could anyone tell me how to store the result of each iteration in an array for the following code

code:

t=4;
r=6;
for it=1:4
output(t,r)=overall_throughput;
output_it(t,r,it)=output(t,r);
end

Answer 1

[Walter Roberson]

t=4;
r=6;
for it=1:4
output(t,r)=overall_throughput;
output_it(t,r,it)=output(t,r);
all_output{it} = output;
all_output_it{it} = output_it;
end

Now a complete record of the results of the iterations is stored in all_output and all_output_it

[Prabha Kumaresan]

When i run the following code i am getting error

Conversion to cell from double is not
possible.

Bmax=3000;
noise=10;
A=[0         0         0         0    0.0447         0
   0         0         0         0    0.0294         0]
B=[0         0         0         0    0.2428         0
   0         0         0         0    0.4424         0]
A=[0         0    1.9728         0         0         0
   0         0    0.0000         0         0         0]
B=[0         0    0.1971         0         0         0
   0         0    0.2703         0         0         0]
A=[0.2373    0.2169         0    0.1979         0    0.2098
   0.0174    0.0186         0    0.0214         0    0.0249]
B=[0.2160    0.7904         0    0.4386         0    0.8620
   0.3774    0.9493         0    0.8335         0    0.9899]
N = length(A) ;
throughput = cell(N,1) ;

for v =1:size(B,2)
for u =1:size(A,1)
for k =1:N
    throughput{k}=throughput(u,v)
    throughput(u,v) =(Bmax.*log2(1+(((A(u,v)).*(B(u,v))/(noise+sum(B(1:u-1,v)).*A(u,v))))))

end
end
end

could you please help me on this.

[Walter Roberson]

On the line

      throughput{k}=throughput(u,v)

you are taking the content of the cell array throughput(u,v) and placing the cell array as the content of throughput(k, 1), resulting in nested cell arrays.

On the line

      throughput(u,v) =(Bmax.*log2(1+(((A(u,v)).*(B(u,v))/(noise+sum(B(1:u-1,v)).*A(u,v))))))

you are taking the numeric value after the "=" and trying to store it as if it were a cell array into location throughput(u,v) .

Perhaps instead of using throughput(u,v) in those places, you might want throughput{u,v}

And it is really not recommended to mix linear indexing {k} with two-dimensional indexing in the same group of code.

If you end up mixing () notation on a cell and {} notation on the cell, then you should double-check the code to be sure you really want to work with the content as being containers, which is () notation, or if you want to work with what is inside the containers, which is {} notation. Sometimes it makes sense to use both in the same block of code, but more often it indicates that cells are being misused.

[Prabha Kumaresan]

ok. could you tell me is there any other way to solve the issue for the following code:

A=[0         0         0         0    0.0447         0
   0         0         0         0    0.0294         0]
B=[0         0         0         0    0.2428         0
   0         0         0         0    0.4424         0]
A=[0         0    1.9728         0         0         0
   0         0    0.0000         0         0         0]
B=[0         0    0.1971         0         0         0
   0         0    0.2703         0         0         0]
A=[0.2373    0.2169         0    0.1979         0    0.2098
   0.0174    0.0186         0    0.0214         0    0.0249]
B=[0.2160    0.7904         0    0.4386         0    0.8620
   0.3774    0.9493         0    0.8335         0    0.9899]

for v =1:size(B,2)
for u =1:size(A,1)
throughput(u,v) =(Bmax.*log2(1+(((A(u,v)).*(B(u,v))/(noise+sum(B(1:u-1,v)).*A(u,v)))))) 
end
end

[Walter Roberson]

throughput{u,v} = ...

[Prabha Kumaresan]

I have modified the line to

throughput{u,v} =(C.*log2(1+(((A{u,v}).*(B{u,v})/(w+sum(B(1:u-1,v)).*A{u,v})))))

but still i am getting error. Cell contents reference from a non-cell array object.