PySpark create new column with mapping from a dict

Question

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'}

Solution 1

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

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

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))'>

Comments

[Shaido]

The problem here is that this will not create a new column, it will replace the values in the original one.

[SummerEla]

Couldn't you just first copy the old values into a new column first, then use this function?

[Andrew Hummel]

zero323 - Having trouble understanding this method. withColumn takes a colName string as first arg, but you're passing a map value instead of key? df.withColumn("value", mapping_expr.getItem(col("key"))) Also result from create_map looks like this and calling getItem() like above is not working for me: Column<b'map(key_a, val_a, key_b, val_b)'> Any thoughts?

[Kieran Cooney]

Replace also requires the new values to be the same type as the original column.

[Markus]

I like the shortness of this solution. If you want an extra column just copy the column using .withColumn("newColumn", "column_to_copy") or so - The example just provides the minimum code you need to know to do this yourselfs :) Sometimes I think comments on SO are just used to be pedantic..

[LePuppy]

Thank you so much for this solution ! It works like a charm. Also if you wish to do the same behaviour with a list instead of the fict : from pyspark.sql.functions import array For example: values = [1, 2, 3, 4, 5] indexing_expr = array(*[lit(x) for x in values]) # The * is important df.withColumn("value", indexing_expr[col("index")])

[safex]

if someone could benchmark this solution against the accepted one, that would be very helpful. I implemented the replace but I does slow down my code considerably which puzzles me.