in algorithm a~b , in MATLAB...??????

Hi everybody, In an algorithm I am working on, I came to a check box with content:
a~b
where the '~' would be read as "a approximates b" which can be defined probably by absolute numerical difference. so if a~b=TRUE (i.e the difference b/n a and b is very small)further computations are done if not( i.e. a~b=FALSE) then termination! , if 'YES' it continues with other computations, and if 'NO' it terminates. How can I write this in MATLAB?

1 个评论

In that context, it would be read as "a approximates b". But like the others say below, you need to define what approximation means in the context.

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 采纳的回答

Assuming ~ mean not (or not equal) in your algorithm you could use ~= or ~isequal(). I prefer the isequal approach since it handles arrays and mismatched sizes ...
if isequal(a, b)
disp('NO')
else
disp('YES')
end
Although it seems a little backwards to me that a~b leads to YES.

5 个评论

Thanx Dani, just some more clarifications and corrections. in my algorithm the ~ looks more of closeness between a and b; so if a~b=TRUE (i.e the difference b/n a and b is very small)further computations are done if not( i.e. a~b=FALSE) then termination!
In that case replace isequal(a, b) with abs(a-b) > tol. Where tol is whatever tolerance your want. This requires a and b to be scalars.
@Meh: Please define "very small". Do you mean the absolute or relative numerical difference? Or are "a" and "b" strings and you search for the edit-distance? Or are "a" and "b" lines in 3D and you search for the shortest connection?
It would be very helpful, if you explain the term "a~b" exactly by editing the original question - not as comment of an answer. Thanks.
@Meh: After you have clarified that "~" should mean "approximation", it got clear, that this amswer does not concern the question. Why do you accept it then as a solution?
It would be helpful for the forum, if you explain what "approximation" means qualitatively for your case. Then an answer would be very easy.
It might also be helpful if I edited my answer to reflect the comment I made about comparing the difference to a tolerance. I am hesitant to do that since the answer has already been accepted.

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