计算python字典中某个值的出现次数?

问题描述

如果我有这样的东西:

D = {'a': 97, 'c': 0 , 'b':0,'e': 94, 'r': 97 , 'g':0}

如果我想例如将0"的出现次数计算为一个值而不必迭代整个列表，这是否可能以及如何实现?

If I want for example to count the number of occurrences for the "0" as a value without having to iterate the whole list, is that even possible and how?

推荐答案

如 THIS ANSWER 中所述，使用 operator.countOf() 是方式去，但你也可以在 sum() 函数中使用生成器，如下所示:

As mentioned in THIS ANSWER using operator.countOf() is the way to go but you can also use a generator within sum() function as following:

sum(value == 0 for value in D.values())
# Or the following which is more optimized 
sum(1 for v in D.values() if v == 0)

或者作为一种稍微优化和功能性更强的方法，您可以通过将整数的 __eq__ 方法作为构造函数传递来使用 map 函数.

Or as a slightly more optimized and functional approach you can use map function by passing the __eq__ method of the integer as the constructor function.

sum(map((0).__eq__, D.values()))

基准测试:

In [15]: D = dict(zip(range(1000), range(1000)))

In [16]: %timeit sum(map((0).__eq__, D.values()))
49.6 s ± 770 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [17]: %timeit sum(v==0 for v in D.values())
60.9 s ± 669 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [18]: %timeit sum(1 for v in D.values() if v == 0)
30.2 s ± 515 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [19]: %timeit countOf(D.values(), 0)
16.8 s ± 74.1 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

请注意，虽然在这种情况下使用 map 函数可能会更优化，但为了对这两种方法有更全面和一般的想法，您还应该对相对较大的数据集运行基准测试.然后，您可以根据您拥有的数据结构和数量使用最合适的方法.

Note that although using map function in this case may be more optimized, but in order to have a more comprehensive and general idea about the two approaches you should run the benchmark for relatively large datasets as well. Then, you can use the most proper approach based on the structure and amount of data you have.

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