Question

我有数据框：

ID,used_at,active_seconds,subdomain,visiting,category
123,2016-02-05 19:39:21,2,yandex.ru,2,Computers
123,2016-02-05 19:43:01,1,mail.yandex.ru,2,Computers
123,2016-02-05 19:43:13,6,mail.yandex.ru,2,Computers
234,2016-02-05 19:46:09,16,avito.ru,2,Automobiles
234,2016-02-05 19:48:36,21,avito.ru,2,Automobiles
345,2016-02-05 19:48:59,58,avito.ru,2,Automobiles
345,2016-02-05 19:51:21,4,avito.ru,2,Automobiles
345,2016-02-05 19:58:55,4,disk.yandex.ru,2,Computers
345,2016-02-05 19:59:21,2,mail.ru,2,Computers
456,2016-02-05 19:59:27,2,mail.ru,2,Computers
456,2016-02-05 20:02:15,18,avito.ru,2,Automobiles
456,2016-02-05 20:04:55,8,avito.ru,2,Automobiles
456,2016-02-05 20:07:21,24,avito.ru,2,Automobiles
567,2016-02-05 20:09:03,58,avito.ru,2,Automobiles
567,2016-02-05 20:10:01,26,avito.ru,2,Automobiles
567,2016-02-05 20:11:51,30,disk.yandex.ru,2,Computers

我需要做

group = df.groupby(['category']).agg({'active_seconds': sum}).rename(columns={'active_seconds': 'count_sec_target'}).reset_index()

但我想添加与

相关的条件

df.groupby(['category'])['ID'].count()

如果category的计数小于5，我想删除此类别。我不知道，我怎么能在那里写下这个条件。

Answer 1

作为EdChum commented，您可以使用filter：

您也可以通过sum简化汇总：

df = df.groupby(['category']).filter(lambda x: len(x) >= 5)

group = df.groupby(['category'], as_index=False)['active_seconds']
          .sum()
          .rename(columns={'active_seconds': 'count_sec_target'})
print (group)

      category  count_sec_target
0  Automobiles               233
1    Computers                47

reset_index的另一个解决方案：

group = df.groupby(['category'])['active_seconds'].sum().reset_index(name='count_sec_target')
print (group)
      category  count_sec_target
0  Automobiles               233
1    Computers                47

熊猫：有条件的群体

1 个答案: