x = 0:0.1:10... What's going on, really?
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John Doe
el 1 de Mayo de 2014
Comentada: Image Analyst
el 9 de Jul. de 2023
Please have a look at the following example:
A = 0:0.1:0.4;
find(A == 0.3)
ans =
Empty matrix: 1-by-0
find(A == 0.1+0.1+0.1)
ans =
4
This is in my opinion expected behavior, as 0.1 can't be represented accurately with floating point numbers. What's bugging me is the following:
A = 5.8:0.1:6
A =
5.8000 5.9000 6.0000
find(A == 5.9)
ans =
2
%%Found it!
A = 5.8:0.1:6.1
A =
5.8000 5.9000 6.0000 6.1000
find(A == 5.9)
ans =
Empty matrix: 1-by-0
%%Didn't find it!
find(A == 5.8+0.1)
ans =
2
%%Found it again!
For the record, linspace results in the same results.
A = linspace(5.8, 6.0, 3)
A =
5.8000 5.9000 6.0000
find(A == 5.9)
ans =
2
A = linspace(5.8, 6.1, 4)
A =
5.8000 5.9000 6.0000 6.1000
find(A == 5.9)
ans =
Empty matrix: 1-by-0
find(A == 5.8+0.1)
ans =
2
Notice that the only difference is that the vectors are going to 6.1 instead of 6.0. So, what's going on? Are the following two actually the same: x = [a:b:c] and y = linspace(a,c,(c-a)/b+1)?
A = 5.8:0.1:6.1
A =
5.8000 5.9000 6.0000 6.1000
B = linspace(5.8,6.1,4)
B =
5.8000 5.9000 6.0000 6.1000
A == B
ans =
1 1 1 1
It might appear that way... But the answer is of course no, they're not the same!
x = -0.1:0.1:0.3
x =
-0.1000 0 0.1000 0.2000 0.3000
y = linspace(-0.1,0.3,5)
y =
-0.1000 0 0.1000 0.2000 0.3000
x == y
ans =
1 1 0 0 1
So, what happens when you do A = 5.8:0.1:6? How are the numbers created? And how can the following be explained?
A = 5.8:0.1:6;
B = 5.8:0.1:6.1;
A(2)-B(2)
ans =
8.8818e-016
eps(5.9)
ans =
8.8818e-016
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Respuesta aceptada
José-Luis
el 1 de Mayo de 2014
Editada: José-Luis
el 1 de Mayo de 2014
Well, just look at how linspace is implemented:
edit linspace
It might explain some of the behavior you see. On an additional note, comparing double value is risky business. Not only the way you perform a computation (as you so thoroughly illustrate) but the order in which you perform the operations will matter.
Even different compilers might give different results, since they might order operations differently. This means that you can potentially get different results in Matlab if you use different platforms.
EDIT:
As for the colon, here is how it is implemented according to the documentation:
j:i:k is the same as [ j,j+i,j+2i, ...,j+m*i ], where m = fix((k-j)/i), for integer values. For information on the definition of j:i:k with floating-point values, see Technical Solution 1-4FLI96. This syntax returns an empty matrix when i == 0, i > 0 and j > k, or i < 0 and j < k.
2 comentarios
Jeffrey
el 25 de Jul. de 2014
The technical solution 1-4FLI96 referenced in the above comment is no longer available. The solution was republished to MATLAB Central Answers here:
Más respuestas (2)
Image Analyst
el 1 de Mayo de 2014
What's going on is explained in the FAQ , along with some workarounds for comparing floating point numbers. http://matlab.wikia.com/wiki/FAQ?title=FAQ&cb=7710#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F
1 comentario
Ayesha iqbal
el 9 de Jul. de 2023
If v= 1:3:12 is a vector then v(3)=....?
1 comentario
Image Analyst
el 9 de Jul. de 2023
This does not look like an answer to @John Doe so it seems like you asking this simple question for yourself. You should know that you can do
v = 1:3:12
element3 = v(3)
but since you don't, you can learn such fundamental concepts by investing 2 hours of your time here:
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