Assign vector to vector with gaps

Stuff doesn't matter - was tired - start reading at Edit1 - sry

Hello,

I have a matrix of row size m and a matrix of row size n where m > n and x = m - n. I also got a vector k of length x with different row indices of m. Now I want to assign n to m by not touching entries m(k) important is that these rows stay unchanged and in place.

Is there an easy one-liner? like:

m(m ~= x) = n;

or

m(m ~= m(x)) = n;

but seems not to work :(

Hope you can help me.

Thanks in advance.

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HERE!:-> @Edit1: Oh, uff was late yesterday and yeah what ever I wrote there is pretty confusing and wrong... so again:

Matrix A of size m * _ where m is the number of rows and matrix B of size n * _ where n is the number of rows. Still true is: m > n and x = m - n and V is a vector with x rows.

Example:

A = [1,1,1; 2,2,1; 3,3,4; 8,3,6; 8,3,0; 8,6,3];
B = [4,2,5; 4,3,2; 6,3,2; 8,0,9];
V = [2; 4];

Result in A:

A = [4,2,5; 2,2,1; 4,3,2; 8,3,6; 6,3,2; 8,0,9];

just for much bigger matrices. Sorry for my unclear post yesterday hope you are still willing to help me.

[Azzi Abdelmalek]

You can make your question clear by posting an example

[Image Analyst]

Your terminology is pretty sloppy. Are m and n the width of the matrix (number of columns which is the "row size" or "row length") or the names of the matrices? Are your matrices really column vectors? Or row vectors?

This is what I have so far but can't go further because of ambiguities in your terminology.

m=10
n=6
x = m-n
vec1 = randi(99, 1, m) % First "matrix" (sample data)
vec2 = randi(99, 1, n) % Second "matrix" (sample data)
k = randperm(m, x) % indexes of m

Answer 1

[goerk]

 A = [1,1,1; 2,2,1; 3,3,4; 8,3,6; 8,3,0; 8,6,3];
 B = [4,2,5; 4,3,2; 6,3,2; 8,0,9];
 V = [2; 4];

 % create 'inversion' of V (there may be better ways) 
 notV = 1:size(A,1);  % vector from 1 to number of rows
 notV(V) = [];  % delete rows which are in V

 % over write the rows that are not in V
 A(notV,:)=B

[Thorsten]

Another way to create notV would be

notV = setdiff(1:size(A,1), V);

Then it's a one-liner (may be faster if you work on big matrices)

A(setdiff(1:size(A,1), V), :) = B;

[goerk]

Thanks, that's a nice short solution.

[baxbear]

Thank you very much :D