我有PypSparkDataframe
data = {'Passenger-Id': {0: 1, 1: 2, 2: 3, 3: 4, 4: 5},'Age': {0: 22, 1: 38, 2: 26, 3: 35, 4: 35}}
df_pd = pd.DataFrame(data, columns=data.keys())
df = spark.createDataFrame(df_pd)+------------+---+
|Passenger-Id|Age|
+------------+---+
| 1| 22|
| 2| 38|
| 3| 26|
| 4| 35|
| 5| 35|
+------------+---+这很管用
df.filter(df.Age == 22).show()但下面不起作用，因为-在列名中
df.filter(df.Passenger-Id == 2).show()attributeerror:“dataframe”对象没有属性“passenger”
我在spark sql也面临同样的问题，
spark.sql("SELECT Passenger-Id FROM AutoMobile").show()
spark.sql("SELECT automobile.Passenger-Id FROM AutoMobile").show()低于错误
analysisexception:无法解析' Passenger '给定的输入列：[automobile.age，automobile.passenger id]
正如一些资料中建议的那样，尝试用单引号表示列名，现在只打印查询中提到的列
spark.sql("SELECT 'Passenger-Id' FROM AutoMobile").show()+------------+
|Passenger-Id|
+------------+
|Passenger-Id|
|Passenger-Id|
|Passenger-Id|
|Passenger-Id|
|Passenger-Id|
+------------+
1条答案
按热度按时间eh57zj3b1#
既然您在列名中有hiphen，我建议您使用
col()函数来自sql.functions```import pyspark.sql.functions as F
df.filter(F.col('Passenger-Id')== 2).show()
+------------+---+
|Passenger-Id|Age|
+------------+---+
| 2| 38|
+------------+---+
df.createOrReplaceTempView("AutoMobile")
spark.sql("SELECT * FROM AutoMobile where
Passenger-Id=2").show()