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Questions about lexical scope

3 messages · Lu Chi-Hsien Joseph, Laurent Gautier

Original
Lu Chi-Hsien Joseph
# 
Dear R-users,

The numerical integration example given in Gentleman and Ihaka (2000), 
"Lexical Scope and Statistical Computing," JCGS, 9, 491-508,
is very interesting and helpful in understanding how lexical scope 
is about.
However, I got some questions that I just can't figure out.
First all, allow me to copy the two functions given by the authors:

midpoint <- function(f, a, b, n=100) {
    h <- (b - a)/n
    (b - a)*mean(sapply(seq(a + h/2, b - h/2, length=n), f))
}

bv.integrate <- function(f, a, b, n=100, rule=midpoint) {
    g <- function(y) {
        fx <- function(x) f(x, y)
        rule(fx, a[2], b[2], n)
    }
    rule(g, a[1], b[1])
}

I modified the function name from "integrate" to "bv.integrate".
To integrate f(x,y)=sin(x + y) over the unit square,
it works:
[1] 0.773651
(compare with .7736445 calculated from 2sin(1)*(1 - cos(1)))

Then I write a function using trapezoidal rule as follows:

trapezoidal <- function(f, a, b, n=100) {
    h <- (b - a)/n
    i <- 1:n
    sum(f(a + (i - 1)*h), f(a + i*h))*h/2
}

I checked with
[1] 0.4596939
[1] 0.4596977

and
[1] 0.9972922
[1] 0.9973002

Then, this is what I got by plugged in "trapezoidal" for the "rule" 
in bv.integrate():
[1] 0.007080675

Hundred time smaller! 
Answer can be "improved" by increasing n, but always 100 times smaller.

Anything wrong with the way I wrote trapezoidal()?
I just can't figure out any difference, in terms of input/output,
between midpoint() and trapezoidal().

Then I tried to use integrate() as the rule in bv.integrate(),
by defining a wrap-up function:

intg <- function(f, a, b, n=100) integrate(f, a, b, n)$value

to have the value as the only output.
But, I got
Error in integrate(f, a, b, n) : evaluation of function gave a result 
of wrong length

I tried to use debugger() to find out what's wrong,
but not succeeded.
There must be something I missed in understanding the lexical scope,
however, what's going on with the function I wrote?
What should be the way to correct them?

Certainly, I know integrate() should be the one to use.
These are questions from my preparation of programming exercise 
of using R in my statistical computation class.

I'll be very appreciated for anyone who might help me on these
questions.

Best regards,

C. Joseph Lu
Department of Statistics
National Cheng-Kung University
Tainan, Taiwan, ROC
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3 days later
Laurent Gautier
# 
Dear Joseph,


Did you get any answer about your problem ?

I have been toying with your examples and get a bit scared of what
is happening with lexical scoping...


I edited your function bv.integrate to have some informations about
what was going on.

bv.integrate <- function(f, a, b, n=100, rule=midpoint) {
  assign("count", 0, envir=.GlobalEnv)
  g <- function(y) {
    assign("count", get("count", envir=.GlobalEnv)+1, envir=.GlobalEnv
    fx <- function(x) f(x, y)
    rule(fx, a[2], b[2], n)
  }
  cat("# times in g:", get("count", envir=.GlobalEnv), "\n")
  rule(g, a[1], b[1])
}

my R sesssion gave the following:
# times in g: 0 
[1] 0.007080675
[1] 2
# times in g: 0 
[1] 0.773651
[1] 100


For some reason the function 'g' defined in bv.integrate is applied only
2 times when the rule is trapezoidal and 100 times when the rule is
midpoint (?!)... and the variable 'count' I defined in '.GlobalEnv' has
different values (depending on whether I am in '.GlobalEnv' or in
'bv.integrate'....


I must have missed something obvious somewhere... but where ?




L.
Laurent Gautier
# 
As I was pointed out by Mr.R. himself(*), the call to
'cat' is done *before* the call to the function 'rule'
(hence the variable "count" should be zero... and
is zero...)

I should probably consider using bigger fonts (or bigger glasses)...


L.


(*): thanks Robert
On Sat, Aug 10, 2002 at 01:32:44PM +0200, Laurent Gautier wrote: