Pandas 如何基于另一列的匹配规则向DataFrame追加对应数据？

我希望通过校验两个DataFrame是否包含相同的文档编号，将一个DataFrame中的数值追加到另一个DataFrame的对应位置。

示例数据 #

ip_df 数据 #

CardName     DocNum    DocDate      DocTotal    DocNum2    PaidToDate   Balance
0       CompanyA  800100001 2021-03-01  10000.000000  920000000  10000.000000  0.000000
1       CompanyA  800100002 2021-03-01  20000.000000  920000000  20000.000000  0.000000
2       CompanyA  800100003 2021-03-01  30000.000000  920000000  30000.000000  0.000000
3       CompanyA  800100004 2021-03-01  40000.000000  920000000  40000.000000  0.000000
4       CompanyA  800100005 2021-03-01  50000.000000  920000000  50000.000000  0.000000
..      ...        ...        ...           ...        ...           ...       ...
94      CompanyY  800100006 2021-03-01  60000.000000  920000005  60000.000000  0.000000
95      CompanyY  800100007 2021-03-01  70000.000000  920000005  70000.000000  0.000000
96      CompanyY  800100008 2021-03-01  80000.000000  920000005  80000.000000  0.000000
97      CompanyZ  800100009 2021-03-01  90000.000000  920000006  90000.000000  0.000000
98      CompanyZ  800100010 2021-03-01  11000.000000  920000006  11000.000000  0.000000

[99 rows x 7 columns]

ar_df 数据 #

BPCode       Balance Currency    DueDate  BPName  TransId       Ref1      Payment Received  Bank Charge
0     XXXXXXXX  10000.000000      USD 2020-09-29  CompanyA   503378  800100001          0            0
1     XXXXXXXX  20000.000000      USD 2021-03-01  CompanyA   543103  800100002          0            0
2     XXXXXXXX  30000.000000      USD 2021-03-01  CompanyA   543171  800100003          0            0
3     XXXXXXXX  40000.000000      USD 2021-03-01  CompanyA   544205  800100004          0            0
4     XXXXXXXX  50000.000000      USD 2021-03-01  CompanyA   544222  800100005          0            0
...        ...           ...      ...        ...  ...      ...        ...               ...          ...
3763  XXXXXXXX  60000.000000      USD 2021-03-02  CompanyY   548612  800100006          0            0
3764  XXXXXXXX  70000.000000      USD 2021-03-02  CompanyY   547727  800100007          0            0
3765  XXXXXXXX  80000.000000      USD 2021-03-30  CompanyY   553819  800100008          0            0
3766  XXXXXXXX  90000.000000      USD 2021-04-01  CompanyZ   547707  800100009          0            0
3767  XXXXXXXX  11000.000000      USD 2021-04-29  CompanyZ   556102  800100010          0            0

[3768 rows x 9 columns]

现有尝试与问题 #

我最初尝试的实现逻辑如下，但不确定是否正确：

for row in ip_df:
    if ip_df.row['DocNum'] == ar_df.row['Ref1']:
        ap_df.row['Payment Received'] = ip_df.row['PaidToDate']

• 不确定上述实现方式是否正确
• 查阅过相关资料，但大多解决方案仅针对标量值匹配场景，不符合当前需求

实际需求 #

校验ip_df中每行的DocNum是否存在于ar_df的Ref1列中，若匹配成功，则将ar_df对应行的Payment Received字段设置为ip_df对应行的PaidToDate值。

期望输出效果 #

BPCode       Balance Currency    DueDate  BPName  TransId       Ref1      Payment Received  Bank Charge
0     XXXXXXXX  10000.000000      USD 2020-09-29  CompanyA   503378  800100001          10000.000000 0
1     XXXXXXXX  20000.000000      USD 2021-03-01  CompanyA   543103  800100002          20000.000000 0
2     XXXXXXXX  30000.000000      USD 2021-03-01  CompanyA   543171  800100003          30000.000000 0
3     XXXXXXXX  40000.000000      USD 2021-03-01  CompanyA   544205  800100004          40000.000000 0
4     XXXXXXXX  50000.000000      USD 2021-03-01  CompanyA   544222  800100005          50000.000000 0
...        ...           ...      ...        ...  ...      ...        ...               ...          ...
3763  XXXXXXXX  60000.000000      USD 2021-03-02  CompanyY   548612  800114258          60000.000000 0
3764  XXXXXXXX  70000.000000      USD 2021-03-02  CompanyY   547727  800113975          70000.000000 0
3765  XXXXXXXX  80000.000000      USD 2021-03-30  CompanyY   553819  800115292          80000.000000 0
3766  XXXXXXXX  90000.000000      USD 2021-04-01  CompanyZ   547707  800113957          90000.000000 0
3767  XXXXXXXX  11000.000000      USD 2021-04-29  CompanyZ   556102  800115741          11000.000000 0

[3768 rows x 9 columns]

内容的提问来源于stack exchange，提问作者Beelz