Comparing two array and combining them with condition - matlab

Question

I have these matrices:
I1 = [60 30 15 35 20 -25 30 5 45 25 -10 40 10];
I2 = [60 30 60 35 20 60 30 60 45 25 60 40 60];
A= 0:12 ;
I want this:
Ir=[60 30 15 NaN 60 35 20 -25 NaN 60 30 5 NaN 60 45 25 -10 NaN 60 40 10 NaN 60]
Ar= [0 1 2 2 2 3 4 5 5 5 6 7 7 7 8 9 10 10 10 11 12 12 12]
How:
When I1 and I2 are same, proceed. When different, use elements of both I1 and I2 and insert an NaN between them.
and Ar is such that use element of A and proceed when elements of I1 and I2 are same. But when different, repeat the value of A 3 times. 1st for I1 and then for NaN and then for I2.
Can't figure out a way to do this. How do I do this?

Answer 1

A more vectorized way to do this would be to create an augmented matrix where the top row is I1, the bottom row is I2 and the middle row is full of NaN. After, create a logical matrix which is the same size as this augmented matrix you just created and you would adjust the columns of this matrix where you'd set the bottom two rows of this logical matrix to 0 for each corresponding column in this matrix where I1 == I2. Once you're done, use this logical matrix and index the augmented matrix. The advantage of this indexing is that it accesses the elements in column major format, which is exactly what you're after. We will only sample the top row of the augmented matrix unless the elements in I1 and I2 don't equal each other. If that's the case, we sample the entire column which includes both I1, nan and I2. Because we accessed by columns, your desired output is a row so you'll need to transpose the result when you're done.
To create the index vector, you would do the same thing but you would create a matrix of IDs where we have three rows where each row is the index ID array A. You'll also need to transpose this result after you index into A:
aug = [I1; nan(1,numel(I1)); I2];
V = true(size(aug));
V(2:3, I1 == I2) = false;
Ir = aug(V).';
ID = repmat(A, 3, 1);
Ar = ID(V).';
We get:
>> format compact
>> Ir
Ir =
Columns 1 through 17
60 30 15 NaN 60 35 20 -25 NaN 60 30 5 NaN 60 45 25 -10
Columns 18 through 23
NaN 60 40 10 NaN 60
>> Ar
Ar =
Columns 1 through 17
0 1 2 2 2 3 4 5 5 5 6 7 7 7 8 9 10
Columns 18 through 23
10 10 11 12 12 12

Answer 2

I can provide you with pseudo-code.
for each value in A
if I1 at [current A value] equals I2 at [current A value]
add I1 at [current A value] to end of Ir
add [current A value] to end of Ar
else
add I1 at [current A value] to end of Ir
add NaN to Ir
add I2 at [current A value] to end of Ir
add [current A value] to Ar 3 times
If your problem was conceptual hopefully this is enough.
Let me know if you have any further questions.

Answer 3

You can use two indices and solve the problem. The first index ii goes over the input vectors, and the second index k grows as the computation proceeds. A simple code using for-loop and if-statement looks like this:
I1 = [60 30 15 35 20 -25 30 5 45 25 -10 40 10];
I2 = [60 30 60 35 20 60 30 60 45 25 60 40 60];
Ir = [];
Ar = [];
k = 1;
for ii=1:length(I1)
if I1(ii)==I2(ii)
Ir(k) = I1(ii);
Ar(k) = ii-1;
k = k + 1;
else
Ir(k:k+2) = [I1(ii) NaN I2(ii)];
Ar(k:k+2) = ii-1;
k = k + 3;
end
end
Will result in:
Ir =
60 30 15 NaN 60 35 20 -25 NaN 60 30 5 NaN 60 45 25 -10 NaN 60 40 10 NaN 60
Ar =
0 1 2 2 2 3 4 5 5 5 6 7 7 7 8 9 10 10 10 11 12 12 12