How to make specific elements of a matrix zero before nth element?

Hi I have a matrix of 24x365. I have an array which reperesents the index of an element in each column.I want to convert all elements to zero in each column which are before that index value.

For example consider a 4x4 matrix.

A=[1 2 3 4

5 6 7 8

9 1 3 4

8 7 5 3]

B=[1,3,2,4] %Index array

so the new matrix should be c which is answer matrix.so the first element in B is the index for column 1.Its value is 1 and as there is no element before 1 so the whole column 1 remains the same in answer matrix C.Now for column 2,the index of the element is 3 so the first and second elemtents of column 2 will become zero in answer matrix C similarly for third column B(3)=2 so column 3 will have its first element as zero. Same condition applies for 4th column that before the 4th element in 4th column every element converts to zero.

Answer matrix:

C=[1 0 0 0

5 0 7 0

9 1 3 0

8 7 5 3]

How can I do that?

[madhan ravi]

You need to explain more

[Rabia Zulfiqar]

Dear Madhan ravi, I have edited my question and added a little bit more explanation.I hope it's cleared now:) but still if you eed more explanation just let me know.

Answer 1

[Matt J]

[m,n]=size(A);

C=A.*((1:m).'>=B);

[Tommy]

Nice! Neither boring nor ugly. I guess using indexing rather than multiplication, converting to linear indices, and checking the upper bound were all unnecessary...

[Matt J]

Thanks Tommy, although I don't se the difference in your version, other than that for some reason you seem to prefer conjugate transpose (1:m)' over regular tranpose (1:m).'

[Tommy]

If you're referring to what my comment previously said, I edited it out after I saw your updated answer. Sorry for any confusion.

And I suppose that's a (bad?) habit of mine.

[Matt J]

No problem. Just one of the side effects of real-time editing...

Answer 2

[Tommy]

Boring:

C = A;
for i = 1:size(A,2)
    C(1:B(i)-1,i) = 0;
end

Ugly:

C = A;
[N,M] = size(C);
C(~any((1:N*M)'>=N*(0:M-1)+B & (1:N*M)'<=N:N:N*M,2)) = 0;

[Matt J]

The for-loop will in many cases be the fastest solution, since it requires the least memory accessing and allocation.