Number format

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Poul Reitzel
Poul Reitzel on 10 Apr 2012
Dear Matlab Community
I am facing a problem to convert a row of numbers to a string in a specific format. The problem is best illustrated by an example:
I convert the following vector
a = [-0.0127 -0.1711 -0.0327 -0.0015 0.0035 -0.0186]
to a string using num2str(a). However, I want the numbers in engineering format, i.e. written as a product of a number and an exponent. I achieve this by writing
num2str(a,'%6.3E'), producing:
-1.266E-002 -1.711E-001 -3.274E-002 -1.492E-003 3.493E-003 -1.864E-002
But I want the format to be like this:
-0.1266E-003 -0.1711E-002 -0.3274E-003 -0.1492E-004 0.3493E-004 -0.1864E-003
Is there a way to do this?
Thanks in advance! Poul

Answers (2)

Thomas
Thomas on 10 Apr 2012
a = [-0.0127 -0.1711 -0.0327 -0.0015 0.0035 -0.0186]
format shortE
a
a =
-1.2700e-02 -1.7110e-01 -3.2700e-02 -1.5000e-03 3.5000e-03 -1.8600e-02
Try it: this works
a = [-0.0127 -0.1711 -0.0327 -0.0015 0.0035 -0.0186]
for i=1:length(arg)
sgn(i) = sign(arg(i));
expnt(i) = fix(log10(abs(arg(i))));
mant(i) = sgn(i) * 10^(log10(abs(arg(i)))-expnt(i));
if abs(mant(i)) < 1
mant(i) = mant(i);
expnt(i) = expnt(i)-2;
end
end
d=[mant;expnt];
out=sprintf('%fe%+04d\n',d)
Ans
out =
-0.127000e-003
-0.171100e-002
-0.327000e-003
-0.150000e-004
0.350000e-004
-0.186000e-003
  1 Comment
Poul Reitzel
Poul Reitzel on 10 Apr 2012
The deal is I want the class to be 'char', while having the numbers written with a zero followed by a dot followed by the digits and an exponent, e.g.
-0.1266E-003 -0.1711E-002 -0.3274E-003 -0.1492E-004 0.3493E-004 -0.1864E-003

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Jan
Jan on 10 Apr 2012
I suggest to search in the forum. Then you find: Answers: engineering-notation
function Str = EngineersStyle(x)
Exponent = 3 * floor(log10(x) / 3);
y = x ./ (10 .^ Exponent);
D = [y(:), Exponent(:)].';
Str = sprintf('%fe%+04d ', D);
Str(end) = [];
end
I cannot test this currently.
  2 Comments
Poul Reitzel
Poul Reitzel on 10 Apr 2012
I will have a look at it tomorrow at work, thank you!
Poul Reitzel
Poul Reitzel on 11 Apr 2012
It is not quite (per my question) what I was looking for, but I will take it from here!

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