how to set 1000 separator for large numbers in GUI MATLAB
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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 Commenti
Juan Saenz-Diez
il 18 Mag 2020
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!
Walter Roberson
il 19 Mar 2026
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.
Risposte (5)
Teja Muppirala
il 24 Mag 2011
import java.text.*
v = DecimalFormat;
in = 123456789;
out = char(v.format(in))
Taken from here:
Jan
il 24 Mag 2011
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 Commenti
Oleg Komarov
il 24 Mag 2011
I get:
>> S = sprintf('%16g', pi * 1e12)
S =
3.14159e+012
Also, 31'....'89.793 doesn't look correct.
Jan
il 24 Mag 2011
@Oleg: Thanks. Fixed typos.
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 ;-)
Jan
il 18 Mar 2026
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
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
F('0.123456789 123456789.123456789') % decimal point
F("123456789 123456789.123456789") % string
F('0 NaN Inf') % special cases
F('-1 +12 -123 +1234 -12345 +123456 -1234567 +12345678 -123456789') % sign
F('1e12345 123456789e-12345 123456789.123456789e-12345') % e-notation
Oleg Komarov
il 24 Mag 2011
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
Matt Fig
il 24 Mag 2011
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 Commenti
Oleg Komarov
il 24 Mag 2011
One liner:
fliplr(regexprep(fliplr(H),'(\d+\.)?\d{3}(?=\d)', '$0,'))
Matt Fig
il 25 Mag 2011
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.
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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