Insert a string after each number of words with conditions

Alan Kilborn

Yes, in playing around with your original regex, I didn’t worry about the resulting space at the end of the line, as I have my “save” shortcut mapped to “trim trailing spaces” + “save”. The ONLY way files should be saved (for me!).

Alan Kilborn

@guy038

And it is great that you have Admin rights here and can edit old posts, but I’m neutral on this. I think that old posts should not be edited and clarifying posts should just be added on. It is difficult to follow sometimes when history is CHANGED rather than simply CORRECTED/CLARIFIED later. :-D

guy038

Hi, Alan,

Yes, you’re right about it : I should have created a new post with the corrections, for a better history ! It’s just that my updated post was, still, quite long and I thought it would be more clear to, simply, change my initial post. But, I do understand your point of view !

Cheers,

guy038

hu ma

@Scott-Sumner
Thanks for the info!

guy038

Hello, @hu-ma and All

To complete the @scott-sumner post, about the two syntaxes of the searched groups, in replacement :

\N, with 0 < N < 10
$N, with 0 <= N < 2,147,483,648

There is the other practical syntax, below :

${N}, with 0 <= N < 2,147,483,648

Indeed, let’s imagine the original text:

abcd
1234
WXYZ

and the first S/R :

SEARCH ^.(..).

REPLACE $100|

You obtain the simple text :

|
|
|

Why ?! Just because, in replacement, the regex engine is looking for the group $100, which, obviously, does not exist ! So, the regex engine rewrites a zero-length string, for the non-existent group 100, followed by the literal character | !

Now, compare, with the second S/R, below :

SEARCH ^.(..).

REPLACE ${1}00|

This time, you, correctly, get the text, below :

bc00|
2300|
XY00|

=> All the changed lines begin by the second and third characters of the original lines of text ( $1 ), and are, simply, followed by the string 00|

Best Regards,

guy038

abuali huma

Sorry for pumping up old thread, but my issue is related to this one.

Cutting to the thread…
Look at Result#2 with desired arrangement

-Example# 2
Greetings My Liege! As your personal advisor [NEWLINE] , I am qualified to assist you in all[NEWLINE] matters related to ruling our civilization.[NEWLINE] I am at your service.

--------
+Seeked arrangement
I am at your service. [NEWLINE] matters related to ruling our civilization.[NEWLINE], I am qualified to assist you in all[NEWLINE]Greetings My Liege! As your personal advisor

I asked before for a way to rearrange the groups between [NEWLINE] to be backward… Now I’m asking for the same but in more automated way…

Because not all lines have the same amount of Groups, I want to arrange all the lines that contains Groups between [NEWLINE] to be backward arrangement.

-Example#3 Contains SIX groups
One [NEWLINE] two [NEWLINE] three [NEWLINE] four [NEWLINE] five [NEWLINE] six

-------
+Seeked arrangement
 six[NEWLINE] five [NEWLINE] four [NEWLINE] three [NEWLINE] two [NEWLINE]One

While using the same regex or python script

-Example#4 Contains 4 groups 
I want [NEWLINE] this [NEWLINE] to be  [NEWLINE] last

------
+Seeked arrangement
 last[NEWLINE] to be [NEWLINE] this [NEWLINE]I want

guy038

Hi, @abuali-huma,

I found a general method, which uses three consecutive S/R. We’ll need two dummy characters, NOT used in the current file. I, personally, chose the # and @ characters, but any other may be used !

