熊猫：通过数据框汇总数据

Question

我有数据框：

ID,"url","app_name","used_at","active_seconds","device_connection","device_os","device_type","device_usage"
1ca9bb884462c3ba2391bf669c22d4bd,"",VK Client,2016-01-01 00:00:13,5,3g,ios,smartphone,home
b8f4df3f99ad786a77897c583d98f615,"",VKontakte,2016-01-01 00:01:45,107,wifi,android,smartphone,home
1ca9bb884462c3ba2391bf669c22d4bd,"",Twitter,2016-01-01 00:02:48,20,3g,ios,smartphone,home
1ca9bb884462c3ba2391bf669c22d4bd,"",VK Client,2016-01-01 00:03:08,796,3g,ios,smartphone,home
b8f4df3f99ad786a77897c583d98f615,"",WhatsApp Messenger,2016-01-01 00:03:32,70,wifi,android,smartphone,home
b8f4df3f99ad786a77897c583d98f615,"",VKontakte,2016-01-01 00:04:42,27,wifi,android,smartphone,home
b8f4df3f99ad786a77897c583d98f615,"",VKontakte,2016-01-01 00:05:30,5,wifi,android,smartphone,home
b8f4df3f99ad786a77897c583d98f615,"",WhatsApp Messenger,2016-01-01 00:05:36,47,wifi,android,smartphone,home
b8f4df3f99ad786a77897c583d98f615,"",VKontakte,2016-01-01 00:06:23,20,wifi,android,smartphone,home
a703114aa8a03495c3e042647212fa63,"",Instagram,2016-01-01 00:06:41,118,3g,android,smartphone,home
1637ce5a4c4868e694004528c642d0ac,"",Camera,2016-01-01 00:06:43,16,wifi,android,smartphone,home
1637ce5a4c4868e694004528c642d0ac,"",VKontakte,2016-01-01 00:07:00,45,wifi,android,smartphone,home
a703114aa8a03495c3e042647212fa63,"",VKontakte,2016-01-01 00:08:40,99,3g,android,smartphone,home
1637ce5a4c4868e694004528c642d0ac,"",VKontakte,2016-01-01 00:10:05,1,wifi,android,smartphone,home

我需要计算每个app_name对每个ID份额。 但是我不能下一步：将每个应用程序对每个ID的总和除以对所有应用程序对ID和下一个100的总和。（以找到百分比）我这样做：

short = df.groupby(['ID', 'app_name']).agg({'app_name': len, 'active_seconds': sum}).rename(columns={'active_seconds': 'count_sec', 'app_name': 'sum_app'}).reset_index()

但是当我尝试时，它只会将数量返回给每个应用程序

short = df.groupby(['ID', 'app_name']).agg({'app_name': len, 'active_seconds': sum / df.ID.app_name.sum() * 100}).rename(columns={'active_seconds': 'count_sec', 'app_name': 'sum_app'}).reset_index()

它返回一个错误

我该如何解决？

Answer 1

您需要的IIUC：

short = df.groupby(['ID', 'app_name'])
          .agg({'app_name': len, 
                'active_seconds': lambda x: 100 * x.sum() / df.active_seconds.sum()})
          .rename(columns={'active_seconds': 'count_sec', 'app_name': 'sum_app'})
          .reset_index()

print (short)

                                 ID            app_name  count_sec  sum_app
0  1637ce5a4c4868e694004528c642d0ac              Camera   1.162791        1
1  1637ce5a4c4868e694004528c642d0ac           VKontakte   3.343023        2
2  1ca9bb884462c3ba2391bf669c22d4bd             Twitter   1.453488        1
3  1ca9bb884462c3ba2391bf669c22d4bd           VK Client  58.212209        2
4  a703114aa8a03495c3e042647212fa63           Instagram   8.575581        1
5  a703114aa8a03495c3e042647212fa63           VKontakte   7.194767        1
6  b8f4df3f99ad786a77897c583d98f615           VKontakte  11.555233        4
7  b8f4df3f99ad786a77897c583d98f615  WhatsApp Messenger   8.502907        2

另一个解决方案：

#you need another name of df, e.g. short1
short1 = df.groupby(['ID', 'app_name'])
           .agg({'app_name': len, 'active_seconds': sum})
           .rename(columns={'active_seconds': 'count_sec', 'app_name': 'sum_app'})
           .reset_index()
short1.count_sec = 100 * short1.count_sec / df.active_seconds.sum()
print (short1)
                                 ID            app_name  count_sec  sum_app
0  1637ce5a4c4868e694004528c642d0ac              Camera   1.162791        1
1  1637ce5a4c4868e694004528c642d0ac           VKontakte   3.343023        2
2  1ca9bb884462c3ba2391bf669c22d4bd             Twitter   1.453488        1
3  1ca9bb884462c3ba2391bf669c22d4bd           VK Client  58.212209        2
4  a703114aa8a03495c3e042647212fa63           Instagram   8.575581        1
5  a703114aa8a03495c3e042647212fa63           VKontakte   7.194767        1
6  b8f4df3f99ad786a77897c583d98f615           VKontakte  11.555233        4
7  b8f4df3f99ad786a77897c583d98f615  WhatsApp Messenger   8.502907        2