Is there a format in MATLAB to display numbers such that commas are automatically inserted into the display?

I would like to display large numbers with commas separating the digits for readability purposes. For example, I would like MATLAB to display one billion as 1,000,000,000 rather than 1000000000. How can I do this?

 Accepted Answer

You can enhance the readability of large numbers by formatting them with commas to separate digits using the "java.text.DecimalFormat" class. See the below code for an example of this:
function numOut = addComma(numIn)
import java.text.*
jf=java.text.DecimalFormat; % comma for thousands, three decimal places
numOut= char(jf.format(numIn)); % omit "char" if you want a string out
end
Then, calling the function will result in a comma separated number.
addComma(10000.12)
% 10,000.12

More Answers (2)

A simple way is to add this two line function:
function numOut = addComma(numIn)
jf=java.text.DecimalFormat; % comma for thousands, three decimal places
numOut= char(jf.format(numIn)); % omit "char" if you want a string out
end
Hope that helps! --ted
Using pattern
vec = 123456789;
txt = string(vec);
pat1 = lookBehindBoundary(digitsPattern); % (?<=\d)
pat2 = asManyOfPattern(digitsPattern(3),1); % (\d{3})+
pat3 = lookAheadBoundary(pat2+lineBoundary("end")); % (?=(\d{3})+$)
pat4 = pat1+pat3; % (?<=\d)(?=(\d{3})+$)
replace(txt,pat4,",")
ans = "123,456,789"

3 Comments

vec = 1234.56789;
txt = string(vec);
pat1 = lookBehindBoundary(digitsPattern); % (?<=\d)
pat2 = asManyOfPattern(digitsPattern(3),1); % (\d{3})+
pat3 = lookAheadBoundary(pat2+lineBoundary("end")); % (?=(\d{3})+$)
pat4 = pat1+pat3; % (?<=\d)(?=(\d{3})+$)
replace(txt,pat4,",")
ans = "1234.5,679"
How about this?
str = arrayfun(@(x) string(num2str(x)), (0:500:2000)');
str2 = [str;str + ".12345"]
str2 = 10×1 string array
"0" "500" "1000" "1500" "2000" "0.12345" "500.12345" "1000.12345" "1500.12345" "2000.12345"
pat1 = lookBehindBoundary(digitsPattern); % (?<=\d)
pat2 = asManyOfPattern(digitsPattern(3),1); % (\d{3})+
pat3 = lookAheadBoundary(pat2+lineBoundary("end")); % (?=(\d{3})+$)
pat4 = pat1+pat3; % (?<=\d)(?=(\d{3})+$)
isDecimal = contains(str2,".");
str3 = split(str2(isDecimal),".");
str2(isDecimal) = str3(:,1);
str4 = replace(str2,pat4,",");
str4(isDecimal) = str4(isDecimal) + "." + str3(:,2)
str4 = 10×1 string array
"0" "500" "1,000" "1,500" "2,000" "0.12345" "500.12345" "1,000.12345" "1,500.12345" "2,000.12345"
That seems to work much better. Note that this
str0 = arrayfun(@(x) string(num2str(x)), (0:500:2000)');
can be replaced with the much simpler and more efficient:
str1 = string(0:500:2000).';
isequal(str0,str1)
ans = logical
1

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