store values from a for loop in a column vector?
显示 更早的评论
I have this function and have to store the values of b(x) over the interval
x=0.01:0.01:5
b0=10;
K=2;
m=3;
for x=0.01:0.01:5
b= b0.*(x.^(m)/(K.^(m)+ x.^(m)));
end
I don't know how to store every value in a column vector, only the value of b(5). any ideas on how to correct this code?
采纳的回答
James Tursa
2017-2-10
编辑：James Tursa
2017-2-10
No need for the for-loop, just a one-liner:
b = b0.*(x.^(m)./(K.^(m)+ x.^(m))); % <-- changed the / to a ./
5 个评论
Margaret Winding
2017-2-10
James Tursa
2017-2-10
Please show the code you are using for fminbnd.
Margaret Winding
2017-2-11
James Tursa
2017-2-11
The first argument of fminbnd should be a function handle so that fminbnd can evaluate the function internally to find the solution. So you could use something like this:
b = @(x)b0.*(x.^(m)./(K.^(m)+ x.^(m))); % <-- create the function handle
[xmax, ymax]= fminbnd(b, 0, 5); % <-- call fminbnd
Or if you really wanted -b make that part of the function handle:
b = @(x)-b0.*(x.^(m)./(K.^(m)+ x.^(m))); % <-- create the function
Margaret Winding
2017-2-11
更多回答（2 个）
James' answer solves the problem efficiently. But if you or anybody else requires a loop, the pre-allocation is important:
x = 0.01:0.01:5
b0 = 10;
K = 2;
m = 3;
b = zeros(size(x)); % Pre-allocate!
for ix = 1:numel(x) % Use index according to vector x
b(ix) = b0 * (x(ix) ^ m / (K ^ m + x(ix) ^ m));
end
Letting an array grow iteratively wastes a lot of resources: In the first iteration a scalar double is reserved and assigned. In the second iteration, a vector of two doubles is reserved, the former contents is copied and the last value is assigned. If you have 1000 elements, Matlab has to reserve sum(1:1000) elements by this way, which are 500'500 and copy almost the same number of doubles. For 1 million elements, we are talking about 2TB of data already, although the final result occupies 8MB RAM only (8 byte per double).
may be you want to keep the for loop because there may be more omitted lines in the loop
x=0.01:0.01:5
b0=10;
K=2;
m=3;
for x=0.01:0.01:5
b= [b;b0.*(x.^(m)/(K.^(m)+ x.^(m)))];
end
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