# Looking for a faster way of finding the first element larger than a given number in a sorted array

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Tom van den Bosch on 3 Aug 2021
Edited: Bruno Luong on 7 Oct 2022
Hi,
I have a vector
a = [1 1 1.01 1 1.05 1 .... 1]
which is basically all numbers >=1. I also have a vector b that is the cummulative sum of this vector (and thus sorted), so each increment in this vector is at least 1 and sometimes (often) more than 1.
I also have a random number r between 0 and the total sum of a. Now, I need to find the first element k in b that is larger than r.
I am currently using a simple find comment:
k = find(b>r, 1)
However, I feel like it should be possible for this to be faster. Since I know that k should be at least floor(r) as increments in the cumsum vector are at least 1, I have tried
n = floor(r)
k = find(b(n:end)>r, 1) + n-1
But this seems to be slower despite skipping searching the first n enties of b. Is there an build-in option in matlab for starting the search at a certain index? If not, is there any other faster way? This function takes about 65% of my total computing time. a Is a vector of size 10.000 but it changes very often and I have to make this call millions of times.
##### 6 CommentsShowHide 5 older comments
Tom van den Bosch on 4 Aug 2021
Thanks for the suggestions all!
I tried sum(b<r) but this was about 30% slower.
I am unsure how the matlab C/C++ compiler works, although I do not doubt that the fastest possible method is through a C++. However, the accepted answer by dpb is sufficient for now as this step is no longer a bottleneck and only takes about 7% of total runtime.

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### Accepted Answer

dpb on 3 Aug 2021
With the JIT compiler, loops are pretty quick any more...in fact, with the starting point as the beginning loop index, that may, indeed, be the fastest option. I'd use a counted loop and break though, not the while I think, although you can compare...
for k=floor(r):numel(b)
if b(k)>r, break, end
end
##### 3 CommentsShowHide 2 older comments
dpb on 8 Aug 2021
I had a machine failure so have been offline since -- glad to hear that seems to help; I had one additional idea that might be better than the straight loop which would be a binary search starting at the known minimum location. I haven't tried coding it to test; I just got MATLAB reloaded on the new machine and have a lot more to go to get it rebuilt but I just came to check out briefly before going on...

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### More Answers (2)

Bruno Luong on 3 Aug 2021
Edited: Bruno Luong on 3 Aug 2021
You might try
k = discretize(r,b);
if b(k) == r
k = k+1;
end
##### 1 CommentShowHide None
Tom van den Bosch on 4 Aug 2021
Thanks! I tried but it seems to be about 3-4 times as slow.

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Zhunping Xue on 7 Oct 2022
If it's an ordered cumsum vector and about 10,000 entries in size, you probably can simply implement the binary search algorith (you can find it anywhere, say, https://en.wikipedia.org/wiki/Binary_search_algorithm). Matlab's find option is a linear search and does not take advantage of the fact that your vector is ordered. It should come out to be orders faster than what you're doing!
James Tursa had the best solution I can think of, a binary search with mex -- but you don't have to do it in mex.
My really, really primitive code (I'm a biologist, don't laugh):
function m=find_ordered_array(array, value)
% finds value in ordered array, else returns the closest value larger than
% the value sought for
% if(value<array(1) || value > array(end))
% disp('value outside array');
% m=-1;
% return;
% else
% l=1;
tol=1e-15;
h=length(array);
if(array(1)>=value)
m=1;
return;
elseif(array(h)<=value)
m=h;
return;
else
l=1;
h=length(array);
m=floor((l+h)/2);
while(l<h-1 && abs(array(m)-value)>tol)
if(array(m)>value)
h=m;
else
l=m;
end
m=floor((l+h)/2);
end
if(abs(array(m)-value)<tol)
elseif(abs(array(h)-value)<tol)
m=h;
elseif(abs(array(l)-value)<tol)
m=l;
else
m=l;
end
m=m+1;
end
% end
##### 1 CommentShowHide None
Bruno Luong on 7 Oct 2022
Edited: Bruno Luong on 7 Oct 2022
@Zhunping Xue discretize in my solution is binary search

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