Inserting a 1000 separator
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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 个评论
Miroslav Balda
2013-2-8
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
2013-2-8
Image Analyst
2013-2-8
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?
采纳的回答
Image Analyst
2013-2-8
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 个评论
Ursel Thomßen
2020-9-15
编辑：Ursel Thomßen
2020-9-15
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
2020-9-15
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);
更多回答（4 个）
Jan
2013-2-8
0 个投票
Please take the time to search in the forum at first before posting a new question:
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,",")
1 个评论
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,",")
Joe
2026-3-13，4:16
For what it's worth, I know this is an older thread, but for anyone who comes across this down the road, I have a simple function which easily solved this guy's issue shortening the decimal which I integrated into Image Analyst's CommaFormat function:
function n=dec(m,d)
format long g % allows for longer decimal point arithmatic, g - cuts out trailing 0's
m=m*10^d; % shift decimal over to the right
n=round(m)/10^d; % round off extra decimal places, and shift decimal back left
endfunction
From this, all I added was a line before the integerPart and decimalParts are separated:
value = dec(value,2); % new line, value has 2 remaining decimal digits
[integerPart, decimalPart]=strtok(num2str(value),'.'); % previously line 5
Hope this helps. It helped me.
1 个评论
Walter Roberson
2026-3-13，5:00
"format long g" only affects the display of data using disp() or display() or by having an expression without a semi-colon. It also affects the display of table() variables and struct() . It never affects any computation.
Assuming that you have already converted to text:
A = '11201453.21';
F = @(t) regexprep(t,'(?<!(\.|[eE][-+]?)\d*)\d{1,3}(?=(\d{3})+(e|E|\.|\>))', '$&,');
B = F(A)
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