# overwriting certain lines in a matrix with previous lines that satisfy a condition.

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Søren on 18 Nov 2014
Commented: Kelly Kearney on 18 Nov 2014
Hi.
I have a vector which marks rows that needs to be overwritten in a matrix.
The marked rows should be replaced with the previously row that isn't marked.
Line 1 will never be marked so no need to take account for the potential error.
The matrix is big and will always have enough rows that markedrows refers to.
Ex:
Vector: markedrows = [3,6,7,14,15,16,17] (markedrows will always be in ascending order)
So the deal is, according to markedrows, in the matrix:
• The new line 3 should be the same as line 2
• The new line 6 should be the same as line 5
• The new line 7, however, shouldn't be line 6 cause line 6 is marked too, it should instead be line 5.
Following this pattern would make line 14, 15, 16 and 17 all turn into line 13.
I hope this makes sence and that you can help :).

Guillaume on 18 Nov 2014
Simply do:
for row = markedrows
m(row, :) = m(row-1, :);
end
row 14 is replaced by row 13, row 15 by row 14 which is now equal to row 13, etc.
Søren on 18 Nov 2014
I must have done a typo the first time I ran it then, it works perfectly now.
Thanks alot to both of you, you've been very helpful!

Kelly Kearney on 18 Nov 2014
marked = [3,6,7,14,15,16,17];
good = setdiff(1:20, marked);
prev = arrayfun(@(x) good(find(good < x,1,'last')), marked)
prev =
2 5 5 13 13 13 13
Kelly Kearney on 18 Nov 2014
Guillaume's method is more efficient if you just need to replace the rows and be done with it, but if you want a record of which rows were replaced with which, this is an alternative.
And just run
m(marked,:) = m(prev,:);
to do the actual replacement.

Andrew Reibold on 18 Nov 2014
Edited: Andrew Reibold on 18 Nov 2014
This is not the most efficient memory-wise, but it is a working solution and I tried to use basic commands. I just made up a matrix m to demonstrate. See the comment for change that will let you increment the other direction if you want.
m = round(rand(2,20)'*10) %a random matrix of numbers 1-10
markedrows = [3,6,7,14,15,16,17];
new_m = m;
for row = 1:size(m,1)
nrow = row;
while find(markedrows == nrow);
nrow = nrow-1; %make this MINUS a PLUS to swap direction.
new_m(row,:) = m(nrow,:);
end
end
new_m %outputs to command window for comparison
Andrew Reibold on 18 Nov 2014
Edited: Andrew Reibold on 18 Nov 2014
Short answer - Yes, there are better solutions. :D
For now, go check Guillaumes solution again. I actually think his solution is superior in many ways now that I correctly realize what it is doing.
It would be WAY better for large matrices because it only spends time where necessary.