["python","apache-spark","dictionary","pyspark","apache-spark-sql"]
PySpark create new column with mapping from a dict

Using Spark 1.6, I have a Spark DataFrame column (named let's say col1) with values A, B, C, DS, DNS, E, F, G and H. I want to create a new column (say col2) with the values from the dict here below. How do I map this? (e,g. 'A' needs to be mapped to 'S' etc.) dict = {'A': 'S', 'B': 'S', 'C': 'S', 'DS': 'S', 'DNS': 'S', 'E': 'NS', 'F': 'NS', 'G': 'NS', 'H': 'NS'}
Inefficient solution with UDF (version independent): from pyspark.sql.types import StringType from pyspark.sql.functions import udf def translate(mapping): def translate_(col): return mapping.get(col) return udf(translate_, StringType()) df = sc.parallelize([('DS', ), ('G', ), ('INVALID', )]).toDF(['key']) mapping = { 'A': 'S', 'B': 'S', 'C': 'S', 'DS': 'S', 'DNS': 'S', 'E': 'NS', 'F': 'NS', 'G': 'NS', 'H': 'NS'} df.withColumn("value", translate(mapping)("key")) with the result: +-------+-----+ | key|value| +-------+-----+ | DS| S| | G| NS| |INVALID| null| +-------+-----+ Much more efficient (Spark >= 2.0, Spark < 3.0) is to create a MapType literal: from pyspark.sql.functions import col, create_map, lit from itertools import chain mapping_expr = create_map([lit(x) for x in chain(*mapping.items())]) df.withColumn("value", mapping_expr.getItem(col("key"))) with the same result: +-------+-----+ | key|value| +-------+-----+ | DS| S| | G| NS| |INVALID| null| +-------+-----+ but more efficient execution plan: == Physical Plan == *Project [key#15, keys: [B,DNS,DS,F,E,H,C,G,A], values: [S,S,S,NS,NS,NS,S,NS,S][key#15] AS value#53] +- Scan ExistingRDD[key#15] compared to UDF version: == Physical Plan == *Project [key#15, pythonUDF0#61 AS value#57] +- BatchEvalPython [translate_(key#15)], [key#15, pythonUDF0#61] +- Scan ExistingRDD[key#15] In Spark >= 3.0 getItem should be replaced with __getitem__ ([]), i.e: from pyspark.sql.functions import col, create_map, lit from itertools import chain mapping_expr = create_map([lit(x) for x in chain(*mapping.items())]) df.withColumn("value", mapping_expr[col("key")])