Python collections.Counter 中所有计数的总和
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Sum of all counts in a collections.Counter
提问by Baz
What is the best way of establishing the sum of all counts in a collections.Counterobject?
确定collections.Counter对象中所有计数总和的最佳方法是什么？
I've tried:
我试过了：
sum(Counter([1,2,3,4,5,1,2,1,6]))
sum(Counter([1,2,3,4,5,1,2,1,6]))
but this gives 21instead of 9?
但这给出了21而不是9？
采纳答案by NPE
The code you have adds up the keys (i.e. the unique values in the list: 1+2+3+4+5+6=21).
您拥有的代码将键（即列表中的唯一值：）相加1+2+3+4+5+6=21。
To add up the counts, use:
要将计数相加，请使用：
In [4]: sum(Counter([1,2,3,4,5,1,2,1,6]).values())
Out[4]: 9
This idiom is mentioned in the documentation, under "Common patterns".
文档中的“常见模式”下提到了这个习语。
回答by Martijn Pieters
Sum the values:
对值求和：
sum(some_counter.values())
Demo:
演示：
>>> from collections import Counter
>>> c = Counter([1,2,3,4,5,1,2,1,6])
>>> sum(c.values())
9
回答by martyn
sum(Counter([1,2,3,4,5,1,2,1,6]).values())
