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鉴于我知道有多少个计数器,可以很容易地硬核 |
运算符以获得计数器的并集,例如:
>>> from collections import Counter
>>> x = Counter([('a'), ('a', 'bc'), ('a', 'bc'), ('xyz', 'hooli')])
>>> y = Counter([('a'), ('a'), ('a'), ('asd', 'asd')])
>>> z = Counter([('a'), ('a'), ('a'), ('a'), ('a', 'bc')])
>>> x | y | z
Counter({'a': 4, ('a', 'bc'): 2, ('asd', 'asd'): 1, ('xyz', 'hooli'): 1})
但是如果我有一个计数器列表作为函数的输入,我该如何获得联合?
我是否必须遍历每个计数器然后执行合并?例如
>>> xyz = [x,y,z]
>>> def counter_union(counters):
... union = Counter()
... for c in counters:
... union = union | c
... return union
...
>>> counter_union(xyz)
Counter({'a': 4, ('a', 'bc'): 2, ('asd', 'asd'): 1, ('xyz', 'hooli'): 1})
还有其他方法吗?