Find and Replace characters in a column range using Regex

Bob Johnson · Oct 25, 2019, 10:49 PM

I need to check columns 50 thru 55 for a - and if found replace with a . My file has 1M lines.

I have figured out how to start it for column 50 but how do i repeat this for the next 4 columns.

Find what: ^.{49}\K-
Replace with: .

Terry R · Oct 26, 2019, 2:41 AM

@Bob-Johnson said in Find and Replace characters in a column range using Regex:

how do i repeat this for the next 4 columns

The simplest way would be to use your original regex, but change the 49 to 50, then 51 etc. So in effect you run the regex 5 times in total, increasing the number by 1 each time.

It is possible to create another regex to cater for 1,2,3,4 or 5 - in a row that only needs running once, but I don’t think it’s worth the effort, especially if what you are editing is not a regular occurrence.

Terry

guy038 · Oct 26, 2019, 6:01 PM

Hello, @bob-johnson, @terry-r and All,

OK, Bob. Luckily your S/R does not change the columns order, as you just want to replace 1 char with a single char, too. So, given the sample text below :

                                                 Column
                                               4455555555
                                               8901234567
0       ABCDE                     VWXYZ        -                12345    # - at column 48
0       ABCDE                     VWXYZ         -               12345    # - at column 49

0       ABCDE                     VWXYZ          -BCDEFG        12345    # - at column 50
0       ABCDE                     VWXYZ          A-CDEFG        12345    # - at column 51
0       ABCDE                     VWXYZ          AB-DEFG        12345    # - at column 52
0       ABCDE                     VWXYZ          ABC-EFG        12345    # - at column 53
0       ABCDE                     VWXYZ          ABCD-FG        12345    # - at column 54
0       ABCDE                     VWXYZ          ABCDE-G        12345    # - at column 55

0       ABCDE                     VWXYZ                -        12345    # - at column 56
0       ABCDE                     VWXYZ                 -       12345    # - at column 57

The following regex S/R :

SEARCH (?-s)^.{49}.{0,5}\K-
REPLACE .
Tick the Wrap around option
Click on the Replace All button
And choose, of course, the Regular expression search mode

It should modify the text, as expected :

                                                 Column
                                               4455555555
                                               8901234567
0       ABCDE                     VWXYZ        -                12345    # - at column 48
0       ABCDE                     VWXYZ         -               12345    # - at column 49

0       ABCDE                     VWXYZ          .BCDEFG        12345    # - at column 50
0       ABCDE                     VWXYZ          A.CDEFG        12345    # - at column 51
0       ABCDE                     VWXYZ          AB.DEFG        12345    # - at column 52
0       ABCDE                     VWXYZ          ABC.EFG        12345    # - at column 53
0       ABCDE                     VWXYZ          ABCD.FG        12345    # - at column 54
0       ABCDE                     VWXYZ          ABCDE.G        12345    # - at column 55

0       ABCDE                     VWXYZ                -        12345    # - at column 56
0       ABCDE                     VWXYZ                 -       12345    # - at column 57

Notes :

The first part (?-s) forces the regex engine to consider that dot ( . ) represents a single standard char only ( Not an EOL one )
The special syntax \K, cancels any previous search, already matched, at current location of the regex engine and, in your case, just consider the last - character

Best Regards,

guy038