设a并且b始终具有相同的键,您可以使用字典理解来实现此目的,如下所示:
a = Counter({'b': 4, 'c': 2, 'a': 1})
b = Counter({'b': 8, 'c': 4, 'a': 2})
c = Counter({k:a[k]*b[k] for k in a})
print(c)
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产量
Counter({'b': 32, 'c': 8, 'a': 2})
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