I am trying to get the last string after '/'.
The column can look like this: "lala/mae.da/rg1/zzzzz" (not necessary only 3 /), and I'd like to return: zzzzz
In SQL and Python it's very easy, but I would like to know if there is a way to do it in PySpark.
Solving it in Python:
original_string = "lala/mae.da/rg1/zzzzz"
last_char_index = original_string.rfind("/")
new_string = original_string[last_char_index+1:]
or directly:
new_string = original_string.rsplit('/', 1)[1]
And in SQL:
RIGHT(MyColumn, CHARINDEX('/', REVERSE(MyColumn))-1)
For PySpark I was thinking something like this:
df = df.select(col("MyColumn").rsplit('/', 1)[1])
but I get the following error: TypeError: 'Column' object is not callable and I am not even sure Spark allows me to do rsplit at all.
Do you have any suggestion on how can I solve this?