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python - 计算列表字典中最流行的项目python

问题描述

我有一本看起来像这样的字典：

dict_users = {
    "user1": ["item1", "item2", "item3", "item1", "item2", "elem3", "thing4", "thing5", "thing6"],
    "user2": ["elem5", "elem8", "elem2", "elem3", "elem8", "elem5", "thing7", "thing1", "thing9"],
    "user3": ["thing9", "thing7", "thing1", "thing4", "elem3", "elem9", "thing3", "thing5", "thing2"],
}

现在从这里开始，我想构建一个新字典，将用户与列表中使用最多的项目结合起来，因此在这种情况下，示例的输出将是：

dict_counted = {
'user1': 'item'
'user2': 'elem'
'user3': 'thing'
}

我现在有这样的事情：

users = ['user1', 'user2', 'user3']

dictOfRatios = dict.fromkeys(users)

for key, value in dict_users.items():
    for value in dict_sers[key]:
        if value.startswith("item"):
            itemlist = list(value)
            for user in dictOfRatios:
                dictOfRatios[user] = len(itemlist)
                
print(dictOfRatios)

但是输出并不如预期，它甚至给出了错误的数字......在这种情况下，匹配的标准可以是从 i、e、t 到完成 item、elem、thing 的任何内容。

标签： python

解决方案

python计数器是你需要的

from collections import Counter
import re

dict_users = {
'user1': ['item1', 'item2', 'item3', 'item1', 'item2', 'elem3', 'thing4', 'thing5', 'thing6'],
'user2': ['elem5', 'elem8', 'elem2', 'elem3', 'elem8', 'elem5', 'thing7', 'thing1', 'thing9'],
'user3': ['thing9', 'thing7', 'thing1', 'thing4', 'elem3', 'elem9', 'thing3', 'thing5', 'thing2']
}

users = {user: Counter() for user in dict_users.keys()}

for us, lst in dict_users.items():
    user_counter = users[us]
    for el in lst:
        item_name = re.split("\d",el)[0]
        user_counter[item_name] += 1

dict_counted = {user: counter.most_common(1)[0][0] for user, counter in users.items()}
print(dict_counted)

输出：

{
 'user1': 'item',
 'user2': 'elem',
 'user3': 'thing'
}

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