# How to add rows to the matrix

5 views (last 30 days)
EK on 17 Dec 2022
Commented: EK on 18 Dec 2022
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 on 17 Dec 2022
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"?

Torsten on 17 Dec 2022
Edited: Torsten on 18 Dec 2022
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
Torsten on 18 Dec 2022
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 on 18 Dec 2022
Thank you so much! You are amazing!

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