CODEWITHSUNDEEP
pythonpandasapache-spark-sqlpyspark
Siddharth Satpathy
Siddharth Satpathy

Reputation: 3043

Multiply column of PySpark dataframe with scalar

I want to multiply a column (say x3) of a PySpark dataframe (say df) with a scalar (say 0.1). Below is an example of a dataframe that I have:

df = sqlContext.createDataFrame(
    [(1, "a", 1551.0), (3, "B", 1925.0)], ("x1", "x2", "x3"))

df.show()

+---+---+----+
| x1| x2|  x3|
+---+---+----+
|  1|  a| 5.0|
|  3|  B|21.0|
+---+---+----+

Below is what I am trying at present:

df_new = df.withColumn( "norm_x3", 0.1*F.col( "x3") )
df_new = df_new.select( [c for c in df_new.columns if c not in {'x3'}] )

The method which I am trying above gives the expected output which is:

+---+---+-------+
| x1| x2|norm_x3|
+---+---+-------+
|  1|  a|    0.5|
|  3|  B|    2.1|
+---+---+-------+

Is there a more elegant and short way of doing the same thing? Thanks.

Upvotes: 0

Views: 4059

Answers (2)

aranelladen
aranelladen

Reputation: 101

The most elegant way would be simply using drop:

df_new = df.withColumn("norm_x3", 0.1*F.col( "x3")).drop("x3")

Alternatively, you can also use withColumnRenamed, but is less preferable because you're overloading "x3" and could cause confusion in the future: 

df_new = df.withColumn("x3", 0.1*F.col( "x3")).withColumnRenamed("x3", "norm_x3")

Upvotes: 4

akuiper
akuiper

Reputation: 214927

Here's one way to do it in one line:

df.select([(df[c] * 0.1).alias('norm_x3') if c == 'x3' else df[c] for c in df.columns]

Or:

df.selectExpr('*', 'x3 * 0.1 as normal_x3').drop('x3')

Upvotes: 3

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