Put the separator every thousands
8 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
Luca Re
el 10 de Feb. de 2024
Comentada: Image Analyst
el 11 de Feb. de 2024
T=100000000;
T1 = regexprep(string(T),'(\d+)(\d{3})$',"$1'$2")
I would like to have the thousands separator for every 1000
the correct result would be:
1'000'000'000
2 comentarios
Stephen23
el 10 de Feb. de 2024
Editada: Stephen23
el 10 de Feb. de 2024
"the correct result would be: 1'000'000'000"
Why should the "correct result" be one billion when the input value is only one hundred million?
100000000 % your input value
1000000000 % your "correct result" with quotes removed
Your "correct result" is ten times larger than the input value: is that intentional or is it ... incorrect ?
Respuesta aceptada
Stephen23
el 10 de Feb. de 2024
Editada: Stephen23
el 10 de Feb. de 2024
Here are some more interesting testcases (with both one hundred million as well as one billion):
S = ["100000000";"1000000000";"123456789.123456789";"-123456789";"0";pi;"1e23456"]
S = regexprep(S,"(?<![eE\.]\d*)\d{1,3}(?=(\d{3})+\>)","$&'")
Más respuestas (1)
Image Analyst
el 10 de Feb. de 2024
I use the attached function I wrote. Adapt as needed, like change commas to apostrophes if you want.
2 comentarios
Stephen23
el 10 de Feb. de 2024
CommaFormat('-123456') % ouch
Image Analyst
el 11 de Feb. de 2024
@Stephen23 thanks for pointing that out. I've corrected it to properly handle cases where the input is negative or a string or a character array instead of a number (double, etc.). New code is attached.
It's definitely more lines than your one liner regexp though. However I'm not as adept with regexp as you -- I never would have figured out that cryptic sequence of regexp characters as you did.
Ver también
Categorías
Más información sobre Characters and Strings en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!