Reputation: 21
I have code like below:
N=10;
R=[1 1 1 1 1 0 0 0 0 0;1 1 1 1 1 1 1 1 1 1];
p=[0.1,0.2,0.01];
B = zeros(N , N);
B(1:N,1:N) = eye(N);
C=[B;R];
for q=p(1:length(p))
Rp=C;
for i=1:N
if(rand < p)
Rp(i,:) = 0;
end
end
end
from this code I vary the value of p
. So for different value of p
, i am getting different Rp
. Now I want to get the total number of "1"'s from each Rp
matrix. it means may be for p1 I am getting Rp1=5, for p2, Rp=4.
For example
Rp1=[1 0 0 0 0;0 1 0 0 0;0 0 0 0 0],
Rp2=[1 0 0 0 0;0 1 0 0 0;1 0 0 0 0],
Rp3=[0 0 0 0 0;0 1 0 0 0;0 0 0 0 0],
So total result will be 2,3,1.
I want to get this result.
Upvotes: 2
Views: 413
Reputation: 21563
Assuming Rp is your matrix, then simply do one of the following:
If your matrix only contains zeros and ones
sum(Rp(:))
Or if your matrix contains multiple values:
sum(Rp(:)==1)
Note that for two dimensional matrices sum(Rp(:))
is the same as sum(sum(Rp))
I think your real question is how to save this result, you can do this by assigning it to an indexed varable, for example:
S(count) = sum(Rp(:));
This will require you to add a count variable that increases with one every step of the loop. It will be good practice (and efficient) to initialize your variable properly before the loop:
S = zeros(length(p),1);
Upvotes: 2
Reputation: 9864
If the matrix contains only 0
and 1
you are trying to count the nonzero values and there is a function for that called nnz
n = nnz(Rp);
As I mentioned in the comments you should replace
if(rand < p)
with
if(rand < q)
Then you can add the number of nonzero values to a vector like
r = [];
for q=p(1:length(p))
Rp=C;
for i=1:N
if(rand < p)
Rp(i,:) = 0;
end
end
r = [r nnz(Rp)];
end
Then r
will contain your desired result. There are many ways to improve your code as mentioned in other answers and comments.
Upvotes: 2
Reputation: 1908
If you need to count the 1's in any matrix M
you should be able to do sum(M(:)==1)
Upvotes: 0