How to rank a vector with repeats without MATLAB unique?

Hi everybody,

I need to rank a vector and currently i'm using this two commands:

data_s=sort(data); [~,rank]=ismember(data,data_s);

For a vector data=[1 1 1 2 2 2 3 3 3] i get i=[1 1 1 4 4 4 7 7 7], but instead i would like to get i=[1 1 1 2 2 2 3 3 3], without using MATLAB function 'unique'. Anybody has any suggestion?

Thanks in advance.

[the cyclist]

Are you sure that that is what you want? If two runners tie for first place, and there is a third runner, then the ranking would typically be reported as

[1 1 3]

and not

[1 1 2]

So, the result you have now is typically what people would expect from a ranking system.

[Guillaume]

"without using MATLAB function unique" Why?

It's possible to get the same output as unique using any of the set functions ismember, setdiff, union, etc., because internally they all call unique. But in that case, why not call unique directly?

If the reason is because it's an assignment, then using any of these functions just to avoid unique is probably not going to earn you a good mark. I would think that you'd be expected to come up with your own unique algorithm.

[pedrogon]

First of all thank you for your attention,

I cannot use function unique, because i'm trying to implement a piece of code in a DSP using MATLAB 2010b and i don't know why it keeps giving me an error saying "floating constant is out of range". So that's why i was trying to implement an alternative to unique.

Answer 1

[Jan Orwat]

Do you want to replicate following behaviour?

[~, ~, rank] = unique(data);
rank = rank.';

There are thousands of ways to do that. For example:

[foo, bar] = sort(data);
rank(bar) = cumsum([1, diff(foo)~=0]);

But, as the cyclist wrote, are you sure?

[pedrogon]

Thank you for your answer, it is also a good alternative. In the algorithm that i'm trying to implement in a DSP, unique function should be used, so for the reasons that i listed above i need to replicate the function behaviour.

Answer 2

[Jan]

 data_s = sort(data);
 result = cumsum([1, diff(data_s) ~= 0]);

[pedrogon]

That you for your answer, but your suggestion does not seem to work well. If i use:

data=[1 1 1 3 3 3 2 2 2];

I will get the following:

result=[1 1 1 2 2 2 3 3 3]