CODEWITHSUNDEEP
pythonapache-sparkpysparkapache-spark-sql
Sotos
Sotos

Reputation: 51592

Split string IF delimiter is found

Consider the following data frame,

cols = ['Id', 'col1', 'col2', 'col3']
vals = [('1A','Not his side|:|This side', 'This side', 'Not this either|:|but this'),  
        ('1B','Keep this', 'This one|:|keep this', 'remove|:|keep that')]

dd1 = sqlContext.createDataFrame(vals, cols)

#+---+------------------------+--------------------+--------------------------+
#|Id |col1                    |col2                |col3                      |
#+---+------------------------+--------------------+--------------------------+
#|1A |Not his side|:|This side|This side           |Not this either|:|but this|
#|1B |Keep this               |This one|:|keep this|remove|:|keep that        |
#+---+------------------------+--------------------+--------------------------+

What I need to do is to split the strings at |:| and keep the second word. However, If the string does not contain the delimiter (|:|), then I get null, i.e.:

users1 = [F.split(F.col(x), "\\|:\\|").alias(x) for x in cols]
dd1.select(*users1).show()

#+---------+---------+---------+
#|     col1|     col2|     col3|
#+---------+---------+---------+
#|This side|     null| but this|
#|     null|keep this|keep that|
#+---------+---------+---------+

The result I'm looking for is:

+---+---------+---------+---------+
|Id |col1     |col2     |col3     |
+---+---------+---------+---------+
|1A |This side|This side|but this |
|1B |Keep this|keep this|keep that|
+---+---------+---------+---------+

Upvotes: 3

Views: 1655

Answers (3)

Shaido
Shaido

Reputation: 28322

Use when and otherwise and check if the string contains "|:|". It can be done as follows:

cols = ['col1', 'col2', 'col3']

users1 = [F.when(F.col(x).contains("|:|"), F.split(F.col(x), "\\|:\\|")[1]).otherwise(F.col(x)).alias(x) for x in cols]
dd1.select(F.col('Id'), *users1)

Here the cols only include the columns that you want to split. The final select will include the Id column as well.

Upvotes: 2

Oli
Oli

Reputation: 10406

The selection you defined in your questions does not yield the result you mentioned. I think you forgot to select the element from the array ;)

users1 = [F.split(F.col(x), "\\|:\\|").alias(x) for x in cols]
dd1.select(*users1).show()
+----+--------------------+--------------------+--------------------+
|  Id|                col1|                col2|                col3|
+----+--------------------+--------------------+--------------------+
|[1A]|[Not his side, Th...|         [This side]|[Not this either,...|
|[1B]|         [Keep this]|[This one, keep t...| [remove, keep that]|
+----+--------------------+--------------------+--------------------+

To achieve what you want, you simply need to select the last element of the array. You can use the size function for this:

users2 = [F.col(x).getItem(F.size(F.col(x))-1).alias(x) for x in cols]
dd1.select(*users1).select(*users2).show()
+---+---------+---------+---------+
| Id|     col1|     col2|     col3|
+---+---------+---------+---------+
| 1A|This side|This side| but this|
| 1B|Keep this|keep this|keep that|
+---+---------+---------+---------+

Upvotes: 1

Ramesh Maharjan
Ramesh Maharjan

Reputation: 41957

You can use when and size inbuilt functions as 

users1 = [F.when(F.size(F.split(F.col(x), "\\|:\\|")) > 1, F.split(F.col(x), "\\|:\\|")[1]).otherwise(F.col(x)).alias(x) for x in cols]
dd1.select(*users1).show()

which should give you 

+---+---------+---------+---------+
| Id|     col1|     col2|     col3|
+---+---------+---------+---------+
| 1A|This side|This side| but this|
| 1B|Keep this|keep this|keep that|
+---+---------+---------+---------+

You can modify the answer so that you can use split function only once.

I hope the answer is helpful and should be a good hint on how to proceed.

Upvotes: 1

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