Find Command not working.

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Muhammad Owais
Muhammad Owais 2021-10-24
Hi everyone . I have a matrix with first row as T=[NaN 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9]. When I use the command now if i have a=0.6 and i try to find index of 'a' as V=find(T==a). it gives me an answer as :
1×0 empty double row vector
How to solve this ?

回答(3 个)

Chunru
Chunru 2021-10-24
use abs(difference) to comapare the float numbers.
T=[NaN 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9];
a=0.6;
V=find(abs(T-a)<1e-10)
V = 7

Image Analyst
Image Analyst 2021-10-24
It's probably not exactly 0.6. Try ismembertol().

Walter Roberson
Walter Roberson 2021-10-24
T = [NaN 0.1:0.1:0.9]
T = 1×10
NaN 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000
a = 0.3;
Visually, it looks like 0.3 is in T
find(T == a)
ans = 1×0 empty double row vector
But find() says it is not there.
Is it there? Let's look at the bit representation
fprintf('a:\n'); arrayfun(@pX, a);
a: 3FD3333333333333
So for the literal, the bit representation ends in all hex 3's.
Is that bit representation present in T?
fprintf('\nT:\n'); arrayfun(@pX, T);
T: FFF8000000000000 3FB999999999999A 3FC999999999999A 3FD3333333333334 3FD999999999999A 3FE0000000000000 3FE3333333333333 3FE6666666666666 3FE999999999999A 3FECCCCCCCCCCCCD
No. There is a very similar bit representation that ends in a 4 rather than a 3. The last 4 bits of the actual value are 0011 (decimal 3), but the closest that T has is something for which the last 4 bits are 0100 (decimal 4).
When you calculate something two different ways, the bits are not necessarily going to come out identical.
function pX(N)
fprintf('%016X\n', typecast(N, 'uint64'));
end
  1 个评论
Image Analyst
Image Analyst 2021-10-24
A thorough explanation. There is also an explanation in the FAQ:
This is stuff you would have learned in your numerical analysis/linear algebra class at the university if you had taken such a class.

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