MATLAB Answers

Finding integers in an array.

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NS
NS on 10 Jun 2011
Hey Guys,
I have an column matrix that basically consists of NaNs and some integers in between them. Are there any functions in MATLAB that will help me find 1. the first integer value in the array, 2. its location 3. store the value
Thanks, NS

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Accepted Answer

Matt Fig
Matt Fig on 10 Jun 2011
V = [nan;nan;9;nan;6;7;nan;9] % An example to work with...
idx = find(V==V,1); % Location. nan never equals nan...
val = V(idx); % Value at Location.

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Andrei Bobrov
Andrei Bobrov on 11 Jan 2013
R = find(abs(rem(M,1)) < eps(100));
Jan
Jan on 11 Jan 2013
@Ricardo: Please do not post a question as a comment to an answer of another question. New questions need a new thread. Thanks.
@Andrei: eps(100) might be wrong, if the magnitude of the data is 1e8. Either eps(M) could be better, or:
find(M==floor(M))
Ricardo Pereira
Ricardo Pereira on 11 Jan 2013
Jan Siman: Sorry. You're right. I should have posted this question on another post.
Your answer works for me.
R = find(M==floor(M))
Thanks a lot

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More Answers (2)

Andrei Bobrov
Andrei Bobrov on 10 Jun 2011
l = isnumeric(V)&rem(V,1)==0;
ii = find(l,1,'first'); % 2
ValInt = V(ii); % 1 and 3

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Matt Fig
Matt Fig on 10 Jun 2011
BTW andrei, I think you are meaning the call to ISNUMERIC to act as a nan filter? Your code works just fine with only a call to REM.
Andrei Bobrov
Andrei Bobrov on 10 Jun 2011
Matt! You're right as always
NS
NS on 10 Jun 2011
Thanks for the help Matt and Andrei.
Andrei, your code works well. I dont know what I did wrong yesterday. I dont think I can accept two answers at the same time here. :(
Thanks for the help once again.

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Yella
Yella on 10 Jun 2011
%An example to work with x = [NaN 1.7 1.6 1.5 NaN 1.9 1.8 1.5 5.1 1.8 Inf 1.4 2.2 1.6 1.8];
for i=1:1:length(x)
if (isnan(x(i)))
continue;
else
p=i;
break;
end
end
x_withno_nan = x(isfinite(x))
x_first=x_withno_nan(1)
It may be length but i made easily understandable for a beginner like me.

  2 Comments

NS
NS on 10 Jun 2011
Thanks for the help Yella. :)
Matt Fig
Matt Fig on 10 Jun 2011
Yella, there is nothing wrong with using a FOR loop for this problem!
I would only add that your code can be simplified:
for ii = 1:numel(x)
if floor(x(ii))==x(ii) % Only integers - use x(ii)==x(ii) or ~isnan(x(ii))
idx = ii; % The index of the first non-nan
val = x(ii); % The value at idx.
break
end
end

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