# How to change the field width of an exponential notation?

9 views (last 30 days)
Oscar Espinosa on 4 May 2020
Edited: Stephen23 on 7 May 2020
Hi, Im trying to change the field width and the number of digits of the exponential, I will like to looks like enginnering notation:
3.45866636216603e+002
and what I get its:
3.45866636216603e+02
The formatSpec have the next form:
%21.14e
So, I dont know what Im doing wrong, its suppose to the total numbers before e its 21, 14 of them are located after the point. But doesnt seem to be like that.
Stephen23 on 4 May 2020
Edited: Stephen23 on 7 May 2020
"...its suppose to the total numbers before e its 21"
The fieldwidth does not specify the number of digits, it specifies the total number of printed characters, including leading spaces, leading zeros, the significant digits, the decimal radix, the exponent character, the exponent digits, and any sign characters. All of these are included in the field width. If the number representation is shorter than the fieldwidth the output string will be padded with leading spaces or leading zeros.

Stephen23 on 4 May 2020
Edited: Stephen23 on 6 May 2020
Given >=2 digits, ensure >=3 digits:
>> val = 3.45866636216603e2;
>> str = sprintf('%.14e',val)
str =
3.45866636216603e+02
>> str = regexprep(sprintf('%.14e',val),'(?<=\D)\d\d\$','0\$&')
str =
3.45866636216603e+020
Given >=1 digits, ensure >=3 digits:
>> str = regexprep(sprintf('%.14e',val),{'(?<=\D)\d\$','(?<=\D)\d\d\$'},'0\$&')
str =
3.45866636216603e+020
A general solution that can provide any number of digits:
>> num = 3; % required number of digits
>> fun = @(s)sprintf('%0*u',num,sscanf(s,'%u'));
>> str = regexprep(sprintf('%.14e',val),'\d+\$','\${fun(\$&)}')
str =
3.45866636216603e+020
Oscar Espinosa on 5 May 2020
That really work it out. Thank you Stephen.

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