# Most populated range of floating point numbers in array

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lvn on 8 Sep 2014
Commented: lvn on 9 Sep 2014
histc can be used on a floating point array to find the bin with the largest number of elements. However these bins are fixed, and for a fixed width of bin, might not be optimal.
Example
a=[0 0.01 0.4 0.45 0.55 0.56 0.60]
histc(a,[0 0.5 1])
ans =
4 3 0
So the most frequent bin is [0, 0.5]. However, I am interested in a function that finds the range of at most 0.5 wide, with the most elements, so in this case [0.4, 0.6] which contains 5 elements. Does anybody know an elegant way of doing this?

Roger Stafford on 8 Sep 2014
Perhaps you won't consider this for-loop solution elegant, but it should be computationally efficient. I will assume that your array 'a' is already in ascending order, as in your example. If not, you should sort it first before using the following code:
d = 0.5; % <-- or whatever you choose
n = length(a);
i1 = 1;
for i2 = 1:n
if a(i2)-a(i1) <= d
m = i2;
else
i1 = i1+1;
end
end
The interval [a(m-n+i1),a(m)] is of width less than or equal to d and contains the maximum number of points among such intervals. If there are other such intervals with the same number of points, this is the first one encountered.
lvn on 9 Sep 2014
Fantastic! My solution was a very slow double for loop over all begin and end combinations, and for my purposes just too slow. Yours works like a charm. Thank you very much!

Honglei Chen on 8 Sep 2014
You can just do
histc(a,[0 0.4 0.6 1])
if you know for sure you want the interval [0.4 0.6]. Otherwise, you can use hist
y = hist(a,[0 0.5 1])
which specifies the center
lvn on 8 Sep 2014
Thanks for the answer, the problem is that I do not know in advance the boundaries of the interval (from one run to another, the vector attains different values). Only the maximum width of the interval is known.