Compare Matrices within tolerance range

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Anna Lin
Anna Lin 2015-6-12
编辑： Paul Hoffrichter 2020-12-10
Hi. I have a matrix A that is 8323 by 3 and matrix B that is 3 by 3. Elements in both these matrices have decimals. I want to find rows in A that is within 5% tolerance of rows in B. I am pretty sure you have to use for loop and if statements to find it. However, i am not sure what to write. I made 2 matrices C and D of same size as B since i calculated the max and min values that correspond to B values.
for y=1:3
for x=1:n1
if Points1(:,y)<=none1(x,y) && Points1(:,y)>=none3(x,y)
%not sure what to write
And as always, thanks.
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Image Analyst
Image Analyst 2015-6-13
I guess she meant "the thousands place", which would be the third place after the decimal. I wish it were thousands of decimal places - maybe in another 50 years it will be.
Jan
Jan 2015-6-13
编辑：Jan 2015-6-13
@Anna Lin: points3, Points1, A, B, C, D, none1, none3 - the bunch of variables is at least confusing. What is the wanted result? Do you want a [8323 x 1] logical vector with a true, if the corresponding row of A is inside one of the three intervals defined by the rows 0.09*B and 1.05*B? Or do you want a [8823 x 3] array?

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回答（2 个）

Guillaume
Guillaume 2015-6-13
Starting in R0215a, there is the function ismembertol that does more or less what you want:
iswithintol = ismembertol(A, B, 0.05, 'ByRows', true)
will find rows of A that differ from the rows of B by no more than 5% of the max of A and B. There are options to change the tolerance criteria.
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Paul Hoffrichter
Paul Hoffrichter 2020-12-10
I was looking up a solution for this problem when I saw ismembertol above. But consider:
A = [1 2 3];
B = [3 1 2];
In my case, and possibly in the OP, A and B are out of tolerance since I want each element of A to be within tolerance of the corresponding element of B. But we get:
ismembertol(A,B)
ans =
1×3 logical array
1 1 1

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Paul Hoffrichter
Paul Hoffrichter 2020-12-10
编辑：Paul Hoffrichter 2020-12-10
For a tolerance match between corresponding elements of the rows, start off with this:
a = a(:);
b = b(:);
%% Check for equivalence of each element, within the tolerance
yn = abs(a - b) <= eps(max(abs(a), abs(b)));
yn = all(yn); % scalar logical output
Reference: https://www.mathworks.com/matlabcentral/fileexchange/36734-isequalfp-check-equality-within-floating-point-precision
Next step is to modify above to include desired tolerance.

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