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 件のコメント
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
2013 年 2 月 8 日
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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その他の回答 (2 件)
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:
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 件のコメント
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,",")
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