In a sorted vector ,how do I compare first element to other elements and then find if there are any duplicates , if yes take their indexes and replace it with 0

A = [ 1, 1 , 2, 3, 4]

I want to first compare first element to all other elements, if there is any duplicate (like 1, 1 in this example) I need to replace it with 0.

Answer 1

[Stephen23]

For a sorted row vector. Find duplicates of the first element, as you requested:

>> A = [1,1,2,3,4];
>> X = A(1)==A;
>> X(1) = false;
>> A(X) = 0
A =
   1   0   2   3   4

Or perhaps you meant to find all duplicates, not just of the first element:

>> A = [1,1,2,3,4];
>> X = [false,diff(A)==0];
>> A(X) = 0
A =
   1   0   2   3   4

[Sagar Saxena]

Thanks Stephen, One more question is :

Instead of replacing the duplicate to 0 . lets say in this example there are 2 duplicates (1,1) so i want to replace both the duplicates number with (1/2,1/2) instead of 0.

so, in general if there are n duplicates , it should be replaced by 1/n.

TIA

[Stephen23]

>> A = [1,1,1,2,3,3,4];
>> X = A(1)==A;
>> X(1) = false;
>> A(X) = 1/nnz(X)
A =
   1.0   0.5   0.5   2.0   3.0   3.0   4.0

[Sagar Saxena]

Thanks for prompt response Stephen. Just one thing is missing here: We have 3 1's so first element should alse be replaced.

So if input is : A = [1,1,1,2,3,3,4];

then output should be : >> A = [1/3,1/3,1/3,2,3,3,4];

TIA

[Sagar Saxena]

I figured it out . Just replaced false with true and it worked . Thanks for your help :)

[Sagar Saxena]

Hi Stephen, In the above code once you have the final output how can you make rest of the elements as 0 which have no duplicates. So it should look like this.

Input : A = [1,1,1,2,3,4];

output: A= [1/3 1/3 1/3 0 0 0]

TIA

[Stephen23]

Use ~X as an index.

[Sagar Saxena]

Its not working , could you let me know exactly which line should I change it. i am taking about this code :

A = [1,1,1,2,3,3,4]; X = A(1)==A; X(1) = true; A(X) = 1/nnz(X)

Input : A = [1,1,1,2,3,4];

output: A= [1/3 1/3 1/3 0 0 0]

[Stephen23]

Look at the logical vector X. Did you see how I used it as an index into A, in order to allocate the value 0 or 1/nnz(X) ? You can do much the same thing with ~X. Look at the code. Think about it. Experiment.

You will come up with something like this:

A = [1,1,1,2,3,3,4];
X = A(1)==A;
A(X) = 1/nnz(X);
A(~X) = 0

[Sagar Saxena]

ohh yes, got it now. Thanks

[Sagar Saxena]

Hi Stephen,

What if in the above code A is a 2d array in 2 dimension for ex:

A(:,:,1 )= 320 x 320)

A(:,:,2) = (320 x320)

So now if I want the same algo to be applied in both the dimensions for this 2d array , what changes do i need to make in the above code (for making it multidimensional ) using loops for row and columns and doing it in both dimensions. I tried but getting errors...TIA

[Sagar Saxena]

Hey Stephen, could you please help with this one.

[TADA]

a small example of what this sorted matrix would look like would help

[Sagar Saxena]

Hi TADA ,

You can assume it to be a random matrix of size lets say 100 x 100 . So we need to sort it first and then the abovce code by stepehen is helpful.

in the above code , instead of 1d now I need to apply same algo in multi dimension . for ex let say the matrix would like something like this :

100 x 100 x 3 so ,it should work in 3 dimensions

TIA

[Stephen23]

@Sagar Saxena: your comment is not very clear: you keep mentioning that you have a "2d array", but then your example shows a 3D array:

A(:,:,1) = (320 x 320)
A(:,:,2) = (320 x 320)

So how many dimensions does your array have?

Please provide a small example of the input and output arrays.

[Sagar Saxena]

Okay sorry for confusion.

so I have 3 dimension in total , i am dealing with random values so its looks like this

A(: , : , 1) = 320 x 320 (random values)

A(: , : , 2)= 320 x 320

A(: , : , 3) = 320 x320

Hope its clear now.

[Sagar Saxena]

Hi Stephen ,

Look at the following code. Just for example I have taken 3x3x2 array

So when you execute this and go to statement with the green arrow (X(1)= true) , it does it only for the first dimension but not on other dimension . So , I want this to be applied in both the dimension . Thats is what I am looking for. Could you now please look into it.

Thanks

[Stephen23]

"it does it only for the first dimension but not on other dimension . So , I want this to be applied in both the dimension ."

Your statement is not clear. The comparison

X(1)==true

Sets the first element of the array X to true. The first elements is simply equivalent to X(1,1,1,1,...), following the rules of linear indexing. It is not clear what you mean by the "other dimension" your array has three dimensions, not two. And none of your operations (except for the sort and Window_2D) really specify anything about dimensions. I cannot guess what "both dimensions" you are referring to, when your array has three dimensions, not two. If you do not explain this then you make it very hard for me to help you.

In any case, your use of eval to concatenate matrices is slow, complex, and should be avoided. It is much simpler and more efficient to use cat with a cell array:

C = {[9,3,7;6,3,8;3,7,9],[8,4,8;2,6,7;2,9,5]};
A = cat(3,C{:})

Read this to know why your eval approach is a bad way to write code and is not recommended:

https://www.mathworks.com/matlabcentral/answers/304528-tutorial-why-variables-should-not-be-named-dynamically-eval

I cannot guess what you want. The best thing to do is to show what the expected output is: you have shown input arrays, now please show what the expected output array should be. I will try and figure the algorithm out from that.

[Stephen23]

PS: taking a wild guess here, that you want to perform this for each page (i.e. along the third dimension) independently:

>> C = {[9,3,7;6,3,8;3,7,9],[8,4,8;2,6,7;2,9,5]};
>> A = cat(3,C{:})
A(:,:,1) =
   9   3   7
   6   3   8
   3   7   9
A(:,:,2) =
   8   4   8
   2   6   7
   2   9   5
>> DM = sort(sort(A,1),2)
DM(:,:,1) =
   3   3   7
   3   6   8
   7   9   9
DM(:,:,2) =
   2   4   5
   2   6   7
   8   8   9
>> X = DM(1,1,:)==DM;
>> [~,~,P] = ind2sub(size(X),find(X));
>> V = accumarray(P,1,[],@(v)1/sum(v));
>> A(X) = V(P);
>> A(~X) = 0
A(:,:,1) =
   0.33333   0.33333   0.00000
   0.33333   0.00000   0.00000
   0.00000   0.00000   0.00000
A(:,:,2) =
   0.50000   0.00000   0.00000
   0.50000   0.00000   0.00000
   0.00000   0.00000   0.00000
   

If that is not the output that you expect then please show me what output you need to get.

[Sagar Saxena]

Thanks Stephen, This works perfectly fine for what I need and you are right instead of using dimension I should be using the term page. Let me try puting this peice of logic into my actual code. Will let you know if I am stuck any further.

But this really helped, Thanks a lot!

Answer 2

[TADA]

how about unique(A)?

Answer 3

[madhan ravi]

A = [ 1, 1, 2, 3, 4];
[~,c,~]=(unique(A,'first'));
idx=(ismember((1:numel(A)),c));
A(~idx)=0

c is the index of unique values in A the left out indices contain duplicate values. The above method works for arbitrary vectors too not only for sorted vector.

Answer 4

[Bruno Luong]

For array with finite elements

A.*[1,diff(A)~=0]