function find() sometimes doesn't work properly

Question

Pavel M 2024-7-18，19:32

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https://ww2.mathworks.cn/matlabcentral/answers/2138581-function-find-sometimes-doesn-t-work-properly

编辑： Stephen23 2024-7-20，8:40

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?

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Stephen23 2024-7-18，20:29

编辑：Stephen23 2024-7-20，8:40

在 MATLAB Online 中打开

"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.

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Answer 1

Star Strider 2024-7-18，19:53

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在 MATLAB Online 中打开

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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.

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Answer 2

DGM 2024-7-18，19:46

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2138581-function-find-sometimes-doesn-t-work-properly#answer_1487676

移动：Steven Lord 2024-7-18，19:48

在 MATLAB Online 中打开

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

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Walter Roberson 2024-7-18，19:55

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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