Just a bit late, but here are two regexes S/R which could achieve the goal !

Note that, regarding the initial timestamps, I will use the convention [M]MM:SS.mmm, where :

• [M]MM represents the number of minutes, from 00 to 119/179, with two or three digits

• SS represents the number of seconds, from 00 to 59, with two digits

• mmm represents the number of milliseconds, from 000 to 999, with three digits

Case A) If your file contains timestamps syntaxes, from 00:00.000 to 119:59.999, only ( so 0 <[M]MM < 2 hours ) , one solution could be :

• SEARCH A   (?<!:)(?:([0-5])|(6)|(7)|(8)|(9)|(10)|(11))(\d:\d{2}\.\d{3})(?=\s)

• REPLACE A (?{1}00:01):(?1\1)(?{2}0)(?{3}1)(?{4}2)(?{5}3)(?{6}4)(?{7}5)\8

Case B) If your file contains timestamps syntaxes, from 00:00.000 to 179:59.999, only ( so 0 < [M]MM < 3 hours ), a longer S/R is :

• SEARCH B   (?<!:)(?:([0-5])|((6)|(7)|(8)|(9)|(10)|(11))|((12)|(13)|(14)|(15)|(16)|(17)))(\d:\d{2}\.\d{3})(?=\s)

• REPLACE B (?{1}00)(?{2}01)(?{9}02):(?1\1)(?{3}0)(?{4}1)(?{5}2)(?{6}3)(?{7}4)(?{8}5)(?{10}0)(?{11}1)(?{12}2)(?{13}3)(?{14}4)(?{15}5)$16

As usual :

• Check the Wrap around option

• Select the Regular expression search mode

• Click on the Replace All button

Best Regards

guy038

P. S.

For instance :

• With the regexes A, the initial text, below :

00:00.000
23:52.984
39:43.529
59:59.999
60:00.000
78:08.168
91:38.524
103:05.216
111:41.465
119:59.999

becomes :

00:00:00.000
00:23:52.984
00:39:43.529
00:59:59.999
01:00:00.000
01:18:08.168
01:31:38.524
01:43:05.216
01:51:41.465
01:59:59.999

• With the regexes B, the following text :

00:00.000
23:52.984
39:43.529
59:59.999
60:00.000
78:08.168
91:38.524
103:05.216
111:41.465
119:59.999
120:00.000
147:33.150
160:00.058
179:59.999

becomes :

00:00:00.000
00:23:52.984
00:39:43.529
00:59:59.999
01:00:00.000
01:18:08.168
01:31:38.524
01:43:05.216
01:51:41.465
01:59:59.999
02:00:00.000
02:27:33.150
02:40:00.058
02:59:59.999

As valuable as Scott’s (and Claudia’s) posts were, we have some really good new posters about scripting (example Eko, and Peter is developing as a Python person), so let’s not be too sad if they decide not to return.

>>> here’s a song <<< for @Scott-Sumner 😪😉😂

sometimes (but very few) we tend to have a little “after work chat” between regulars in public, which can be a bit off topic from time to time. i hope you didn’t mind.

I don’t understand all of this but what I got makes me laughing … :-D

singing: ahaaaa, ahahahaaa … all the things i could do … ;-)

:-D - always reminds me of this

I’m trying to avoid function lookups as those are expensive …

yes, i’m a bit short on money too at the moment … and don’t even dare to give me an (s.h) for this comment 😉

Indeed, there are always more formats that might exist. I’ve only seen colon-separated in .srt files, so I think that keeping it generic enough that it won’t mess up an existing .srt, even if it does have some with hours and some without.

BTW: I had forgotten why I included the [:\d] rather than just : in my negative lookbehind: without the \d in the character class, 1:15:00.000 (which shouldn’t match) would partially match on 5:00.000, which would be even worse.

And running a test with 1:15:00.000, even with your simpler expression, works correctly (ie, doesn’t try to change it) – ahh, that’s because the minutes are less than 60. I guess unless there’s a strange 1:65:00.000, yours won’t be a problem. I guess yours is generic enough.

YES - this is a bug it should > 59 - omg.
About mixed data you are right but this is always the question what if it looked like
hh.mm.ss.msec …

1. if min > 60: = if the data is 60:00.000, it wouldn’t change. Make it if min >=60:
2. If the OP (or someone else) has mixed data, or had partially changed them, and came back later and tried the same script, weird stuff will happen. I’d recommend: editor.rereplace('(?<![:\d])\d+:\d+\.\d+',change_format), which adds a negative lookbehind to not match if there’s a colon or another digit before the \d+

so it is still a little bit faster - to be honest, haven’t expected it.

looped 1000 times over the same text
16.7720000744 <-- return m.group(0)
16.6819999218 <- return ‘{}:{}’.format(*parts[:])

Okay…so I didn’t follow any of that, but I look forward to the come back. :)

actually I’m trying to avoid function lookups as those are expensive, especially when it involves
Python->C->Python conversion. But I must admit, in this case I don’t think that I gain any performance improvement, it might be even slower. Let’s test it. Will come back.

Maybe I misunderstood the intent of the return line I called out. I was thinking that it is simply putting back together the original text but I didn’t look at the OP’s data or problem description all that closely. I guess you would have returned m.group(0) if that were the case.

Anyway, a replace function could have some logic that in certain cases it would not change the original text. Returning None (or even, gasp, falling off the end of the function without returning anything) could be that signal. I certainly did not try editor.rereplace() in that manner, maybe it already works that way.

I don’t understand the none return - I mean, why would you want to replace something with None
where None means don’t replace anything? Which case do you have in mind?

That’s it.

Well…you didn’t explain the else part, and that’s where I have a question. I would do it this way and I don’t know why you did it differently:

return '{}:{}'.format(*parts)

I just had the thought that it would be nice if editor.rereplace() would take a replace function returning None as a signal to not do any replacement.

this is quite easy.
editor is a object exposed by pythonscript plugin which can manipulate the scintilla component used by notepad++.
rereplace is the method which allows to have the first parameter being a regular expression and
the second parameter being a function which is called for each match of the regex.

The function change_format gets the match object in variable m
As no regular expression matching group has been defined everything should be accessible in group 0.
parts = m.group(0).split(':')
split what has been reported in match object by colon and return a list in variable parts
min = int(parts[0])
take the first element in that list and convert it to an integer, save it in variable min
if min > 60:
if min greater than 60 we need to do something, if not else branch gets executed
real_minute = min % 60
real_minute will get the remainder of division by 60
hours = min / 60
hours the result of division by 60
return '{:02}:{:02}:{}'.format(hours, real_minute, parts[1])
return the new string.
:02 means that we want to have it in two digit format 1->01 but 10->10 …
That’s it.

Now I need to analyze this and figure it out how it was done.

I guess this python script should do the job

def change_format(m):
    parts = m.group(0).split(':')
    min = int(parts[0])
    if min > 60:
        real_minute = min % 60
        hours = min / 60
        return '{:02}:{:02}:{}'.format(hours, real_minute, parts[1])
    else:
        return '{}:{}'.format(*parts[:])
    
    
editor.rereplace('\d+:\d+\.\d+',change_format)