How to change/convert the format of a timestamp?

Alan Kilborn

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

Eko palypse

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[:])

PeterJones

My two comments would be:

if min > 60: = if the data is 60:00.000, it wouldn’t change. Make it if min >=60:
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+

Eko palypse

@PeterJones

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 …

PeterJones

@Eko-palypse ,

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.

Meta Chuh

@Eko-palypse

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 😉

Eko palypse

@Meta-Chuh

:-D - always reminds me of this

Meta Chuh

@Eko-palypse

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

Eko palypse

@Meta-Chuh

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

Meta Chuh

@Eko-palypse
i also didn’t understand many of weird al yankovic’s insider jokes, but he made a lot of 80’s songs parodies, a funny one was “fat”, a parody of michael jacksons “bad” … or at least it used to be funny to me when i was a kid ;-)

Meta Chuh

btw: my apologies to you @Dana-Wright if you had to read everything after your “Worked like a charm! Thank you very much!” and eko’s explanation.

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.

Meta Chuh

one more song and then it’s enough for today:

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

Alan Kilborn

@Meta-Chuh

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.

guy038

Hello, @dana-wright, @eko-palypse, @alan-kilborn, @meta-chuh, @peterjones and All,

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 :

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 :

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

Meta Chuh

@guy038
it’s never too late, if people care … and thankfully many do 😃
thumbs up 👍

PeterJones

Thank you, @guy038. I had been trying the notation similar to (?(1)00:01) in the replace, rather than (?{1}00:01), which is why I wasn’t able to get the conditional to work right.