累计计算列表中每个元素的出现次数答案

【问题标题】：Count cumulatively the occurrences of each element of a list累计计算列表中每个元素的出现次数
【发布时间】：2020-08-07 10:46:54
【问题描述】：

我有一个清单：

selection_list=[3, 2, 3, 2, 2, 2]

我想累计计算每个元素的出现次数，我希望输出为：

['1/2', '1/4' , '2/2', '2/4', '3/4', '4/4'] # 1/2 means: first occurrence of two

dico={'selection':[], 'Number':[]}
l=[]
keys=[]
for i in range(len(selection_list)):
        dico['selection'].append(selection_list[i])
        #counting the total occurrences of each element
        occurrences = collections.Counter(selection_list)
for i in occurrences.keys():
        keys.append(i)
    #indexes of each element of the list 
for j in range(len(keys)):
       l.append([i for i,val in enumerate(selection_list) if val==keys[j]])
        

for i in range(len(l)):
       for j in l[i] :
       dico['Number'].insert(int(j), str(len(l[i]))+'-'+ str(len(l[i])) )

我得到这个输出：

dico={'selection': [2, 3, 2, 3, 3, 3], 'UnfoldingNumber': ['2-2', '4-4', '2-2', '4-4', '4-4', '4-4']}

我错过了什么？

【问题讨论】：

标签： python python-3.x list indexing find-occurrences

【解决方案1】：

这是一个简单的解决方案示例：

from collections import Counter

selection_list=[3, 2, 3, 2, 2, 2]

numerator = {i:0 for i in set(selection_list)}
denominator = Counter(selection_list)

result = []
for v in selection_list:
    numerator[v] += 1
    result.append(str(numerator[v]) + '/' + str(denominator[v]))

print(result)

【讨论】：