Reputation: 1
I have a set of data which looks like this
XXXYYY,20151001,09:16,17370.40,17390.60,17362.40,17381.00,50,0
What I want to do is, deleting everything after the value 17381.00 i.e delete ",50,0".I have 7500 rows of such data and column number after 17381.00 is 61 in all rows.
What i am looking for is to able to delete everything from column 61.
The expected output is XXXYYY,20151001,09:16,17370.40,17390.60,17362.40,17381.00
Upvotes: 0
Views: 473
Reputation: 521259
You may try the following find and replace, in regex mode:
Find: ^((?:[^,]*,){59}[^,]*),.*$
Replace: $1
Here is an explanation of the regex pattern:
^
(
(?:[^,]*,){59} match and capture 'xxx,' 59 times
[^,]* match and capture 'xxx' (no comma)
)
, match a ,
.* consume the remainder of the line
$
Then, we replace with just the capture group $1
, effectively removing columns 61 onwards.
Upvotes: 1