How to replace k-th diagonal by vector?

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Clarisha Nijman
Clarisha Nijman 2018-10-2
回答： Clarisha Nijman 2018-10-3
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?

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采纳的回答

Guillaume
Guillaume 2018-10-2
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.
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Clarisha Nijman
Clarisha Nijman 2018-10-3
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
Guillaume 2018-10-3
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

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更多回答（3 个）

Bruno Luong
Bruno Luong 2018-10-3
编辑：Bruno Luong 2018-10-3
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
>>
Adam
Adam 2018-10-2
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!
  2 个评论
Clarisha Nijman
Clarisha Nijman 2018-10-2
Can you pls explain your answer a little bit more?
Guillaume
Guillaume 2018-10-2
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.

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Clarisha Nijman
Clarisha Nijman 2018-10-3
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!!!

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