function find() sometimes doesn't work properly
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Pavel M
le 18 Juil 2024 à 19:32
Modifié(e) : Stephen23
le 20 Juil 2024 à 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?
1 commentaire
Stephen23
le 18 Juil 2024 à 20:29
Modifié(e) : Stephen23
le 20 Juil 2024 à 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.
Réponse acceptée
Star Strider
le 18 Juil 2024 à 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)
.
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DGM
le 18 Juil 2024 à 19:46
Déplacé(e) : Steven Lord
le 18 Juil 2024 à 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 commentaire
Walter Roberson
le 18 Juil 2024 à 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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