如何获取嵌套列表python中特定元素的计数

Question

count_freq   data
3            [['58bcd029', 2, 'expert'], 
              ['58bcd029', 2, 'user'], 
             ['58bcd029', 2, 'expert']]
2            [['58bcd029', 2, 'expert'], 
             ['58bcd029', 2, 'expert']]
1            [['1ee429fa', 1, 'expert']]

所以我想从数据框的每一行和每个列表中获取“专家”和“用户”的计数。 在获得专家和用户的数量后，我想将各自的 id 存储在另一个列表中。 我曾尝试将它们转换为字典并使用键进行计算，但它不起作用。 任何人都可以帮助我这样做吗？

我想要这种格式的数据框：

count_freq   count_expert  ids                     count_user ids
3            2             ['58bcd029','58bcd029'] 1          ['58bcd029']
2            2             ['58bcd029','58bcd029'] 0          []
1            1             ['1ee429fa']            0          []

Answer 1

一种解决方案可能是：

import pandas as pd

data = pd.DataFrame({
    'col': [[['58bcd029', 2, 'expert'],
             ['58bcd029', 2, 'user'],
             ['58bcd029', 2, 'expert']],
            [['58bcd029', 2, 'expert'],
             ['58bcd029', 2, 'expert']],
            [['1ee429fa', 1, 'expert']]]
})

print(data)
                                                 col
0  [[58bcd029, 2, expert], [58bcd029, 2, user], [...
1     [[58bcd029, 2, expert], [58bcd029, 2, expert]]
2                            [[1ee429fa, 1, expert]]



data['count_expert'] = data['col'].apply(lambda x: [item for sublist in x for item in sublist].count('expert'))
data['count_user'] = data['col'].apply(lambda x: [item for sublist in x for item in sublist].count('user'))
data['ids_expert'] = data['col'].apply(lambda x: list(set([sublist[0] for sublist in x if sublist[2] == 'expert'])))
data['ids_user'] = data['col'].apply(lambda x: list(set([sublist[0] for sublist in x if sublist[2] == 'user'])))


# For the purpose of illustration, I just selected these rows, but `col` is also there.
print(data[['count_expert', 'count_user', 'ids_expert', 'ids_user']])

   count_expert  count_user  ids_expert    ids_user
0             2           1  [58bcd029]  [58bcd029]
1             2           0  [58bcd029]          []
2             1           0  [1ee429fa]          []