If we have A=[100 -0.1 0]; B=[30 0.2 -2]; t1='text 1'; t2=text 2'
how to use fprintf so that the output saved in a file will look like that
100 -1.000E-0001 0.000E-0000 'text 1'
30 2.000E-0001 -2.000E-0000 'text 2'
I put together a "one-liner" (spread across several lines for better readability) that takes an array, a single number format, and a delimiter and returns the desired string. And while you found the leading blank-space flag, I prefer the +
flag, though the function will work with both:
A=[-0.1 0];
B=[0.2 -2];
minLenExp = 4;
extsprintf = @(num,fmt,delim) ...
strjoin(cellfun(...
@(toks)[toks{1},repmat('0',1,max([0,minLenExp-length(toks{2})])),toks{2}],...
regexp(sprintf(fmt,num),'([+-\s][\.\d]+[eE][+-])(\d+)','tokens'),...
'UniformOutput',false),delim);
Astr = extsprintf(A,'%+.4E',' ');
Bstr = extsprintf(B,'%+.4E',' ');
disp([Astr;Bstr]);
Running this yields:
>> foo
-1.0000E-0001 +0.0000E+0000
+2.0000E-0001 -2.0000E+0000
( foo
is just what the script file is called.)
Here's a more general approach that searches for the exponential format instead of assuming it:
A=[100 -0.1 0].';
B=[30 0.2 -2];
extsprintf = @(fmt,arr) ...
regexprep(...
sprintf(fmt,arr),...
regexprep(regexp(sprintf(fmt,arr),'([+-\s][\.\d]+[eE][+-]\d+)','match'),'([\+\.])','\\$1'),...
cellfun(@(match)...
cellfun(...
@(toks)[toks{1},repmat('0',1,max([0,minLenExp-length(toks{2})])),toks{2}],...
regexp(match,'([+-\s][\.\d]+[eE][+-])(\d+)','tokens'),...
'UniformOutput',false),...
regexp(sprintf(fmt,arr),'([+-\s][\.\d]+[eE][+-]\d+)','match')));
fmt = '%3d %+.4E %+.4e';
disp(extsprintf(fmt,A));
disp(extsprintf(fmt,B));
Outputs
>> foo
100 -1.0000E-0001 +0.0000e+0000
30 +2.0000E-0001 -2.0000e+0000
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