Directly using Index of max value without storing

Hi,

I want to use the index of a maximum value in a vector directly without storing it as a variable. i.e:

I have a vector X = [1 3 2 9 2 1], and a vector Y = [10 20 30 50 60 70] and I want to extract the value in Y corresponding to the max value in X. I know I can use:

[M,I] = max(X); then

Value = Y(I);

But I can't work out how to do this in a single line without storing the value of "I", to acheive something like:

Value = Y(...index of max value of X...)

Any help greatly appreciated.

Thanks,

Matt

[Stephen23]

@Matthew Knights: what output do you want if there are two identical maximum values ? E.g.:

 X = [1 3 2 9 2 9]
 Y = [10 20 30 50 60 70]
 value = ???

[Matthew Knights]

Good question.

My problem is such that this won't occur for my application, so it is something I hadn't considered.

But for argument's sake I guess in an ideal world you would want both Y values given... But I see how this makes things more complicated.

Answer 1

[Guillaume]

This is one of the downside of matlab's syntax. It is not possible to chain expressions that use anything but the 1st return value. You have to use a temporary variable.

[Matthew Knights]

Thanks Guillaume.

I guess that's a fair sacrifice for Matlab's positive qualities.

It's always reassuring to find that the question you ask has a genuine reason behid it rather than just missing something glaringly obvious.

Many thanks,

Matthew

Answer 2

[Andrei Bobrov]

out = Y(max(X)==X)

[Guillaume]

That will not return the same, if there are several values equal to the max. Arguably, it could be better, but it's certainly not equivalent.

[Andrei Bobrov]

unique(Y(max(X)==X))

[Guillaume]

Certainly not! There's no guarantee that the value in Y would be the same for identical max values in X.

This would work:

Y(find(max(X) == X, 1))

as long as X and Y are vectors. Reproducing the same behaviour for matrices would be harder.

[Matthew Knights]

Thank you for the suggestions. My application is fairly simple so these will be applicable regardless of limitations for more complex problems. But I'm always glad to understand the whole scope for future use.

Many thanks