在不使用集合的情况下对事件进行计数 [英] Counting occurrences without using collections.Counter

问题描述

我正在尝试检索列表中最频繁和不频繁的元素。

I am trying to retrieve the most frequent and less frequent elements in a list.

frequency([13,12,11,13,14,13,7,11,13,14,12,14,14])

我的输出是：

([7], [13, 14])

我尝试过：

import collections
s = [13,12,11,13,14,13,7,11,13,14,12,14,14]
count = collections.Counter(s)
mins = [a for a, b in count.items() if b == min(count.values())]
maxes = [a for a, b in count.items() if b == max(count.values())]
final_vals = [mins, maxes]

但是我不知道不想使用 collections 模块并尝试一种更加面向逻辑的解决方案。

您能帮我没有收藏吗？

But I don't want to use the collections module and try a more logic oriented solution.
Can you please help me to do it without collections?

推荐答案

您可以使用 try 和除了方法和 dict 来模拟 Counter 。

You could use a try and except approach with a dict to emulate a Counter.

def counter(it):
    counts = {}
    for item in it:
        try:
            counts[item] += 1
        except KeyError:
            counts[item] = 1
    return counts

或者您可以使用 dict.get ，默认值为 0 ：

def counter(it):
    counts = {}
    for item in it:
        counts[item] = counts.get(item, 0) + 1
    return counts

您应该执行 min（） 和 max（），以避免重复计算该数量（该函数现在为 O（n）而不是 O（n ^ 2）：

And you should do the min() and max() outside the comprehensions to avoid repeatedly calculating that quantity (the function is now O(n) instead of O(n^2):

def minimum_and_maximum_frequency(cnts):
    min_ = min(cnts.values())
    max_ = max(cnts.values())
    min_items = [k for k, cnt in cnts.items() if cnt == min_]
    max_items = [k for k, cnt in cnts.items() if cnt == max_]
    return min_items, max_items

然后将按预期工作：

>>> minimum_and_maximum_frequency(counter([13,12,11,13,14,13,7,11,13,14,12,14,14]))
([7], [13, 14])

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