How to replace elements in an array with the indices of the array's sorted, unique values?

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This is what I have (below). I am wondering if there is a better way of accomplishing the aforementioned task. Also, is there a name for this process? I don't have a formal computer science education so I'm curious.
A = randi(9,5,2); % Initialize matrix.
uA = unique(A); % Unique values of A.
B = A; % Output.
for idx = 1:length(uA) % Index to unique values, replace with
B(B == uA(idx)) = idx; % their sorted indices.
end

採用された回答

Stephen23
Stephen23 2017 年 10 月 23 日
編集済み: Stephen23 2017 年 10 月 23 日
All you need is the third output of unique followed by reshape:
>> [~,~,Y] = unique(A);
>> reshape(Y,size(A))
For example:
>> B % from running your code
B =
4 5
2 3
1 4
4 7
5 6
>> [~,~,Y] = unique(A);
>> reshape(Y,size(A))
ans =
4 5
2 3
1 4
4 7
5 6
  2 件のコメント
Stephen23
Stephen23 2017 年 10 月 23 日
@Dominik Mattioli: I hope that it helps!
It is always worth reading the documentation for every function that you use: there is a lot of useful info in there!

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その他の回答 (2 件)

Walter Roberson
Walter Roberson 2017 年 10 月 23 日
That code is wrong: you are changing B "in place" and comparing against all of it, so you could end up changing the same location multiple times. For example, suppose one of the unique values was 1/2 and another was 1, and those got index values 1 and 2 respectively. You change the 1/2 entries to 1. But that creates a 1 that is then there to be matched against uA(2) so you would mistake the 1 that was the index with the 1 that was the value.
You should just use
[uA, B] = unique(A);
  1 件のコメント
Dominik Mattioli
Dominik Mattioli 2017 年 10 月 23 日
Very good point, thank you. The other two "Steve"s pointed out the easy solution to my naive question.

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Steven Lord
Steven Lord 2017 年 10 月 23 日
Do you want the ic third output from unique, or a version of that output that you've used reshape on to make it the same shape as A?

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