不使用 collections.Counter 来计算出现次数

Question

我正在尝试检索列表中最频繁和不太频繁的元素。

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]

但我不想使用该collections模块并尝试更面向逻辑的解决方案。
你能帮我在没有集合的情况下做到这一点吗？

Answer 1

您可以使用 atry和except方法dict来模拟Counter。

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

或者您也可以使用dict.get默认值0：

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

并且您应该在推导式之外执行min()和以避免重复计算该数量（该函数现在而不是：max()O(n)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

这将按预期工作：

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