# function find() sometimes doesn't work properly

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Pavel M
on 18 Jul 2024 at 19:32

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?

##### 1 Comment

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)

f(31494)

The same value? Nope, different values.

So far everything seems to be working exactly as documented and expected.

### Accepted Answer

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

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)

.

##### 0 Comments

### More Answers (1)

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)

f(31494)

##### 1 Comment

Walter Roberson
on 18 Jul 2024 at 19:55

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