I am trying to filter a dataframe in pyspark using a list. I want to either filter based on the list or include only those records with a value in the list. My code below does not work:
# 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 in l)
# expected: (0,1), (0,1), (0,2), (1,2)
# include only records with these scores in list l
records = df.where(df.score in l)
# expected: (1,10), (1,20), (3,18), (3,18), (3,18)
Gives the following error: ValueError: Cannot convert column into bool: please use '&' for 'and', '|' for 'or', '~' for 'not' when building DataFrame boolean expressions.
3 Answers
what it says is "df.score in l" can not be evaluated because df.score gives you a column and "in" is not defined on that column type use "isin"
The code should be like this:
# 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)
Note that where() is an alias for filter(), so both are interchangeable.
based on @user3133475 answer, it is also possible to call the isin() method from F.col() like this:
import pyspark.sql.functions as F
l = [10,18,20]
df.filter(F.col("score").isin(l))
I found the join implementation to be significantly faster than where for large dataframes:
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"])
