Select 筛选pyspark数据帧

Select 筛选pyspark数据帧,select,apache-spark,pyspark,spark-dataframe,apache-spark-2.0,Select,Apache Spark,Pyspark,Spark Dataframe,Apache Spark 2.0,我试图根据一些规则从pyspark数据帧中选择一些值。在pyspark中获得异常 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 = '1

我试图根据一些规则从pyspark数据帧中选择一些值。在pyspark中获得异常

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()
下面是我遇到的例外情况。你能给我一些建议吗

文件“”,第1行 df.选择(df.card_key,F.when((df.tran_赞助商='GAMES')和(df.location_代码='9145'),'ENTERTAINMENT')。当((df.tran_赞助商='XYZ')和(df.tran_赞助商='XYZ')和(df.l_代码isin('6001','6002','6003','6004','6005','6006','6007','6008','6009','6010','6011','6012','6012','3','6014')),财务)。否则(df.tran_赞助商)。show() ^ SyntaxError:无效语法


我刚刚发现,isin没有问题,问题在于赋值运算符:(

它工作得很好,如果有人在调查,谢谢你的努力

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()