我试图检索列表中最常见和较不频繁的元素.

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]

但我不想使用集合模块并尝试更加面向逻辑的解决方案.
没有收藏,你可以帮我做吗？

解决方法:

您可以使用带有dict的try和except方法来模拟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

你应该在理解之外做min()和max()以避免重复计算该数量(函数现在是O(n)而不是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])