Inserting a 1000 separator

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joseph Frank
joseph Frank am 8 Feb. 2013
Bearbeitet: Stephen23 am 14 Nov. 2023
Hi, I am building a table in which I need to insert numbers with a comman 1000 separator and two decimal points. For example: A=11201453.21 % should be A=1,1201,453.21 Any hint about how to do it? Best,
  3 Kommentare
joseph Frank
joseph Frank am 8 Feb. 2013
yes the resulting expression is not a matlab command. I want to generate A={'1,1201,453.21'} in matlab
Image Analyst
Image Analyst am 8 Feb. 2013
My code gives close to that:
A=11201453.21
stringVersionOfA = CommaFormat(A)
cellVersionOfString = {stringVersionOfA}
In the command window:
A =
11201453.21
stringVersionOfA =
11,201,453.21
cellVersionOfString =
'11,201,453.21'
Can you explain why your second group has 4 numbers (1201) instead of 3? And is that the reason why the code I posted earlier does not meet your requirements?

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Image Analyst
Image Analyst am 8 Feb. 2013
Here's a function I've used:
%=====================================================================
% Takes a number and inserts commas for the thousands separators.
function [commaFormattedString] = CommaFormat(value)
% Split into integer part and fractional part.
[integerPart, decimalPart]=strtok(num2str(value),'.');
% Reverse the integer-part string.
integerPart=integerPart(end:-1:1);
% Insert commas every third entry.
integerPart=[sscanf(integerPart,'%c',[3,inf])' ...
repmat(',',ceil(length(integerPart)/3),1)]';
integerPart=integerPart(:)';
% Strip off any trailing commas.
integerPart=deblank(integerPart(1:(end-1)));
% Piece the integer part and fractional part back together again.
commaFormattedString = [integerPart(end:-1:1) decimalPart];
return; % CommaFormat
  2 Kommentare
Ursel Thomßen
Ursel Thomßen am 15 Sep. 2020
Bearbeitet: Ursel Thomßen am 15 Sep. 2020
This works perfectly! I even exchanged perdiod and comma and can get German numberformat :-)
... As long as I enter numbers for value. Can anybody tell me, what do I need to change if I want to enter variables (1x1) for value?
Thank you in advance!
Image Analyst
Image Analyst am 15 Sep. 2020
Not sure what you want to enter. It can already take constants
[commaFormattedString] = CommaFormat(1234567)
or variables
value = 12345678;
[commaFormattedString] = CommaFormat(value)
Do you want to enter a string? If so, you have to convert it to a numerical value first.
value = str2double(str);

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Weitere Antworten (2)

Jan
Jan am 8 Feb. 2013
Please take the time to search in the forum at first before posting a new question:

Toshiaki Takeuchi
Toshiaki Takeuchi am 14 Nov. 2023
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"
  1 Kommentar
Stephen23
Stephen23 am 14 Nov. 2023
Bearbeitet: Stephen23 am 14 Nov. 2023
Using the OP's example value:
vec = 11201453.21;
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 = "11201453.21"

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