如何在有重复的列上合并两个 DataFrame，并输出没有重复的行答案

【问题标题】：How do I merge two DataFrame on a column with duplicates, and output without duplicates row如何在有重复的列上合并两个 DataFrame，并输出没有重复的行
【发布时间】：2021-09-27 06:50:21
【问题描述】：

我有两个数据框df1和df2如下图：

df1:

	company	occupation
0	A	Administrator
1	B	Engineer
2	C	Engineer
3	D	Account
4	E	Administrator
5	F	Engineer

df2:

	occupation	description
0	Account	balance
1	Engineer	database
2	Administrator	chores
3	Administrator	calling
4	Engineer	frontend
5	Engineer	backendend

我想要什么：

	company	occupation	description
0	A	Administrator	chores
1	B	Engineer	database
2	C	Engineer	frontend
3	D	Account	balance
4	E	Administrator	calling
5	F	Engineer	backendend

我试过pd.merge(df1,df2,how="inner")，但总是得到重复行：

	company	occupation	description
0	A	Administrator	chores
1	A	Administrator	calling
2	E	Administrator	chores
3	E	Administrator	calling
4	B	Engineer	database
5	B	Engineer	frontend
6	B	Engineer	backendend
7	C	Engineer	database
8	C	Engineer	frontend
9	C	Engineer	backendend
10	F	Engineer	database
11	F	Engineer	frontend
12	F	Engineer	backendend
13	D	Account	balance

代码：

import pandas as pd
df1 = pd.DataFrame({"company":["A","B","C","D","E","F"],"occupation":["Administrator","Engineer","Engineer","Account","Administrator","Engineer"]})
df2 = pd.DataFrame({"occupation":["Account","Engineer","Administrator","Administrator","Engineer","Engineer"],"description":["balance","database","chores","calling","frontend","backendend"]})
df3 = pd.DataFrame({"company":["A","B","C","D","E","F"],"occupation":["Administrator","Engineer","Engineer","Account","Administrator","Engineer"],"description":["chores","database","balance","frontend","calling","backendend"]})
df4 = pd.merge(df1,df2,how="inner")
display(df1)
display(df2)
display(df3)
display(df4)

【问题讨论】：

您想要的输出准确吗？我想您可能希望将df1 中某个职业的出现与df2 中的相应出现相匹配，例如第一工程师被分配“数据库”。如果是，那么您想要的输出可能不准确？
我认为frontend 和balance 可能会互换。
是的，打错了，我修改了

标签： python pandas dataframe join duplicates

【解决方案1】：

让我们尝试使用groupby cumcount 创建一个键列来跟踪位置，然后在occupation 和key 上合并：

df1['key'] = df1.groupby('occupation').cumcount()
df2['key'] = df2.groupby('occupation').cumcount()
df4 = df1.merge(df2, on=['occupation', 'key']).drop('key', axis=1)

df4:

  company     occupation description
0       A  Administrator      chores
1       B       Engineer    database
2       C       Engineer    frontend
3       D        Account     balance
4       E  Administrator     calling
5       F       Engineer  backendend

df4 不丢弃key:

  company     occupation  key description
0       A  Administrator    0      chores
1       B       Engineer    0    database
2       C       Engineer    1    frontend
3       D        Account    0     balance
4       E  Administrator    1     calling
5       F       Engineer    2  backendend

也可以通过直接合并系列而不影响df1或df2：

df4 = df1.merge(
    df2,
    left_on=['occupation', df1.groupby('occupation').cumcount()],
    right_on=['occupation', df2.groupby('occupation').cumcount()]
).drop('key_1', axis=1)

df4:

  company     occupation description
0       A  Administrator      chores
1       B       Engineer    database
2       C       Engineer    frontend
3       D        Account     balance
4       E  Administrator     calling
5       F       Engineer  backendend

【讨论】：

【解决方案2】：

您可以合成需要的合并条件部分。职业在数据框中的位置。

df1 = pd.DataFrame({'company': ['A', 'B', 'C', 'D', 'E', 'F'],
 'occupation': ['Administrator','Engineer','Engineer','Account','Administrator','Engineer']})

df2 = pd.DataFrame({'occupation': ['Account','Engineer','Administrator','Administrator','Engineer','Engineer'],
 'description': ['balance','database','chores','calling','frontend','backendend']})

df1.assign(oid=df1.groupby("occupation", as_index=False).cumcount()).merge(
    df2.assign(oid=df2.groupby("occupation", as_index=False).cumcount()),
    on=["occupation", "oid"],
)

	company	occupation	oid	description
0	A	Administrator	0	chores
1	B	Engineer	0	database
2	C	Engineer	1	frontend
3	D	Account	0	balance
4	E	Administrator	1	calling
5	F	Engineer	2	backendend

【讨论】：

这和我的回答不一样吗？
@HenryEcker 在语义上是等价的。使用cumcount()跟踪位置