Compare two or more cell arrays with a specific tolerence?

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Hi.
I have a cell array with length 3 :
M_t={{[0,1],[0 0 2]},{[0,1.01],[0 0 2]},{[0,1],[0 .001 2]}};
I want to compare these 3 arrays with a specific tolerence. How can i do this?
I have tried intersect before.
Thanks.
  7 个评论
Najmeh Eskandari
Najmeh Eskandari 2019-2-20
suppose that i use ismembertol:
ismembertol([0,1],[0,1.01],1e-02)
logical
1 1
ismembertol([0,0,2],[0,0,2],1e-02)
logical
1 1 1
in this case i want to get result 1 but forexample in case:
ismembertol([0,1],[0,1.5],1e-02)
logical
1 0
ismebertol([0,0,2],[0,0,2],1e-02)
logical
1 1
the final result that i want is:
0
I have several cell arrays with such matrices and want to compare the arrays. If in the way that i described the matrices are equal the result shuld be 1.
-The absolute error is my concern.
Jan
Jan 2019-2-20
编辑:Jan 2019-2-20
In the case
ismebertol([0,0,2],[0,0,2],1e-02)
the output has 3 elements, not "logical1 1". Is this a typo?
ismembertol uses a strange scaling: for the elements u of the array A and the elements v of the array B:
abs(u-v) <= tol*max(abs([A(:);B(:)]))
This means, that the tolerance is multiplied my the maximum absolute value of the arrays. I do not know a case, where this is really useful.
ismembertol([1, 900], [20, 1000], 1)
Is the result [true, true] expected?!
For all of my cases, ismembertol did not offer a useful method to apply the tolerence. What a pity.
By the way, you still did not mention, which kind of comparison you exactly want. "The absolute error" is not a unique definition. It can mean:
all(abs(a(:) - b(:)) < tol) % or <=
any(abs(a(:) - b(:)) < tol) % or <=
sum(abs(a(:) - b(:)) < tol) < numel(A) / 2
% or a method, which does not consider the order or the elements
You have mentioned ismembertol, which does not consider the order of elements. So it is not clear, if [1, 0] and [0, 1] should be considered as equal or not.
It is hard to find out, what you exactly want. I've spent some time to ask you specific question, but I do not get clear statements and in consequence I assume, that I cannot help you.

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回答(1 个)

Rik
Rik 2019-2-20
The code below will iterate through the cell levels. The resulting output is only true if all dimensions are equal and all elements are within tolerance.
M_t={{[0,1],[0 0 2]},{[0,1.01],[0 0 2]},{[0,1],[0 .001 3]}};
clc
assert(ismembertol_nested(M_t(1),M_t(2),0.1),'test failed for equal arrays');
assert(~ismembertol_nested(M_t(1),M_t(3),0.1),'test failed for unequal arrays');
assert(~ismembertol_nested(M_t(1),M_t{2},0.1),'test failed for unequal inputs');
function tf=ismembertol_nested(A,B,tol)
%Un-nest cell arrays until ismembertol can be used on the inner arrays.
%This will return false instead of an error if the shapes are not equal on
%any level.
try
if isa(A,'cell')
tol_cell=num2cell(repmat(tol,size(A)));
tf=all(cellfun(@ismembertol_nested,A,B,tol_cell));
else
ia=ismembertol(A,B,tol);
ib=ismembertol(B,A,tol);
tf=all([ia,ib]);
end
catch
tf=false;
end
end
  2 个评论
Jan
Jan 2019-2-20
编辑:Jan 2019-2-20
@Rik: I'm dissapointed by the way ismembertol applies the tolerances. See my comment above:
abs(u-v) <= tol*max(abs([A(:);B(:)]))
So in ismembertol([1, 900], [20, 1000], 1) the tolerance 1 is multiplied by 1000, the maximum element, such that [1,1] is replied.
In addition the OP did not clarify yet, if the order of the elements matters or not.
Rik
Rik 2019-2-20
That seems strange indeed. I think it would make much more sense if that scale factor would be 1 by default, not that max. So to get the behavior I expected you need to call this
ismembertol(A,B,'DataScale',1)
I can see what the usefullness is if you are using it to check for float rounding errors, but that is not what I suspect most people would be using it for. There doesn't seem to be an easy way to search the FEX for the use of a function, so I can't check that suspicion.

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