Pyspark: Replacing value in a column by searching a dictionary

After a lot of searching and alternatives I think that the simplest way to replace using a python dict is with pyspark dataframe method replace:

deviceDict = {'Tablet':'Mobile','Phone':'Mobile','PC':'Desktop'}
df_replace = df.replace(deviceDict,subset=['device_type'])

This will replace all values with the dict, you can get the same results using df.na.replace() if you pass a dict argument combined with a subset argument. It's not clear enough on his docs because if you search the function replace you will get two references, one inside of pyspark.sql.DataFrame.replace and the other one in side of pyspark.sql.DataFrameNaFunctions.replace, but the sample code of both reference use df.na.replace so it is not clear you can actually use df.replace.

Here is a little helper function, inspired by the R recode function, that abstracts the previous answers. As a bonus, it adds the option for a default value.

from itertools import chain
from pyspark.sql.functions import col, create_map, lit, when, isnull
from pyspark.sql.column import Column

df = spark.createDataFrame([
    ('Tablet', ), ('Phone', ),  ('PC', ), ('Other', ), (None, )
], ["device_type"])

deviceDict = {'Tablet':'Mobile','Phone':'Mobile','PC':'Desktop'}

df.show()
+-----------+
|device_type|
+-----------+
|     Tablet|
|      Phone|
|         PC|
|      Other|
|       null|
+-----------+

Here is the definition of recode.

def recode(col_name, map_dict, default=None):
    if not isinstance(col_name, Column): # Allows either column name string or column instance to be passed
        col_name = col(col_name)
    mapping_expr = create_map([lit(x) for x in chain(*map_dict.items())])
    if default is None:
        return  mapping_expr.getItem(col_name)
    else:
        return when(~isnull(mapping_expr.getItem(col_name)), mapping_expr.getItem(col_name)).otherwise(default)

Creating a column without a default gives null/None in all unmatched values.

df.withColumn("device_type", recode('device_type', deviceDict)).show()

+-----------+
|device_type|
+-----------+
|     Mobile|
|     Mobile|
|    Desktop|
|       null|
|       null|
+-----------+

On the other hand, specifying a value for default replaces all unmatched values with this default.

df.withColumn("device_type", recode('device_type', deviceDict, default='Other')).show()

+-----------+
|device_type|
+-----------+
|     Mobile|
|     Mobile|
|    Desktop|
|      Other|
|      Other|
+-----------+

You can do this using df.withColumn too:

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

deviceDict = {'Tablet':'Mobile','Phone':'Mobile','PC':'Desktop'}

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

df = df.withColumn('device_type', mapping_expr[df['dvice_type']])
df.show()

The simplest way to do it is to apply a udf on your dataframe :

    from pyspark.sql.functions import col , udf

    deviceDict = {'Tablet':'Mobile','Phone':'Mobile','PC':'Desktop'}
    map_func = udf(lambda row : deviceDict.get(row,row))
    df = df.withColumn("device_type", map_func(col("device_type")))

Another way of solving this is using CASE WHEN in traditional sql but using f-strings and using the python dictionary along with .join for automatically generating the CASE WHEN statement:

column = 'device_type' #column to replace

e = f"""CASE {' '.join([f"WHEN {column}='{k}' THEN '{v}'" 
             for k,v in deviceDict.items()])} ELSE {column} END"""

df.withColumn(column,F.expr(e)).show()

+-----------+
|device_type|
+-----------+
|     Mobile|
|     Mobile|
|    Desktop|
|      Other|
|       null|
+-----------+

Note: if you want to return NULL where the keys doesnot match, just change ELSE {column} END to ELSE NULL END in the case statement for variable e

column = 'device_type' #column to replace

e = f"""CASE {' '.join([f"WHEN {column}='{k}' THEN '{v}'" 
             for k,v in deviceDict.items()])} ELSE NULL END"""

df.withColumn('New_Col',F.expr(e)).show()

+-----------+-------+
|device_type|New_Col|
+-----------+-------+
|     Tablet| Mobile|
|      Phone| Mobile|
|         PC|Desktop|
|      Other|   null|
|       null|   null|
+-----------+-------+