PySpark create new column with mapping from a dict
PythonApache SparkDictionaryPysparkApache Spark-SqlPython Problem Overview
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 and I want to create a new column (say col2
) with the values from the dict
here below, how do I map this? (so f.i. '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'}
Python Solutions
Solution 1 - Python
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:
df.withColumn("value", mapping_expr[col("key")]).show()
Solution 2 - Python
Sounds like the simplest solution would be to use the replace function: http://spark.apache.org/docs/2.4.0/api/python/pyspark.sql.html#pyspark.sql.DataFrame.replace
mapping= {
'A': '1',
'B': '2'
}
df2 = df.replace(to_replace=mapping, subset=['yourColName'])
Solution 3 - Python
If you want to create a map col from a nested dictionary you can use this:
def create_map(d,):
if type(d) != dict:
return F.lit(d)
level_map = []
for k in d:
level_map.append(F.lit(k))
level_map.append(create_map(d[k]))
return F.create_map(level_map)
d = {'a': 1, 'b': {'c': 2, 'd': 'blah'}}
print(create_map(d)) # <- Column<b'map(a, 1, b, map(c, 2, d, blah))'>
Solution 4 - Python
you can use the function which convert dictionary into case syntax in Spark SQL
func_mapper = lambda dic,col,default : f"(CASE {col} WHEN " + " WHEN ".join([ f"'{k}' THEN '{v}'" for (k,v) in dic.items() ]) + f" ELSE '{default}' END)"