Hi,
I am having a dataframe column (name of the column is CTOFF)
and I intend to prefix with '0' in case the length of the column is 3.
Unfortunately, I am unable to acheive my goal and wonder whether you can
help me here.
Command which I am executing:
ctoff_dedup_prep_temp =
ctoff_df.withColumn('CTOFF_NEW',when(length(col('CTOFF')) ==
3,'0'+col('CTOFF')))
ctoff_dedup_prep_temp.show()
+------------+----------+----------+--------------------+---------+-----+---------+
|EVNT_SRVC_AR|EVNT_FCLTY|EVNT_TP_CD| NTWRK_PRD_CD|
DY_OF_WK|CTOFF|CTOFF_NEW|
+------------+----------+----------+--------------------+---------+-----+---------+
| HKG| HKC| AR|2,3,7,8,C,D,J,P,Q...|1,2,3,4,5|
1440| null|
| HKG| HKC| AR| C,Q,T,Y|1,2,3,4,5|
730| 730.0|
| HKG| HKC| AR| E,K,C,Q,T,Y|1,2,3,4,5|
600| 600.0|
| HKG| HKC| AR| E,K,C,Q,T,Y|1,2,3,4,5|
900| 900.0|
+------------+----------+----------+--------------------+---------+-----+---------+
The result which I want is:
+------------+----------+----------+--------------------+---------+-----+---------+
|EVNT_SRVC_AR|EVNT_FCLTY|EVNT_TP_CD| NTWRK_PRD_CD|
DY_OF_WK|CTOFF|CTOFF_NEW|
+------------+----------+----------+--------------------+---------+-----+---------+
| HKG| HKC| AR|2,3,7,8,C,D,J,P,Q...|1,2,3,4,5|
1440| 1440|
| HKG| HKC| AR| C,Q,T,Y|1,2,3,4,5|
730| 0730|
| HKG| HKC| AR| E,K,C,Q,T,Y|1,2,3,4,5|
600| 0600|
| HKG| HKC| AR| E,K,C,Q,T,Y|1,2,3,4,5|
900| 0900|
+------------+----------+----------+--------------------+---------+-----+---------+
So I want the '0' to be prefixed but it's getting suffixed as '.0'. Any
clue around why is this happening
Thanks,
Debu
