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How to determine repeated value in row of an array

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I have a data of
1 1 2 2
2 2 3 1.5
3 3 4 2.2
4 4 5 6.8
5 4 6 1
6 5 7 5.1
7 6 8 2.5
8 7 9 3.3
9 8 10 2.8
10 8 11 1.4
11 10 12 3.2
12 11 13 2.7
13 12 14 1.9
14 13 15 4.5
How can i create a code to have it to check on each row if the value is repeated.
As an example:
The value that i wanted to check is on row 4 column row 2 and 3.
The value selected is 4 and 5. These value are then being compared with the value of the row below it.
If the value in the following row which is row 5 have value that are same with the value above which is 4, it will continue this comapring process with the value 4 and 6 with another row until there is no other identical appear.
After identifying all repeative value, the value in column 4 of those repeative value are being added together.
  4 Comments
Ive J
Ive J on 14 Dec 2020
Can you provide an example along with algorithm steps? For instance:
% Step 1
A = [1 1 2 2;
3 2 1 4];
% step 2
% what to do
% step N
Chau Chin Haw
Chau Chin Haw on 15 Dec 2020
%step 1
A =
[1 1 2 2
2 2 3 1.5
3 3 4 2.2
4 4 5 6.8
5 4 6 1
6 5 7 5.1
7 6 8 2.5
8 7 9 3.3]
%Step 2
%Check value in row 4 with the value in row 5
%If there are similar value in row 5 (value 4), comparing continue with row 5 and 6
%If none, end
%Step 3
%Sum all the value in column 5 of the repeative value
Sum = 6.8+1+5.1+2.5+3.3;

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Accepted Answer

Ive J
Ive J on 15 Dec 2020
Edited: Ive J on 15 Dec 2020
Your arrays is N*4 and not N*5, so you may want to modify your example. Also add/modify conditions if necessary.
A = [1 1 2 2
2 2 30 1.5
3 20 4 2.2
4 4 5 6.8
5 4 6 1
6 20 70 5.1
7 6 30 2.5
8 7 9 3.3]
vals = [4 ,20, 30, 70, 5];
sumVals = nan(numel(vals), 1); % NaN for values not satisfying the condition
for i = 1:numel(vals)
checkMe = find(any(A == vals(i), 2)); % find all rows containing vals(i)
if ~isempty(diff(checkMe)) && all(diff(checkMe) == 1)% if vals(i) appreas in consecutive rows and appears at least twice
sumVals(i, 1) = sum(A(checkMe, end)); % sum over last column
end
end
sumVals
10.0000
NaN
NaN
NaN
7.8000

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