熊猫:有条件的groupby

2024-05-11

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

我需要去做

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

但我想添加与以下条件相关的条件

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

如果算作category少于5,我想放弃这个类别。 我不知道,我怎么能在那里写这个条件。


As EdChum 评论 https://stackoverflow.com/questions/39634175/pandas-groupby-with-condition/39634269#comment66572870_39634175, 您可以使用filter http://pandas.pydata.org/pandas-docs/stable/groupby.html#filtration:

您还可以通过以下方式简化聚合sum:

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

另一种解决方案是reset_index http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.reset_index.html:

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

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
本文内容由网友自发贡献,版权归原作者所有,本站不承担相应法律责任。如您发现有涉嫌抄袭侵权的内容,请联系:hwhale#tublm.com(使用前将#替换为@)

熊猫:有条件的groupby 的相关文章

随机推荐