function find() sometimes doesn't work properly

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]

"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.

Answer 1

[Star Strider]

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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.

Answer 2

[DGM]

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

https://matlab.fandom.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F

https://www.mathworks.com/matlabcentral/answers/97069-how-can-i-compare-numbers-for-equality-within-a-tolerance

[Walter Roberson]

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.