Ciro,
It has to do with the way that the
length
function works. It will find and return the largest dimension in a 2-D array. Since you start with an empty array for
xvals
its initial length is 0, and after the first interation of your loop if becomes a 2x1 column vector with length 2. You
xvals
variable then continues to grow by 1 column each iteration of the loop. By the third iteration it is a 2x3 array and
length
returns 3 since the number of columns respresents the largest dimension.
xvals = [];
idx = 1;
beta = 3;
xold = 0.5;
% Trans
fori = 1:10000
xnew = (xold-xold^2)*beta;
xold = xnew;
end
fori=1:5
xnew = (xold-xold^2)*beta;
xold = xnew;
fprintf("Interations: %i,\tlength(xvals)= %i\n",i,length(xvals))
idx = length(xvals)+1;
xvals(1,idx) = beta;
xvals(2,idx) = xnew;
end
If you are intending to focus on the size of array in a fixed dimension, you can you
size
instead of
length
.
size(xvals,1)% Returns the row size
size(xvals,2)%返回列的大小
[r,c] = size(xvals);% Returns both the row and column sizes as a 1x2 array for 2-D arrays
If you are trying to have your
idx
variable increase by 1 each iteration, then you should just use your loop variable,
i
, that is already increasing by 1.
xvals = [];
beta = 3;
xold = 0.5;
% Trans
fori = 1:10000
xnew = (xold-xold^2)*beta;
xold = xnew;
end
fori=1:5
xnew = (xold-xold^2)*beta;
xold = xnew;
fprintf("%d\t%d\n",beta,i);
xvals(1,i) = beta;
xvals(2,i) = xnew;
end
