How to add rows to the matrix

Hallo,

I have a matrix of 6 columns (an example attached here) that are : trials, stimulus IDs, Hit, Miss, CR, FA.

Hit, Miss, CR, FA colomns are coded as a 0 and 1 but the FA column has sometimes numbers larger than 1 that indicates number of repeated trials that are not included in to the matrix. I need correct this and everywhere when FA colomn has number > 1 replace it with repeated rows in a number of repeats. For example, in given example in the trial 6 FA has number of repeats 5. I need to replace it with 5 consequentive rows after 6th trial (trials 7-11) in which stimulus is the same, Hit, Miss, CR, would have 0 and FA column 1. Same is for trial 20 (add 3 repeated trials) and 29 (add 2 repeated trials) etc

Can anyone help with this?

[Jan]

You have attached an Excel file. Is importing this file a part of the problem? What is the wanted output? A Matlab matrix or another Excel file?

Does "raws" mean "rows"?

Answer 1

[Torsten]

Like this ?

A = [1 4 0 0 1 0		
     2 3 0 1 0 0		
     3 3 0 1 0 0		
     4 4 0 0 1 1		
     5 3 0 1 0 0		
     6 4 0 0 1 5		
     7 3 0 1 0 0		
     8 3 1 0 0 0		
     9 4 0 0 1 0		
     10 4 0 0 0 1];
B = [];
irow = 0;
for i = 1:size(A,1)
  B(irow+1,:) = [irow+1,A(i,2:6)];
  irow = irow + 1;
  trials = A(i,6);
  if trials ~= 0 && trials ~= 1
    for j = 1:trials  
      B(irow+j,:) = [irow+j 0 0 0 0 1];
    end
    irow = irow + trials;
  end
end
B(1:10,:)

ans = 10×6

1 4 0 0 1 0 2 3 0 1 0 0 3 3 0 1 0 0 4 4 0 0 1 1 5 3 0 1 0 0 6 4 0 0 1 5 7 0 0 0 0 1 8 0 0 0 0 1 9 0 0 0 0 1 10 0 0 0 0 1

B(11:15,:)

ans = 5×6

11 0 0 0 0 1 12 3 0 1 0 0 13 3 1 0 0 0 14 4 0 0 1 0 15 4 0 0 0 1

[EK]

I am sorry no. I need that it reads excel file goes through FA colomn find the rows with FA>1 and replaced them by repeated rows with FA =1 same stimulus id, Hit, Miss, CR, = 0. For example if FA =4 add 4 rows, if FA=10 add 10 rows and etc

[Torsten]

A = [1 4 0 0 1 0		
    2 3 0 1 0 0		
    3 3 0 1 0 0		
    4 4 0 0 1 1		
    5 3 0 1 0 0		
    6 4 0 0 1 5		
    7 3 0 1 0 0		
    8 3 1 0 0 0		
    9 4 0 0 1 0		
    10 4 0 0 0 1];
B = [];
irow = 0;
for i = 1:size(A,1)
    trials = A(i,6);
    if trials ~= 0 && trials ~= 1
      for j = 1:trials
        B(irow+j,:) = [A(i,1:5),1];
      end
      irow = irow + trials;
    else
      B(irow+1,:) = A(i,1:6);
      irow = irow + 1;
    end
end
B(1:10,:)

ans = 10×6

1 4 0 0 1 0 2 3 0 1 0 0 3 3 0 1 0 0 4 4 0 0 1 1 5 3 0 1 0 0 6 4 0 0 1 1 6 4 0 0 1 1 6 4 0 0 1 1 6 4 0 0 1 1 6 4 0 0 1 1

B(11:14,:)

ans = 4×6

7 3 0 1 0 0 8 3 1 0 0 0 9 4 0 0 1 0 10 4 0 0 0 1

[EK]

I added now an example of the test and corrected matrix

[Torsten]

Why is the row 6 times repeated although FA = 5 ?

[Torsten]

A = [1	4	0	0	1	0
    2	3	0	1	0	0
    3	3	0	1	0	0
    4	4	0	0	0	1
    5	3	0	1	0	0
    6	4	0	0	0	5
    7	3	0	1	0	0
    8	3	1	0	0	0
    9	4	0	0	1	0
    10	4	0	0	0	2
    11	4	0	0	1	0
    12	3	0	1	0	0
    13	3	0	1	0	0
    14	4	0	0	0	1];
B = [];
irow = 0;
for i = 1:size(A,1)
    trials = A(i,6);
    if trials ~= 0 && trials ~= 1
        for j = 1:trials
            B(irow+j,:) = [irow+j,A(i,2:5),1];
        end
        irow = irow + trials;
    else
        B(irow+1,:) = [irow+1,A(i,2:6)];
        irow = irow + 1;
    end
end
B(1:10,:)

ans = 10×6

1 4 0 0 1 0 2 3 0 1 0 0 3 3 0 1 0 0 4 4 0 0 0 1 5 3 0 1 0 0 6 4 0 0 0 1 7 4 0 0 0 1 8 4 0 0 0 1 9 4 0 0 0 1 10 4 0 0 0 1

B(11:end,:)

ans = 9×6

11 3 0 1 0 0 12 3 1 0 0 0 13 4 0 0 1 0 14 4 0 0 0 1 15 4 0 0 0 1 16 4 0 0 1 0 17 3 0 1 0 0 18 3 0 1 0 0 19 4 0 0 0 1

[EK]

Thank you so much! You are amazing!