How to set value in a matrix

I have a double matrix 4x5 matrix

A=[11 12 13 14 15;

21 22 23 24 25;

31 32 33 34 35;

41 42 43 44 45],

B=[3 0 2 4 1]

B indicates to set all the rows of B[i] in A(from row 1 to that position) to 0.

C = [0 12 0 0 0;

0 22 0 0 25;

0 32 33 0 35;

41 42 43 0 45]

Thanks

Answer 1

[Adam Danz]

for i = 1:numel(B)
    A(1:B(i),i) = 0; 
end

[Rik]

Simple and effective. I tried to see which of the two is faster, but timeit warns me it is running too fast for an accurate measurement.

clc
timeit(@fun1)
timeit(@fun2)
timeit(@fun2_compact)
function A=fun1
A=[11 12 13 14 15;
    21 22 23 24 25;
    31 32 33 34 35;
    41 42 43 44 45];
B=[3 0 2 4 1];
for i = 1:numel(B)
    A(1:B(i),i) = 0;
end
end
function A=fun2
A=[11 12 13 14 15;
    21 22 23 24 25;
    31 32 33 34 35;
    41 42 43 44 45];
B=[3 0 2 4 1];
a=-(1:size(A,1))';
L= a+B(:)' >=0;
A(L)=0;
end
function A=fun2_compact
A=[11 12 13 14 15;
    21 22 23 24 25;
    31 32 33 34 35;
    41 42 43 44 45];
B=[3 0 2 4 1];
A(-(1:size(A,1))'+B(:)'>=0)=0;
end

[Shannon]

Thank you so much, Rik! It is awesome!

[Rik]

You're welcome

[Adam Danz]

Thanks for the timing comparisons, Rik!

One way around the limitations to timeit is by increasing the size of the arrays. With the large array inputs below, both of your solutions are faster.

A = randi(10000,1000); 
B=randi(10000,1,1000)-1;

ans =
      0.18254  % fun1 (my loop)
ans =
     0.026578  % fun2
ans =
     0.026723  % funct_compact

Answer 2

[Rik]

This was a fun puzzle. I believe the code below is one of the many possible solutions. It does use implicit expansion, so for pre-R2016b you will have to use bsxfun.

A=[11 12 13 14 15; 
      21 22 23 24 25; 
      31 32 33 34 35; 
      41 42 43 44 45];
B=[3 0 2 4 1];

a=-(1:size(A,1))';
L= a+B(:)' >=0; %the (:)' is to ensure B is a row vector

C=A;C(L)=0

[Adam Danz]

+1; although I like your compact version even better.

[Rik]

Thank you. While I like the one-linerness of the compact version, I think it looks like voodoo witchcraft, so I decided to post this as the answer

[Adam Danz]

Ha! I know what you mean....