map values in a dataframe from a dictionary using pyspark

Question

I want to know how to map values in a specific column in a dataframe.

I have a dataframe which looks like:

df = sc.parallelize([('india','japan'),('usa','uruguay')]).toDF(['col1','col2'])

+-----+-------+
| col1|   col2|
+-----+-------+
|india|  japan|
|  usa|uruguay|
+-----+-------+

I have a dictionary from where I want to map the values.

dicts = sc.parallelize([('india','ind'), ('usa','us'),('japan','jpn'),('uruguay','urg')])

The output I want is:

+-----+-------+--------+--------+
| col1|   col2|col1_map|col2_map|
+-----+-------+--------+--------+
|india|  japan|     ind|     jpn|
|  usa|uruguay|      us|     urg|
+-----+-------+--------+--------+

I have tried using the lookup function but it doesn't work. It throws error SPARK-5063. Following is my approach which failed:

def map_val(x):
    return dicts.lookup(x)[0]

myfun = udf(lambda x: map_val(x), StringType())

df = df.withColumn('col1_map', myfun('col1')) # doesn't work
df = df.withColumn('col2_map', myfun('col2')) # doesn't work

Answer 1

I think the easier way is just to use a simple dictionary and df.withColumn.

from itertools import chain
from pyspark.sql.functions import create_map, lit

simple_dict = {'india':'ind', 'usa':'us', 'japan':'jpn', 'uruguay':'urg'}

mapping_expr = create_map([lit(x) for x in chain(*simple_dict.items())])

df = df.withColumn('col1_map', mapping_expr[df['col1']])\
       .withColumn('col2_map', mapping_expr[df['col2']])

df.show(truncate=False)