How to replace duplicate elements with the elements of the second array?

Hi , I have two arrays , one with duplicate elements and I want to keep the first occurance of the element and replace other repeated values with the elements of the second array. Here is the example

a=[23 34 56 78 100 10 12];
b= [2 2 2 4 6 0 0 11 11 11 6];% contains repeating elements
 
 iwant=[2 23 34 4 6 0 56 78 11 100 10 12]

[Dyuman Joshi]

How does 12 fit into the output you want? Do you want to attach the remaining elements from a to appended to the modified b array?

[lilly lord]

In b, 6 comes twice and therefore replaced by 12.

b= [2 2 2 4 6 0 0 11 11 11 6]

[Dyuman Joshi]

That I understood, but it doesn't make sense. iwant has more elements than b -

b= [2 2 2 4 6 0 0 11 11 11 6]

b = 1×11

2 2 2 4 6 0 0 11 11 11 6

iwant=[2 23 34 4 6 0 56 78 11 100 10 12]

iwant = 1×12

2 23 34 4 6 0 56 78 11 100 10 12

Either iwant is written incorrectly or there's something missing

[lilly lord]

Example

Set=[0 1,2,...100]; b=[number occurs once & repeating numbers]

a=[missing numbers];

i want = [elements of b that occurs once plus repeating elements replced by missing numbers]. ofcourse iwant is larger than a or b.

Size(iwant)= Size(Set) (order of elements will be different.

[Dyuman Joshi]

Let's go through the elements -

a=[23 34 56 78 100 10 12];

b= [2 2 2 4 6 0 0 11 11 11 6];

Starting with 2nd index

index 2, b(2) appears before, replaced by a(1), [2 23 2 4 6 0 0 11 11 11 6]

index 3, b(3) appears before, replaced by a(2), [2 23 34 4 6 0 0 11 11 11 6]

index 4, b(4) doesn't appear before

index 5, b(5) doesn't appear before

index 6, b(6) doesn't appear before

index 7, b(7) appears before, replaced by a(3), [2 23 24 4 6 0 56 11 11 11 6]

index 8, b(8) doesn't appear before

index 9, b(9) appears before, replaced by a(4), [2 23 24 4 6 0 56 11 78 11 6]

index 10, b(10) appears before, replaced by a(5), [2 23 24 4 6 0 56 11 78 100 6]

index 11, b(11) appears before, replaced by a(6), [2 23 24 4 6 0 56 11 78 100 10]

Now, appending remaining elements of a (7 to end) to b, [2 23 24 4 6 0 56 11 78 100 10 12]

which is different from iwant=[2 23 34 4 6 0 56 78 11 100 10 12]

As @Voss has stated in their answer.

It seems that there's a mistake in iwant. I insist you clarify this.

[lilly lord]

Yes u r right Sir

Answer 1

[Voss]

Maybe something like this?

a = [23 34 56 78 100 10 12];
b = [2 2 2 4 6 0 0 11 11 11 6];% contains repeating elements
 
yougot = b;
a_idx = 1;
for ii = 1:numel(b)
    if any(b(1:ii-1) == b(ii))
        yougot(ii) = a(a_idx);
        a_idx = a_idx+1;
    end
end
if a_idx <= numel(a)
    yougot = [yougot a(a_idx:end)];
end
disp(yougot);

     2    23    34     4     6     0    56    11    78   100    10    12

iwant = [2 23 34 4 6 0 56 78 11 100 10 12];
disp(iwant);

     2    23    34     4     6     0    56    78    11   100    10    12

Note that the 11 and the 78 are switched in iwant vs yougot. I think that's a mistake in iwant, according to the description of the algorithm you gave.

[lilly lord]

Thank you very much Sir, You have solved the problem.

[Voss]

You're welcome!

Answer 2

[Fabio Curto]

Hi,

If i get the question correctly, you want to append the remaining unused elements of a. I wrote a basic idea, maybe it can be optimized in some way!

a=[23 34 56 78 100 10 12];
b= [2 2 2 4 6 0 0 11 11 11 6];% contains repeating elements

changed = 1;                % flag initialization
j = 1;                      % a vector index

for i=2:length(b)
    if changed == 1
        temp = b(i-1);
    end
    if b(i) == temp
        b(i) = a(j);
        j = j+1;
        changed = 0;
    else
        changed = 1;
    end
end

for j=j:length(a)
    b = [b a(j)];
end

Hope it is helpful,

[lilly lord]

The output is

[2 23 34 4 6 0 56 11 78 100 6 10 12],

where 6 appears twice.