根据列表对列进行分类，并将结果汇总

Question

假设我有一个数据帧，如下所示：

d = {'name': ['spain', 'greece','belgium','germany','italy'], 'davalue': [3, 4, 6, 9, 3]}
df = pd.DataFrame(data=d)

index name  davalue
0    spain      3
1    greece     4
2    belgium    6
3    germany    9
4    italy      3

我想根据name列中的字符串列表进行聚合和求和。例如，我可能有：southern=['spain', 'greece', 'italy']和northern=['belgium','germany']

我的目标是使用sum进行聚合，并获得：

index name  davalue
0   southern    10
1   northen     15

其中10=3+4+3和15=6+9

我想象的是：

df.groupby(by=[['spain','greece','italy'],['belgium','germany']])

可能存在。这个docs说

A label or list of labels may be passed to group by the columns in self

但我不确定我是否理解这在语法方面意味着什么

Answer 1

一种方法是使用^{}并将结果用作grouper：

import numpy as np

southern=['spain', 'greece', 'italy']
northern=['belgium','germany']

g = np.select([df.name.isin(southern),
               df.name.isin(northern)],
              ['southern', 'northern'],
              'others')

df.groupby(g).sum()

          davalue
northern       15
southern       10

Answer 2

df["regional_group"]=df.apply(lambda x: "north" if x["home_team_name"] in ['belgium','germany'] else "south",axis=1)

您将创建一个新列，以后可以通过该列进行grouppy

df.groupby("regional_group")["davavalue"].sum()

Answer 3

我会建立一个字典和地图：

d = {v:'southern' for v in southern}
d.update({v:'northern' for v in northern})

df['davalue'].groupby(df['name'].map(d)).sum()

输出：

name
northern    15
southern    10
Name: davalue, dtype: int64