一种解决方案可能是：

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'] = data['col'].apply(lambda x: set(sublist[0] for sublist in x))

# For the purpose of illustration, I just selected these rows, but `col` is also there.

print(data[['count_expert', 'count_user', 'ids']])

   count_expert  count_user         ids
0             2           1  {58bcd029}
1             2           0  {58bcd029}
2             1           0  {1ee429fa}