how to set 1000 separator for large numbers in GUI MATLAB

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Natalia P.
Natalia P. el 24 de Mayo de 2011
Editada: Dominique el 29 de Ag. de 2023
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!
  1 comentario
Juan Saenz-Diez
Juan Saenz-Diez el 18 de Mayo de 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!

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Respuestas (4)

Teja Muppirala
Teja Muppirala el 24 de Mayo de 2011
import java.text.*
v = DecimalFormat;
in = 123456789;
out = char(v.format(in))
Taken from here:

Jan
Jan el 24 de Mayo de 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
  3 comentarios
Jan
Jan el 24 de Mayo de 2011
@Oleg: Thanks. Fixed typos.
adams13
adams13 el 3 de Mzo. de 2023
Editada: adams13 el 8 de Mzo. de 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 ;-)

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Oleg Komarov
Oleg Komarov el 24 de Mayo de 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
Matt Fig el 24 de Mayo de 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 comentarios
Matt Fig
Matt Fig el 25 de Mayo de 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.
Dominique
Dominique el 29 de Ag. de 2023
Editada: Dominique el 29 de Ag. de 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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