Put the separator every thousands

T=100000000;
T1 = regexprep(string(T),'(\d+)(\d{3})$',"$1'$2")
T1 = "100000'000"
I would like to have the thousands separator for every 1000
the correct result would be:
1'000'000'000

2 Commenti

Stephen23
Stephen23 il 10 Feb 2024
Modificato: Stephen23 il 10 Feb 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 ?
Yes, I counted the zeros wrong :DI to communicate 1000000000 become should 1'000'000'000

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 Risposta accettata

Stephen23
Stephen23 il 10 Feb 2024
Modificato: Stephen23 circa 20 ore fa
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 = 7×1 string array
"100000000" "1000000000" "123456789.123456789" "-123456789" "0" "3.1416" "1e23456"
S = regexprep(S,'(?<!(\.|[eE][-+]?)\d*)\d{1,3}(?=(\d{3})+(e|E|\.|\>))', '$&,')
S = 7×1 string array
"100,000,000" "1,000,000,000" "123,456,789.123456789" "-123,456,789" "0" "3.1416" "1e23456"

Più risposte (1)

Image Analyst
Image Analyst il 10 Feb 2024

0 voti

I use the attached function I wrote. Adapt as needed, like change commas to apostrophes if you want.

2 Commenti

CommaFormat('-123456') % ouch
ans = '-,123,456'
@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.

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Richiesto:

il 10 Feb 2024

Modificato:

il 19 Mar 2026 alle 15:20

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