how to set 1000 separator for large numbers in GUI MATLAB

Hello! I am trying to find a way to set 1000 separator for the large numbers displayed in my GUI. For example: I would like the number 100000 to be displayed in my GUI like 100.000 or 100,000.
How could I do it? Thank you, in advance!

2 commentaires

Hello! I wonder if this is what Natalia was asking, and whether she got her answer. What I would like to do is what I think Natalia is asking: how to get Matlab to display "normal" numbers with 1000 separators automatically, not in our own code output. Like doing "format bank" but adding the ',' separator to the thousands (as any banking application would do, which is the goal). Anyway, that is what I am looking for. Thanks for helping!
There is no way to get MATLAB to display numbers with 1000 separators automatically. You need to convert the numbers to strings and post-process the numbers, or you need to use Java techniques.

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Réponses (5)

Without Java:
S = sprintf('%.16g', pi * 1e12); % [EDITED: was '%16']
T(1:length(S)) = char(0);
T(strfind(S, '.') - 4:-3:1) = char(39); % [EDITED: -3 -> -4]
S = [S; T];
S = reshape(S(S ~= 0), 1, []);
>> 3'141'592'653'589.793

4 commentaires

I get:
>> S = sprintf('%16g', pi * 1e12)
S =
3.14159e+012
Also, 31'....'89.793 doesn't look correct.
Jan
Jan le 24 Mai 2011
@Oleg: Thanks. Fixed typos.
adams13
adams13 le 3 Mar 2023
Modifié(e) : adams13 le 8 Mar 2023
There is a small bug in this solution. For the integer value, it will not add any thousend separators.
strfind(S, '.')
will produce an empty array.
One can replace the line by
dotPos = strfind(S, '.');
if isempty(dotPos)
dotPos = length(S) + 1;
end
T(dotPos - 4:-3:1) = char(39); % [EDITED: -3 -> -4]
or even better
T(find([S '.'] == '.', 1) - 4:-3:1) = char(39); % [EDITED: -3 -> -4, strfind -> find([... '.'],1)]
Otherwise the solution is charming ;-)
An updated version:
function S = NumWithSep(N, FSpec, Sep)
% N: Number to display
% FSpec: Format specifier for SPRINTF
% SEP: Char used as separator
S = sprintf(FSpec, N);
Fin = strfind(S, '.');
if isempty(Fin)
Fin = length(S) + 1;
end
Ini = find(isstrprop(S, 'digit'), 1);
S(2, Fin - 4:-3:Ini) = Sep;
S = S(S ~= char(0)).';
end

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Stephen23
Stephen23 le 19 Mar 2026
Modifié(e) : Stephen23 le 19 Mar 2026
No FLIPLR required:
F = @(t) regexprep(t,'(?<!(\.|[eE][-+]?)\d*)\d{1,3}(?=(\d{3})+(e|E|\.|\>))', '$&,');
And tested:
F('1 12 123 1234 12345 123456 1234567 12345678 123456789') % integer
ans = '1 12 123 1,234 12,345 123,456 1,234,567 12,345,678 123,456,789'
F('0.123456789 123456789.123456789') % decimal point
ans = '0.123456789 123,456,789.123456789'
F("123456789 123456789.123456789") % string
ans = "123,456,789 123,456,789.123456789"
F('0 NaN Inf') % special cases
ans = '0 NaN Inf'
F('-1 +12 -123 +1234 -12345 +123456 -1234567 +12345678 -123456789') % sign
ans = '-1 +12 -123 +1,234 -12,345 +123,456 -1,234,567 +12,345,678 -123,456,789'
F('1e12345 123456789e-12345 123456789.123456789e-12345') % e-notation
ans = '1e12345 123,456,789e-12345 123,456,789.123456789e-12345'
Another alternative w/o java:
n = pi * 1e12;
c = fix(log10(n)+1);
dec = 3;
fmt = [repmat('%c',1,mod(c,dec)) repmat('''%c%c%c',1,fix(c/dec)) '%s'];
sprintf(fmt, sprintf('%.3f',n))
ans =
3'141'592'653'589.793
Another alternative.
H = '123456.09'; % A string number.
S = regexp(H,'\.','split');
S{1} = fliplr(regexprep(fliplr(S{1}),'\d{3}(?=\d)', '$0,'));
H2 = [S{1},'.',S{2}]
Note that if H will never have a decimal amount, this becomes a simple one-liner. In this case,
H = '123456'; % A string number with no decimal point.
H2 = fliplr(regexprep(fliplr(H),'\d{3}(?=\d)', '$0,'))

3 commentaires

One liner:
fliplr(regexprep(fliplr(H),'(\d+\.)?\d{3}(?=\d)', '$0,'))
Nice Oleg, I looked for that but couldn't find it. Great work!
That lazy quantifier is a miracle worker, I need to spend more time with it.
Dominique
Dominique le 29 Août 2023
Modifié(e) : Dominique le 29 Août 2023
It seems to me that this method works ideally when H is char array.
If the numericals were stored in a string array, then mind converting string to char .
example: table height is 62500
H = string(height(my_table)); % string var
H2 = fliplr(regexprep(fliplr(H),'\d{3}(?=\d)', '$0,')) % returns: 625,00 -> misleading
H = char(H) % char var
H2 = fliplr(regexprep(fliplr(H),'\d{3}(?=\d)', '$0,')) % returns: 62,500 -> expected
H2 = fliplr(regexprep(fliplr(H),'\d{3}(?=\d)', '$0 ')) % returns: 62 500 -> blank space separator
% ! Note:
H = char(height(my_table)); % may lead you to trouble

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le 19 Mar 2026

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