The first S/R :
- Changes any string [NEWLINE], possibly preceded and/or followed with a space character, by the dummy character #
- Adds, also, a # character at the end of any non-blank line
The second S/R is the main S/R, which rewrites the different parts, between the # character, in reverse order.
- Note that this S/R will have to be performed as many times, till the message Replace All: 0 occurrences were replaced occurs, in the Replace dialog
- The general idea, about this S/R, is to switch the beginning and ending parts of the found text, adding a @ character, at the end of the exchanged parts, in order that the next run of this S/R, will avoid these moved parts of text ! Hence, the decreasing number of occurrences found, till zero :-))
The Third S/R :
- Changes the # character, possibly preceded by a @ character, inside text, by the string [NEXLINE], preceded and followed with a space character
- Deletes the # character, possibly preceded by a @ character, when located at the end of the lines

All these S/R will use the Regular expression search mode, the Wrap around option and the Replace All button, of the Replace dialog

So, let’s start with the original text, below :

One [NEWLINE] two [NEWLINE] three [NEWLINE] four [NEWLINE] five [NEWLINE] six [NEWLINE] seven [NEWLINE] eight [NEWLINE] nine [NEWLINE] ten [NEWLINE] eleven [NEWLINE] twelve
One [NEWLINE] two [NEWLINE] three [NEWLINE] four [NEWLINE] five [NEWLINE] six [NEWLINE] seven [NEWLINE] eight [NEWLINE] nine [NEWLINE] ten [NEWLINE] eleven
One [NEWLINE] two [NEWLINE] three [NEWLINE] four [NEWLINE] five [NEWLINE] six [NEWLINE] seven [NEWLINE] eight [NEWLINE] nine [NEWLINE] ten
Other text NOT concerned
by this Search Replacement
One [NEWLINE] two [NEWLINE] three [NEWLINE] four [NEWLINE] five [NEWLINE] six [NEWLINE] seven [NEWLINE] eight [NEWLINE] nine
One [NEWLINE] two [NEWLINE] three [NEWLINE] four [NEWLINE] five [NEWLINE] six [NEWLINE] seven [NEWLINE] eight
One [NEWLINE] two [NEWLINE] three [NEWLINE] four [NEWLINE] five [NEWLINE] six [NEWLINE] seven
Bla bla blah
Bla bla blah
Bla bla blah
One [NEWLINE] two [NEWLINE] three [NEWLINE] four [NEWLINE] five [NEWLINE] six
One [NEWLINE] two [NEWLINE] three [NEWLINE] four [NEWLINE] five
One [NEWLINE] two [NEWLINE] three [NEWLINE] four
Dummy text
inserted, in between !
One [NEWLINE] two [NEWLINE] three
One [NEWLINE] two
One
I want [NEWLINE] this [NEWLINE] to be  [NEWLINE] last

After running the following S/R , once :

SEARCH (?-s)\x20?[NEWLINE]\x20?|(?<=.)$

REPLACE #

You should get the text, below :

One#two#three#four#five#six#seven#eight#nine#ten#eleven#twelve#
One#two#three#four#five#six#seven#eight#nine#ten#eleven#
One#two#three#four#five#six#seven#eight#nine#ten#
Other text NOT concerned#
by this Search Replacement#
One#two#three#four#five#six#seven#eight#nine#
One#two#three#four#five#six#seven#eight#
One#two#three#four#five#six#seven#
Bla bla blah#
Bla bla blah#
Bla bla blah#
One#two#three#four#five#six#
One#two#three#four#five#
One#two#three#four#
Dummy text#
inserted, in between !#
One#two#three#
One#two#
One#
I want#this#to be #last#

After running the following S/R, SEVEN times, one after another :

SEARCH (?-s)([^@#\r\n]+?)#(.+#)?([^@#\r\n]+)#

REPLACE \3@#\2\1@#

The modified text is, now :

twelve@#eleven@#ten@#nine@#eight@#seven@#six@#five@#four@#three@#two@#One@#
eleven@#ten@#nine@#eight@#seven@#six#five@#four@#three@#two@#One@#
ten@#nine@#eight@#seven@#six@#five@#four@#three@#two@#One@#
Other text NOT concerned#
by this Search Replacement#
nine@#eight@#seven@#six@#five#four@#three@#two@#One@#
eight@#seven@#six@#five@#four@#three@#two@#One@#
seven@#six@#five@#four#three@#two@#One@#
Bla bla blah#
Bla bla blah#
Bla bla blah#
six@#five@#four@#three@#two@#One@#
five@#four@#three#two@#One@#
four@#three@#two@#One@#
Dummy text#
inserted, in between !#
three@#two#One@#
two@#One@#
One#
last@#to be @#this@#I want@#

