# function find() sometimes doesn't work properly

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Pavel M on 18 Jul 2024 at 19:32
Edited: Stephen23 on 20 Jul 2024 at 8:40
hello! i have a simple part of code, but in some cases function find doesnt work
f = [10 : 0.001 : 60];
frez = [48.7234 48.4347 46.4930 46.7682 44.9716 45.9232 48.2044 47.7394 55.0087 49.9675];
for i = 1 :length(frez)
x1 = find(f == round(frez(i)-5, 3));
x2 = find(f == round(frez(i)+5, 3));
Sd{i} = {x1 x2};
end
Sd{3}, Sd{8}, Sd{9} have 1 empty value! Why?
round(frez(3)-5, 3)
ans =
41.4930
>> find(f==ans+0.001)
ans =
31495
>> f(31494)
ans =
41.4930
But that code work!
Why 'find' doesnt find index even though it is?
Stephen23 on 18 Jul 2024 at 20:29
Edited: Stephen23 on 20 Jul 2024 at 8:40
"function find() sometimes doesn't work properly"
What is more likely is some binary floating point numbers have different values.
"Why 'find' doesnt find index even though it is?"
Because it isn't:
f = 10:0.001:60; % got rid of the superfluous square brackets
frez = [48.7234 48.4347 46.4930 46.7682 44.9716 45.9232 48.2044 47.7394 55.0087 49.9675];
for i = 1 :length(frez)
x1 = find(f == round(frez(i)-5, 3));
x2 = find(f == round(frez(i)+5, 3));
Sd{i} = {x1 x2};
end
format hex
round(frez(3)-5, 3)
ans =
4044bf1a9fbe76c9
f(31494)
ans =
4044bf1a9fbe76c8
The same value? Nope, different values.
So far everything seems to be working exactly as documented and expected.

Star Strider on 18 Jul 2024 at 19:53
With Floating-Point Numbers you need to use a tolerance, so with find, usually one of the approaches in tthe second loop will work —
f = [10 : 0.001 : 60];
frez = [48.7234 48.4347 46.4930 46.7682 44.9716 45.9232 48.2044 47.7394 55.0087 49.9675];
for i = 1 :length(frez)
x1 = find(f == round(frez(i)-5, 3));
x2 = find(f == round(frez(i)+5, 3));
Sd{i,:} = {x1 x2};
end
for k = 1:numel(Sd)
Sdvec = Sd{k}
end
Sdvec = 1x2 cell array
{[33724]} {[43724]}
Sdvec = 1x2 cell array
{[33436]} {[43436]}
Sdvec = 1x2 cell array
{1x0 double} {[41494]}
Sdvec = 1x2 cell array
{[31769]} {[41769]}
Sdvec = 1x2 cell array
{[29973]} {[39973]}
Sdvec = 1x2 cell array
{[30924]} {[40924]}
Sdvec = 1x2 cell array
{[33205]} {[43205]}
Sdvec = 1x2 cell array
{1x0 double} {[42740]}
Sdvec = 1x2 cell array
{[40010]} {1x0 double}
Sdvec = 1x2 cell array
{[34969]} {[44969]}
f = [10 : 0.001 : 60];
frez = [48.7234 48.4347 46.4930 46.7682 44.9716 45.9232 48.2044 47.7394 55.0087 49.9675];
for i = 1 :length(frez)
x1 = find(f >= round(frez(i)-5, 3), 1, 'first');
x2 = find(f <= round(frez(i)+5, 3), 1, 'last');
Sd{i,:} = [x1 x2];
end
Out = cell2mat(Sd)
Out = 10x2
33724 43724 33436 43436 31495 41494 31769 41769 29973 39973 30924 40924 33205 43205 32740 42740 40010 50001 34969 44969
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.

DGM on 18 Jul 2024 at 19:46
Moved: Steven Lord on 18 Jul 2024 at 19:48
Floating point arithmetic has limited precision
f = [10 : 0.001 : 60];
frez = [48.7234 48.4347 46.4930 46.7682 44.9716 45.9232 48.2044 47.7394 55.0087 49.9675];
format long
round(frez(3)-5, 3)
ans =
41.493000000000002
f(31494)
ans =
41.492999999999995
Walter Roberson on 18 Jul 2024 at 19:55
IEEE Double Precision arithmetic is not able to exactly represent 0.001, as it operates in binary instead of in decimal. The reasons are the same as the reason why finite decimal is not able to exactly represent 1/3. No matter what base you use for finite calculations, there are going to be finite values that cannot be exactly represented.

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