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 件のコメント

Miroslav Balda
Miroslav Balda 2013 年 2 月 8 日
Does the initial command have a fixed structure? How you say where to put dots? The resulting expression is not MATLAB command. If it has to be, The right side shot be enclosured in braces.
joseph Frank
joseph Frank 2013 年 2 月 8 日
yes the resulting expression is not a matlab command. I want to generate A={'1,1201,453.21'} in matlab
Image Analyst
Image Analyst 2013 年 2 月 8 日
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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 採用された回答

Image Analyst
Image Analyst 2013 年 2 月 8 日

2 投票

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 件のコメント

Ursel Thomßen
Ursel Thomßen 2020 年 9 月 15 日
編集済み: Ursel Thomßen 2020 年 9 月 15 日
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 2020 年 9 月 15 日
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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その他の回答 (2 件)

Jan
Jan 2013 年 2 月 8 日

0 投票

Please take the time to search in the forum at first before posting a new question:
Toshiaki Takeuchi
Toshiaki Takeuchi 2023 年 11 月 14 日

0 投票

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 件のコメント

Stephen23
Stephen23 2023 年 11 月 14 日
編集済み: Stephen23 2023 年 11 月 14 日
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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