I have P=[2 3 4 0 5 1 0 2 ]; and I want to create a loop to remove zeros from P and calculate C, description in the code

%% , for example P=[2 0 4 0 5 1 2 0 ]; and after the first iteration P=[2 4 0 5 1 2 0]; and second iteration P=[2 0 4 5 1 2 0 ]; and third one

P=[2 0 4 0 5 1 2 ]; which one will give me the best C .. if I removed the first zero, the second or the thrid one.

%%parameters
clc;
clear;
Pt=5;
Pn=2;
W=rand(4,8);
HT=rand(8,4);
D_Matrix = HT*W_bar;
D_Square = (abs(diag(D_Matrix))').^2;
 P = waterfill(Pt,Pn./D_Square);
 Find_Zero=find(~P); %Find Zero 

if (~isempty(Find_Zero)) 
           while (~isempty(Find_Zero)) 
             First_Z=Find_Zero(1); %First Zero 
             W_bar(:,First_Z) = []; %Delete Col
             HT(First_Z,:) = []; %Delete  Row
             D_Matrix = HT*W_bar; 
             D = (abs(diag(D_Matrix))').^2; 
             P = waterfill(Pt,Pn./D);
             Find_Zero=find(~P); %Find Zero Power
           end

             R = P.*D/Pn;
             C = sum(log2(1+R));  %Capacity
           
          end
    

[Rena Berman]

(Answers Dev) Restored edit

Answer 1

[Jan]

P = [2 0 4 0 5 1 2 0 ];
zeroIndex = find(P == 0);
for k = 1:numel(zeroIndex)
   Q = P;
   Q(zeroIndex(k)) = [];
   ...  Insert your tests of Q here
end