Hello,
Also, t1 and min(xt) do not vary inside the loop so if it enters the
loop it never exits.
And
res1[j] <-(a*h)
res2 <-sum( res1[j])
is equivalent to
res2 <- a*h
so the inner-most loop is not needed at all.
Hope this helps,
Rui Barradas
Em 05-12-2012 04:20, Jim Lemon escreveu:
On 12/05/2012 01:01 AM, anoumou wrote:
Hello all,
I need a help.
I am modeling a disease and a create a R function like that:
...
But i do not get the results,i try by all means but i d'ont
understant the
problem.
Hi anoumou,
Your function provides almost no indication of what two of its five
arguments are supposed to be. If, with a certain degree of optimistic
inference, we suppose "x" to be a data frame organized as shown below
the function. "t", "i" and "CONTAGIEUX" must be the appropriately
named columns of the data frame. This leaves two columns, "Symptomes"
and "Incubation". Say we flip a coin to decide which of these to
assign to "r", and with all of our degrees of freedom gone, we assume
that the other is "h". We are forced to the conclusion that "a" is a
nuisance argument, added to throw us off the scent.
Peering within the function, we notice that the date format is wrong,
braces are unmatched and that our data frame is alphabetically ordered
by name of country to no purpose whatsoever. The best I can do here is
to make the function potentially able to do something if you can work
out what to do with it.
Lambda<-function (x,date1,r,h,a) {
ndate1 <- as.Date(date1, "%Y-%m-%d")
t1 <- as.numeric(ndate1)
x[order(x$i),]
xt <-x[,"t"]
xi <-x[,"i"]
CONTAGIEUX <-x[,"CONTAGIEUX"]
while ( t1 < min(xt) ){
for (i in 1:length(xi) ){
for (j in 1:CONTAGIEUX[length(CONTAGIEUX)]){
res1[j] <-(a*h)
res2 <-sum( res1[j])
}
}
lambda[i] <- r*res2
}
x<-data.frame(x,lambda)
x
}
Jim
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