No visible binding for global variable

Dear Jamil,

Presumably, the variables alloc.dose, r0, r1, and r2 will exist in the
global environment when you call the compiled function lf1.c(), but didn?t
exist when you compiled the function. If so, then lf1.c() should work
properly despite the notes (not errors, or even warnings) that was printed.

On the other hand, relying on global data in a function is poor
programming style ? it would be better to make alloc.dose, etc., arguments
to the function. As well, if you?re compiling the function to improve
efficiency, then you probably also want to compile psi0(), etc.

I hope that this helps,
 John

John Fox
Sen. William McMaster Professor of Social Statistics
McMaster University
Hamilton, Ontario, Canada
Web: http::/socserv.mcmaster.ca/jfox

On Jan 27, 2017, at 10:43 AM, Naser Jamil <jamilnaser79 at gmail.com>

wrote:

Dear R-users,
I would like to seek your suggestion. I have the following code which

runs

smoothly. But when I compile the function (lf1.c), it shows "no visible
binding for global variable" for some of the arguments.

###################################

library(compiler)


psi0<-function(theta1,theta2,theta3,theta4,x){
z1<-exp(theta1+theta2*x)
z2<-exp(theta3+theta4*x)
1/((1+z1)*(1+z2))
                                            }

psi1<-function(theta1,theta2,theta3,theta4,x){
z1<-exp(theta1+theta2*x)
z2<-exp(theta3+theta4*x)
z1/((1+z1)*(1+z2))
                                            }
psi2<-function(theta3,theta4,x) {
z2<-exp(theta3+theta4*x)
z2/(1+z2)
                               }



lf1<-function(w) {
  v<-1
  w1<-w[1]
  w2<-w[2]
  w3<-w[3]
  w4<-w[4]
  for (i in 1:length(alloc.dose)) {
      dose.i<-alloc.dose[i]
      r0.i<-r0[i]
      r1.i<-r1[i]
      r2.i<-r2[i]
      z1<-exp(w1+w2*dose.i)
      z2<-exp(w3+w4*dose.i)
      psi0<-1/((1+z1)*(1+z2))
      psi1<-z1*psi0
      v<-v*(psi0^r0.i)*(psi1^r1.i)*((1-psi0-psi1)^r2.i)
                                  }
  return(v)
                 }


lf1.c<-cmpfun(lf1)

###############################

May I know how to avoid this message? If I leave the code as it is, will
that affect the result anyway?

Thanks in advance.


Regards,
Jamil.

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