在不使用集合的情况下计算出现次数。计数器

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])

我试了一下：

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面向逻辑的解决方案。
你能帮我做没有收集吗？

Answer 1

s = [13,12,11,13,14,13,7,11,13,14,12,14,14]

occurrences = dict()

for item in s:

    occurrences[item] = occurrences.setdefault(item, 0) + 1

mins = [a for a, b in occurrences.items() if b == min(occurrences.values())]
maxs = [a for a, b in occurrences.items() if b == max(occurrences.values())]

final_vals = [mins, maxs]

集合中的default更适合替换计数器来计算列表中项目的出现次数。但是因为你限制了集合的使用。所以setdefault处理KeyError更优雅。

Answer 2

data = [13,12,11,13,14,13,7,11,13,14,12,14,14]
occurrences = {}
for i in data:
     if i in occurrences.keys():
             occurrences[i] += 1
     else:
             occurrences[i] = 1

max_vals = [i for i in occurrences.keys() if occurrences[i] == max(occurrences.values())]
min_vals = [i for i in occurrences.keys() if occurrences[i] == min(occurrences.values())]

Answer 3

您可以对< code>dict使用< code>try和< code>except方法来模拟< code >计数器。

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< code>dict.get，默认值为< code>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（），以避免重复计算这个数量（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])