它说的是“df.score in l”不能被评估，因为df.score给了你一个列，而“in”没有在该列类型上定义，使用“isin”
代码应该是这样的：

# define a dataframe
rdd = sc.parallelize([(0,1), (0,1), (0,2), (1,2), (1,10), (1,20), (3,18), (3,18), (3,18)])
df = sqlContext.createDataFrame(rdd, ["id", "score"])

# define a list of scores
l = [10,18,20]

# filter out records by scores by list l
records = df.filter(~df.score.isin(l))
# expected: (0,1), (0,1), (0,2), (1,2)

# include only records with these scores in list l
df.filter(df.score.isin(l))
# expected: (1,10), (1,20), (3,18), (3,18), (3,18)

请注意，where()是filter()的别名，因此两者可以互换。

基于@user3133475的回答，也可以像这样从col()调用isin()函数：

from pyspark.sql.functions import col

l = [10,18,20]
df.filter(col("score").isin(l))

我发现join的实现对于大型的嵌入式框架来说要比where快得多：

def filter_spark_dataframe_by_list(df, column_name, filter_list):
    """ Returns subset of df where df[column_name] is in filter_list """
    spark = SparkSession.builder.getOrCreate()
    filter_df = spark.createDataFrame(filter_list, df.schema[column_name].dataType)
    return df.join(filter_df, df[column_name] == filter_df["value"])