R help - Feb 2013 - generate variable y to produce excess zero in ZIP analysis

Dear Mr/Mrs

I am Lili Puspita Rahayu, student from magister third level of Statistics in
Bogor Agriculture University.
Mr/
Mrs, now I'm analyzing the Zero inflated Poisson (ZIP), which is a solution
of
the Poisson regression
where the response variable (Y) has zero excess. ZIP now
I was doing did not use real data, but using simulated data in R. Simulations
by generating data on variables x1, x2, x3 with each size 
n = 100, after which generate
data on response variable (Y). However,
when I generate the variable y, after generating variables x1, x2, x3, then the
simulation result in the variable y that does not have a zero excess. Sometimes
just a coincidence there are 23%, 25% the proportion of zero on the variable Y.
This
is because I generate variables x1, x2, x3 with a distribution that has a small
parameter values​​. I've
been consulting with my lecturer, and suggested to generate variable Y that
can control the proportion of zero on ​​ZIP analysis. I've
been trying to make the syntax, but has not succeeded.I would like to ask for
assistance
to R to make the syntax to generate simulated Y variables that can control the
proportion of zeros after generating variables x1, x2, x3 on ZIP analysis.Thus,
I can examine more deeply to determine how much the proportion of zeros on
response variable (Y) that
can be used in the Poisson regression analysis, parametric ZIP and ZIP
semiparametric.

syntax that I made previously by generating variable y without being controlled
to produce zero excess in R :
> b0=1.5
> b1=-log(2)
> b2=log(3)
> b3=log(4)
> n=100
> x1<-rnorm(n, mean=5, sd=2)
> x2<-runif(n, min=1, max=2)
> x3<-rnorm(n, mean=10, sd=15)
> 
> y<-seq(1,n)
> for(i in 1:n)
+ {
+ m<-exp(b0+b1*x1[i]+b2*x2[i]+b3*x3[i])
+ yp<-rpois(1,m)
+ y[i]<-yp
+ }


I am very
grateful for the assistance of R.
I am looking forward to hearing from you. Thank you very much.

Sincerely yours
Lili Puspita Rahayu 
	[[alternative HTML version deleted]]

> Subject: [R] generate variable y to produce excess zero in 
> ZIP analysis
Maybe something like

rzip <-  function(n, lambda, zip=0.0) {
	#zip is the desired proportion of _excess_ zeros
               if(zip>1.0 || zip <0) stop("zip must be in
(0,1)")
               n.zip <- ceiling(zip*n)
               n.pois <- n-n.zip
              sample( c( rpois(n.pois, lambda), rep(0,n.zip) ) )
}

#Example
rzip(50, 3.5, zip=0.5)

Having said that, I'd bet there's a package out there that already does
it better...

 S Ellison
> 
> 
> 
> Dear Mr/Mrs
> 
> I am Lili Puspita Rahayu, student from magister third level 
> of Statistics in Bogor Agriculture University. 
> Mr/
> Mrs, now I'm analyzing the Zero inflated Poisson (ZIP), which 
> is a solution of the Poisson regression where the response 
> variable (Y) has zero excess. ZIP now I was doing did not use 
> real data, but using simulated data in R. Simulations by 
> generating data on variables x1, x2, x3 with each size n = 
> 100, after which generate data on response variable (Y). 
> However, when I generate the variable y, after generating 
> variables x1, x2, x3, then the simulation result in the 
> variable y that does not have a zero excess. Sometimes just a 
> coincidence there are 23%, 25% the proportion of zero on the 
> variable Y. This is because I generate variables x1, x2, x3 
> with a distribution that has a small parameter values??. I've 
> been consulting with my lecturer, and suggested to generate 
> variable Y that can control the proportion of zero on ??ZIP 
> analysis. I've been trying to make the syntax, but has not 
> succeeded.I would like to ask for assistance to R to make the 
> syntax to generate simulated Y variables that can control the 
> proportion of zeros after generating variables x1, x2, x3 on 
> ZIP analysis.Thus, I can examine more deeply to determine how 
> much the proportion of zeros on response variable (Y) that 
> can be used in the Poisson regression analysis, parametric 
> ZIP and ZIP semiparametric.
> 
> syntax that I made previously by generating variable y 
> without being controlled to produce zero excess in R :
> 
> > b0=1.5
> > b1=-log(2)
> > b2=log(3)
> > b3=log(4)
> > n=100
> > x1<-rnorm(n, mean=5, sd=2)
> > x2<-runif(n, min=1, max=2)
> > x3<-rnorm(n, mean=10, sd=15)
> > 
> > y<-seq(1,n)
> > for(i in 1:n)
> + {
> + m<-exp(b0+b1*x1[i]+b2*x2[i]+b3*x3[i])
> + yp<-rpois(1,m)
> + y[i]<-yp
> + }
> 
> 
> I am very
> grateful for the assistance of R.
> I am looking forward to hearing from you. Thank you very much.
> 
> Sincerely yours
> Lili Puspita Rahayu 
> 	[[alternative HTML version deleted]]
> 
> 
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