过滤一个pyspark数据帧

Question

我想根据一些规则从pyspark数据帧中选择一些值。在pyspark获得例外。

from pyspark.sql import functions as F 

df.select(df.card_key,F.when((df.tran_sponsor = 'GAMES') & (df.location_code = '9145'),'ENTERTAINMENT').when((df.tran_sponsor = 'XYZ') & (df.location_code = '123'),'eBOOKS').when((df.tran_sponsor = 'XYZ') & (df.l_code.isin(['123', '234', '345', '456', '567', '678', '789', '7878', '67', '456'])),'FINANCE').otherwise(df.tran_sponsor)).show() 

以下是我遇到的异常。你能提出一些建议吗？

File "", line 1 df.select(df.card_key,F.when((df.tran_sponsor = 'GAMES') & (df.location_code = '9145'),'ENTERTAINMENT').when((df.tran_sponsor = 'XYZ') & (df.location_code = '123'),'eBOOKS').when((df.tran_sponsor = 'XYZ') & (df.l_code.isin(['6001', '6002', '6003', '6004', '6005', '6006', '6007', '6008', '6009', '6010', '6011', '6012', '6013', '6014'])),'FINANCE').otherwise(df.tran_sponsor)).show() ^ SyntaxError: invalid syntax

Answer 1

好吧，我只是想通了，没有问题ISIN的事情，问题是用赋值运算符:(

df.select(df.card_key,F.when((df.tran_sponsor == 'GAMES') & (df.location_code == '9145'),'ENTERTAINMENT').when((df.tran_sponsor == 'XYZ') & (df.location_code == '123'),'eBOOKS').when((df.tran_sponsor == 'XYZ') & (df.l_code.isin(['123', '234', '345', '456', '567', '678', '789', '7878', '67', '456'])),'FINANCE').otherwise(df.tran_sponsor)).show() 

它运作良好，感谢努力，如果有人在寻找到它。