How to extract leading non-zero digit?

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John Booker
John Booker 2011-1-4
评论： Asif Newaz 2019-11-22
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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采纳的回答

Walter Roberson
Walter Roberson 2011-1-20
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))));
  2 个评论
Jan
Jan 2012-6-28
I suggest to accept this one.
John Booker
John Booker 2013-8-1
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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更多回答（3 个）

Ned Gulley
Ned Gulley 2011-1-11
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
Daniel Shub
Daniel Shub 2012-6-28
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.
  1 个评论
Jan
Jan 2012-6-28
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 2012-9-5
Another solution based on regexp (from this question):
regexp(num2str(x), '(?<=(^|\s+)[\-\.0]*)[1-9](?=[\d\.]*)', 'match')
  1 个评论
Asif Newaz
Asif Newaz 2019-11-22
can u explain the 'expression'... i've found regexp quite complicated but very useful

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