Seven consecutive runs of that regex S/R are required, to get the sought text :

Run 1 : 12 occurrences replaced
Run 2 : 10 occurrences replaced
Run 3 : 7 occurrences replaced
Run 4 : 5 occurrences replaced
Run 5 : 3 occurrences replaced
Run 6 : 1 occurrences replaced
Run 7 : 0 occurrences replaced

Note : After each run, you may hit the Find Next button, before hitting the Replace All button, to guess the general process !

The part [^@#\r\n], in the searched regex, represents any single character, different from @, #, \n and \r

Then, after running the last S/R, once :

SEARCH (?-s)(@?#)(?=.)|@?#

REPLACE ?1\x20[NEWLINE]\x20

We obtain our final text :

twelve [NEWLINE] eleven [NEWLINE] ten [NEWLINE] nine [NEWLINE] eight [NEWLINE] seven [NEWLINE] six [NEWLINE] five [NEWLINE] four [NEWLINE] three [NEWLINE] two [NEWLINE] One
eleven [NEWLINE] ten [NEWLINE] nine [NEWLINE] eight [NEWLINE] seven [NEWLINE] six [NEWLINE] five [NEWLINE] four [NEWLINE] three [NEWLINE] two [NEWLINE] One
ten [NEWLINE] nine [NEWLINE] eight [NEWLINE] seven [NEWLINE] six [NEWLINE] five [NEWLINE] four [NEWLINE] three [NEWLINE] two [NEWLINE] One
Other text NOT concerned
by this Search Replacement
nine [NEWLINE] eight [NEWLINE] seven [NEWLINE] six [NEWLINE] five [NEWLINE] four [NEWLINE] three [NEWLINE] two [NEWLINE] One
eight [NEWLINE] seven [NEWLINE] six [NEWLINE] five [NEWLINE] four [NEWLINE] three [NEWLINE] two [NEWLINE] One
seven [NEWLINE] six [NEWLINE] five [NEWLINE] four [NEWLINE] three [NEWLINE] two [NEWLINE] One
Bla bla blah
Bla bla blah
Bla bla blah
six [NEWLINE] five [NEWLINE] four [NEWLINE] three [NEWLINE] two [NEWLINE] One
five [NEWLINE] four [NEWLINE] three [NEWLINE] two [NEWLINE] One
four [NEWLINE] three [NEWLINE] two [NEWLINE] One
Dummy text
inserted, in between !
three [NEWLINE] two [NEWLINE] One
two [NEWLINE] One
One
last [NEWLINE] to be  [NEWLINE] this [NEWLINE] I want

The search part looks for the regex @?#, either, inside the lines ( case group 1 defined ) or at end of lines ( case NO group 1 )

The replacement part means that, IF group 1 exists, the searched text is replaced by the string [NEWLINE], surrounded by space characters, ELSE NO replacement occurs

Et voilà !

Best Regards,

guy038

abuali huma

Thanks very much!
But just to be clear, in the first regex

SEARCH (?-s)\x20?[NEWLINE]\x20?|(?<=.)$

REPLACE #

Removing the value ** \x20 ** will result this
SEARCH (?-s)?[NEWLINE]?|(?<=.)$

Which will result capturing the space “if available” before and after [NEWLINE] string in first and last group?

abuali huma

I found out the removing the \x20 does what I described… Thanks again

abuali huma

I modified the original Search regex, as it catches some Unicode characters with will break the line in a middle of a word. So in the modified regex I replace \W with \x20 (space character)… so far no word breaking issues
Here is the modified one
(?-s).{1,44}(?=\x20)