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PySpark DataFrame | withColumn method

Machine Learning
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PySpark
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PySpark DataFrame
schedule Jul 1, 2022
Last updated
local_offer PySpark
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PySpark DataFrame's withColumn(~) method can be used to:

  • add a new column

  • update an existing column

Parameters

1. colName | string

The label of the new column. If colName already exists, then supplied col will update the existing column. If colName does not exist, then col will be a new column.

2. col | Column

The new column.

Return Value

A PySpark DataFrame (pyspark.sql.dataframe.DataFrame).

Examples

Consider the following PySpark DataFrame:

df = spark.createDataFrame([["Alex", 25], ["Bob", 30], ["Cathy", 50]], ["name", "age"])
df.show()
+-----+---+
| name|age|
+-----+---+
| Alex| 25|
| Bob| 30|
|Cathy| 50|
+-----+---+

Updating column values based on original column values in PySpark

To update an existing column, supply its column label as the first argument:

df.withColumn("age", 2 * df.age).show()
+-----+---+
| name|age|
+-----+---+
| Alex| 50|
| Bob| 60|
|Cathy|100|
+-----+---+

Note that you must pass in a Column object as the second argument, and so you cannot simply use a list as the new column values.

Adding a new column to a PySpark DataFrame

To add a new column AGEE with 0s:

import pyspark.sql.functions as F
df.withColumn("AGEE", F.lit(0)).show()
+-----+---+----+
| name|age|AGEE|
+-----+---+----+
| Alex| 25| 0|
| Bob| 30| 0|
|Cathy| 50| 0|
+-----+---+----+

Here, F.lit(0) returns a Column object holding 0s. Note that since column labels are case insensitive, if you pass in "AGE" as the first argument, you would end up overwriting the age column.

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Published by Isshin Inada
Edited by 0 others
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