On 22-10-2012, at 09:49, Roslina Zakaria wrote:
> Dear r-users,
>
> I would like to convert my Matlab code to R-code, however it dies not work as
> expected. Hope somebody can help me to match Matlab and r codes.
>
> R code:
>
> rr <- function(r,cxn)
> {
> tol <- 1E-4;
>
> for(i in 1:n)
> {
> t1 <- (1+(i-1)*r)*log((1+(i-1)*r))
> t2 <- (i-1)*(1-r)*log(1-r)
> rri <- ((t1+t2)/i*log(i))-cxn
The first ( in the expression for rri is misplaced and should be moved to just
after the / by comparing to the expression for f in the Matlab code.
> rr <- rri > tol
> }
> round(rr,4)
> }
> rr1 <- rr(0.5,0.0242) ; rr1
>
What is this function supposed to do? Incomprehensible.
In the for with i rri is compared to tol and rr becomes a logical.
After the loop is finished rr is equal to the last value of rr for i==n.
So you could just as well have set i to n and then you could dispense with the
for loop.
Why is the function returning round(rr,4) when rr is a logical?
>
> Matlab code:
>
> function F = cxncnr(r)
> n = 4;
> % terms
> t1 = (1+(n-1)*r)*log((1+(n-1)*r));
> t2 = (n-1)*(1-r)*log(1-r);
>
> %f = term - cxn
> f = (t1+t2)/(n*log(n)) - 0.05011007
>
> F = [f];
> % r0 = [0.5] ; r = fsolve(@cxncnr,r0)
>From this code I gather that you are trying to solve an equation with one
>variable.
Some code:
cxncnr <- function(r) {
n <- 4
t1 <- (1+(n-1)*r)*log((1+(n-1)*r))
t2 <- (n-1)*(1-r)*log(1-r)
X <- (t1+t2)/(n*log(n)) - 0.05011007
X
}
# a plot to determine endpoints
curve(cxncnr,from=0.0,to=.5)
sol <- uniroot(cxncnr,c(.01,.50))
sol
cxncnr(sol$root)
Berend
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