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

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Rabia Zulfiqar on 1 Jun 2020
Edited: Tommy on 2 Jun 2020
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.
C=[1 0 0 0
5 0 7 0
9 1 3 0
8 7 5 3]
How can I do that?

madhan ravi on 1 Jun 2020
You need to explain more
Rabia Zulfiqar on 1 Jun 2020
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.

Matt J on 1 Jun 2020
Edited: Matt J on 1 Jun 2020
[m,n]=size(A);
C=A.*((1:m).'>=B);

Show 1 older comment
Matt J on 1 Jun 2020
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 on 1 Jun 2020
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 on 1 Jun 2020
No problem. Just one of the side effects of real-time editing...

Tommy on 1 Jun 2020
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;

#### 1 Comment

Matt J on 1 Jun 2020
The for-loop will in many cases be the fastest solution, since it requires the least memory accessing and allocation.