Directly using Index of max value without storing

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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
  2 个评论
Stephen23
Stephen23 2016-7-26
@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
Matthew Knights 2016-7-26
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.

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采纳的回答

Guillaume
Guillaume 2016-7-26
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.
  1 个评论
Matthew Knights
Matthew Knights 2016-7-26
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

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更多回答(1 个)

Andrei Bobrov
Andrei Bobrov 2016-7-26
out = Y(max(X)==X)
  4 个评论
Guillaume
Guillaume 2016-7-26
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
Matthew Knights 2016-7-26
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

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