Question

鉴于df1和df2，我想获取df3。我要匹配的唯一列/行是 Pop 和 Homes 。我已经包含了数据的 Other 列，以获取数量众多的列的解决方案。

df1
City        Pop  Homes Other
City_1      100      1     0
City_1      100      2     6
City_1      100      2     2
City_1      100      3     9
City_1      200      1     6
City_1      200      2     6
City_1      200      3     7
City_1      300      1     0

df2
City        Pop  Homes Other
City_1      100      1     0
City_1      100      2     6
City_1      100      2     2
City_1      100      8     9
City_1      200      1     6
City_1      200      2     6
City_1      800      3     7
City_1      800      8     0

df3
City        Pop  Homes Other
City_1      100      1     0
City_1      100      2     6
City_1      100      2     2
City_1      200      1     6
City_1      200      2     6

我曾考虑过按城市，流行音乐和房屋进行分组，例如df1.groupby（['City'，'Pop'，'Homes']），但后来我不知道该如何滤除< em> Pop 和 Homes 。

编辑

这是我的代码，因此您可以更轻松地帮助我。

df1_string = """City_1      100      1     0
City_1      100      2     6
City_1      100      2     2
City_1      100      3     9
City_1      200      1     6
City_1      200      2     6
City_1      200      3     7
City_1      300      1     0"""

df2_string = """City_1      100      1     0
City_1      100      2     6
City_1      100      2     2
City_1      100      8     9
City_1      200      1     6
City_1      200      2     6
City_1      800      3     7
City_1      800      8     0"""

df1 = pd.DataFrame([x.split() for x in df1_string.split('\n')], columns=['City', 'Pop', 'Homes', 'Other'])
df2 = pd.DataFrame([x.split() for x in df2_string.split('\n')], columns=['City', 'Pop', 'Homes', 'Other'])

df1_keys = [x for x in df1.groupby(['Pop', 'Homes']).groups.keys()]
df2_keys = [x for x in df2.groupby(['Pop', 'Homes']).groups.keys()]

print(df1_keys)
[('100', '1'), ('100', '2'), ('100', '3'), ('200', '1'), ('200', '2'), ('200', '3'), ('300', '1')]
print(df2_keys)
[('100', '1'), ('100', '2'), ('100', '8'), ('200', '1'), ('200', '2'), ('800', '3'), ('800', '8')]

从这里看来，过滤掉不相等的组对似乎很简单，但是我不能解决这个问题。我尝试过：

df1 = df1[df1.groupby(['Pop', 'Homes']).groups.keys().isin(df2.groupby(['Pop', 'Homes']).groups.keys())]

当它不起作用时，还有其他变化-但是我感觉它已经接近工作了。

解决方案

df1.set_index(['Pop', 'Homes'], inplace=True)
df2.set_index(['Pop', 'Homes'], inplace=True)

df1 = df1[df2.index.isin(df1.index)]

df1.reset_index(inplace=True)

Answer 1

IIUC，如果索引中包含“城市”，“流行”，“家”，则可以使用isin：

df2[df2.index.isin(df1.index)]

输出：

                 Count
City  Pop Homes       
City1 100 20       152
          24       184
      200 41       163
          42       163

Answer 2

为数据框创建多索引，并为交集进行内部联接。

import pandas as pd
import numpy as np


df1_string = """City_1      100      1     0
City_1      100      2     6
City_1      100      2     2
City_1      100      3     9
City_1      200      1     6
City_1      200      2     6
City_1      200      3     7
City_1      300      1     0"""

df2_string = """City_1      100      1     0
City_1      100      2     6
City_1      100      2     2
City_1      100      8     9
City_1      200      1     6
City_1      200      2     6
City_1      800      3     7
City_1      800      8     0"""

df1 = pd.DataFrame([x.split() for x in df1_string.split('\n')], columns=['City', 'Pop', 'Homes', 'Other'])
df2 = pd.DataFrame([x.split() for x in df2_string.split('\n')], columns=['City', 'Pop', 'Homes', 'Other'])

# Dataframes benefit from having indexes that reflect that tabular data
df1.set_index(['City', 'Pop', 'Homes'], inplace=True)
df2.set_index(['City', 'Pop', 'Homes'], inplace=True)

# an inner join on the multiindex will provide the intersaction of the two
result = df1.join(df2, how='inner', on=['City', 'Pop', 'Homes'], lsuffix='_l', rsuffix='_r')

# a join provides all of the joined columns
result.reset_index(inplace=True)
result.drop(['Other_r'], axis=1, inplace=True)
result.columns = ['City', 'Pop', 'Homes', 'Other']

print(result)

如果一对列值未在另一个df2中配对，则删除df1中的行

2 个答案: