How to replace k-th diagonal by vector?

I have a code here that give me errors. In the code I am computing a i-th dot product and want to replace this values on the i-th super diagonal.

lambda=0.42;
n=100;
p=0.005;
P=zeros(1001,1001);

for i=1:1000
   a=i:1000;
   b=a-i;
   v=poisspdf(a,lambda);
   w=binopdf(b,n,p);
   c=dot(v,w);
   d=size(diag(P,i),1);%this is the size of the vector with elements of the kth diagonal
   e=c*ones(d,1);
   diag(P,i)=e;

end

But matlab gives me an error saying:

Subscript indices must either be real positive integers or logicals.

Error in pdfnumber (line 18)
   diag(P,i)=e;

Then I tried a more simple code. But is only replaces the 1 super diagonal for me, while i runs from 1 to 1000.

for i=1:1000
   a=i:1000;
   b=a-i;
   v=poisspdf(a,lambda);
   w=binopdf(b,n,p);
   c=dot(v,w);
   P(i,i+1)=c;
end

What am I doing wrong?

Answer 1

[Guillaume]

You can't use diag to assign to diagonales of a matrix, only to get them. The simplest way to assign to the diagonales is to replace:

d=size(diag(P,i),1);%this is the size of the vector with elements of the kth diagonal
e=c*ones(d,1);
diag(P,i)=e;

by

P(i*size(P, 1)+1:size(P, 1)+1:end) = e;

which simply computes the linear indices of the diagonal elements.

[Clarisha Nijman]

Thanks a lot everyone, I understood it better now, and I will try to apply the suggested code. More suggestions are always welcome.

[Clarisha Nijman]

Hallo,

The suggestion of Guillame works the best

P(i*size(P, 1)+1:size(P, 1)+1:end) = e;

But there is still something more:

It only replaces the super-diagonals. As I do not fully understand the code, I am trying a lot of other combination to replace also the sub-diagonals.

Things as: P(-i*size(P, 1)+1:size(P, 1)+1:end) = e; P(start:i*size(P, 1)+1:size(P, 1)+1) = e; P(begin:i*size(P, 1)+1:size(P, 1)+1) = e;

But alas! I did not succeed. Can you help me also with this?

kind regards,

Clarisha

[Guillaume]

Well, your original code did not access the subdiagonals so I assumed that you were only interested in the super-diagonals.

For subdiagional -i, use:

P(i+1:size(P, 1)+1:1+size(P, 1)*min(size(P, 1)-i,size(P, 2))) = e;  %where i is the positive index of subdiagonal -i. 0 is main diagonal

Answer 2

[Bruno Luong]

You can use the function IDIAG in this FEX

>> P = rand(5,4)

P =

      0.3517    0.2858    0.0759    0.1299
      0.8308    0.7572    0.0540    0.5688
      0.5853    0.7537    0.5308    0.4694
      0.5497    0.3804    0.7792    0.0119
      0.9172    0.5678    0.9340    0.3371

>> P(idiag(size(P),-1)) = (1:4)

P =

      0.3517    0.2858    0.0759    0.1299
      1.0000    0.7572    0.0540    0.5688
      0.5853    2.0000    0.5308    0.4694
      0.5497    0.3804    3.0000    0.0119
      0.9172    0.5678    0.9340    4.0000

>>

Answer 3

[Adam]

P is defined as a matrix of all zeros and you appear to be overwriting the 'diag' function with a variable of the same name which you then try to index into using P. The entirety of P, in fact!

[Clarisha Nijman]

Can you pls explain your answer a little bit more?

[Guillaume]

You cannot assign a value to a function, so

somefunction(atsomevalue) = something

is never correct in matlab. What you are attempting to do is assign a new value to the return value of the diag function, so if we'd made that explicit, your

diag(P, i) = e;

could be decomposed to:

tempvariable = diag(P, i);
tempvariable = e;

Even if matlab allowed that, you can see it still wouldn't work as you would only be changing the value of the temporary, not the diagonale of P.

Since assigning a value to a function is not allowed and since matlab allows you to use the same names for functions and variables, matlab interpret your diag line as an assigment of e to the matrix diag at rows P and column i. Some values in P are 0 which is not a valid row index, hence the error.

Answer 4

[Clarisha Nijman]

Yes Guillaume! It works! Thanks a lot. You are indeed right about the code. I pasted the code partially, and expected that the leading question: How to replace k-th diagonal by vector?, would cover the rest.

But thanks anyway!!!