remove last few characters in PySpark dataframe column

user_8329213 0
Viewed 34836

I am having a PySpark DataFrame. How can I chop off/remove last 5 characters from the column name below -

from pyspark.sql.functions import substring, length
valuesCol = [('rose_2012',),('jasmine_2013',),('lily_2014',),('daffodil_2017',),('sunflower_2016',)]
df = sqlContext.createDataFrame(valuesCol,['name'])
df.show()

+--------------+
|          name|
+--------------+
|     rose_2012|
|  jasmine_2013|
|     lily_2014|
| daffodil_2017|
|sunflower_2016|
+--------------+

I want to create 2 columns, the flower and year column.

Expected output:

+--------------+----+---------+
|          name|year|   flower|
+--------------+----+---------+
|     rose_2012|2012|     rose|
|  jasmine_2013|2013|  jasmine|
|     lily_2014|2014|     lily|
| daffodil_2017|2017| daffodil|
|sunflower_2016|2016|subflower|
+--------------+----+---------+

year column I have created -

df = df.withColumn("year", substring(col("name"),-4,4))
df.show()
+--------------+----+
|          name|year|
+--------------+----+
|     rose_2012|2012|
|  jasmine_2013|2013|
|     lily_2014|2014|
| daffodil_2017|2017|
|sunflower_2016|2016|
+--------------+----+

I don't know how to chop last 5 characters, so that I only have the name of flowers. I tried something like this, by invoking length, but that doesn't work.

df = df.withColumn("flower",substring(col("name"),0,length(col("name"))-5))

How can I create flower column with only flower names?

pyspark substring python
5 Answers

You can use expr function

>>> from pyspark.sql.functions import substring, length, col, expr
>>> df = df.withColumn("flower",expr("substring(name, 1, length(name)-5)"))
>>> df.show()
+--------------+----+---------+
|          name|year|   flower|
+--------------+----+---------+
|     rose_2012|2012|     rose|
|  jasmine_2013|2013|  jasmine|
|     lily_2014|2014|     lily|
| daffodil_2017|2017| daffodil|
|sunflower_2016|2016|sunflower|
+--------------+----+---------+
user_83127020

You can use split function. this code does what you want:

import pyspark.sql.functions as f

newDF = df.withColumn("year", f.split(df['name'], '\_')[1]).\
           withColumn("flower", f.split(df['name'], '\_')[0])

newDF.show()

+--------------+----+---------+
|          name|year|   flower|
+--------------+----+---------+
|     rose_2012|2012|     rose|
|  jasmine_2013|2013|  jasmine|
|     lily_2014|2014|     lily|
| daffodil_2017|2017| daffodil|
|sunflower_2016|2016|sunflower|
+--------------+----+---------+
user_66456170

In this case, since we want to extract alphabetical characters, so REGEX will also work. 

from pyspark.sql.functions import regexp_extract 
df = df.withColumn("flower",regexp_extract(df['name'], '[a-zA-Z]+',0))
df.show()
+--------------+----+---------+
|          name|year|   flower|
+--------------+----+---------+
|     rose_2012|2012|     rose|
|  jasmine_2013|2013|  jasmine|
|     lily_2014|2014|     lily|
| daffodil_2017|2017| daffodil|
|sunflower_2016|2016|sunflower|
+--------------+----+---------+
user_83292130

Adding little tweak to avoid hard coding and identify column length dynamically through location of underscore('_') using instr function.

df = spark.createDataFrame([('rose_2012',),('jasmine_2013',),('lily_2014',),('daffodil_2017',),('sunflower_2016',)],['name'])


df.withColumn("flower",expr("substr(name, 1, (instr(name,'_')-1) )")).\
        withColumn("year",expr("substr(name, (instr(name,'_')+1),length(name))")).show()
user_77390870 
>>> from pyspark.sql.functions import substring, length, expr
>>> df = df.withColumn("flower",expr("substring(name, 0, length(name)-5)"))
>>> df.show()
user_30494190
Related