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How to extract leading non-zero digit?

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I'm working on a research problem related to Benford's Law which states that the distribution of leading digits is not random. This is probably because many things grow logarithmically. I am trying to extract the leading digit from these vectors below:
10 --> 1
13 --> 1
0.3 --> 3
-4 --> 4
-5 --> 5
-0.006 --> 6
Input will be a vector
x = [1 0.3 -2 0.001 -0.0006, 582398, 3020];
Output should be
y = [1 3 2 1 6 5 3];
Any help?

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Accepted Answer

Walter Roberson
Walter Roberson on 20 Jan 2011
The answer is complicated because numbers such as 0.0006 have no exact representation in binary floating point number, and the closest number that can be represented might not have the same leading binary digit.
If one does NOT take that factor in to account, then:
y = floor(abs(x) ./ 10.^floor(log10(abs(x))));

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Jan
Jan on 28 Jun 2012
I suggest to accept this one.
John Booker
John Booker on 1 Aug 2013
Sorry it took me so long to accept. I stole this question from Ned before Answers launched and forgot about it =).

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More Answers (3)

Ned Gulley
Ned Gulley on 11 Jan 2011
You'll probably need to do some kind of textual manipulation. Here's one way to do it.
function y = leadingDigit(x)
s = sprintf('%1.2e\n',abs(x));
y = s(1:(length(s)/length(x)):end)-48;
end

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Daniel Shub
Daniel Shub on 28 Jun 2012
Is this really the first question on Answers? I was going to use our new magic power and start accepting answers. The only problem is I think there is now a better answer to this question on Loren's blog.

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Jan
Jan on 28 Jun 2012
I thought of accepting this answer, because it contains a link to good solutions. But actually the method shown by Walter hits the point (and is found in this blog also). Ned's method is more efficient than STR2DOUBLE, but a clean numerical approach is moire direct for a numerical question.

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Oleg Komarov
Oleg Komarov on 5 Sep 2012
Another solution based on regexp (from this question):
regexp(num2str(x), '(?<=(^|\s+)[\-\.0]*)[1-9](?=[\d\.]*)', 'match')

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Asif Newaz
Asif Newaz on 22 Nov 2019
can u explain the 'expression'... i've found regexp quite complicated but very useful

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