priority of operators in the FOR ( ) statement

-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch 
[mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Patrick Burns
Sent: Tuesday, August 23, 2005 4:12 AM
To: Ravi.Vishnu at outokumpu.com
Cc: r-help at stat.math.ethz.ch
Subject: Re: [R] priority of operators in the FOR ( ) statement

The command that I think is most useful in this situation
is 'browser()'.

Even a couple decades of programming in the S language
hasn't yet solved the problem of my fingers typing code that
doesn't match what I want to happen.  I quite consistently have
a

browser()

call in functions that I write to make sure that what I am assuming
is the same as what R assumes.


Patrick Burns
patrick at burns-stat.com
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and "A Guide for the Unwilling S User")

Ravi.Vishnu at outokumpu.com wrote:

Dear All,
I spent an entire evening in debugging a small, fairly 

simple program in R 

- without success. It was my Guru in Bayesian Analysis, 

Thomas Fridtjof, 

who was able to diagonose the problem. He said that it took 

a long time 

for him also to locate the problem.
This program illustrates in some ways the shortcomings of the error 
messages that R responds with. In this case, it was quite 

misleading and 

directs attention to a location far removed the actual 

problem statement.

Without any more introductory comments, let me directly discuss the 
essential problem. I am enclosing the entire program after a brief 
discussion.

The problem arises from the following statement (nr is an integer 
constant) :
for ( i in 1:nr-1) {.......}
The unexpected problem (at least for me) is that R reads the above 
statement as (i in (1:nr)-1) {.....}. This makes i be 

initially as zero 

which leads to an error because the for loop in R starts from 1. The 
problem is easily fixed by writing the for loop as  ( i in 1:(nr-1)) 
{.......}. This would be an easy problem to fix if R 

directly indicates 

what the problem is. Instead, it gives mystifying error 

messages which are 

totally misleading. For example, to the program given below, 

I got the 

following error message (these point to commands elsewhere 

in the program) 

:
Error in if ((x >= 0) & (x < s2)) return(x/2) else if ((x >= 

s2) & (x <  : 

       missing value where TRUE/FALSE needed

I would like clarifications on the following points :
1. I am just curious to know if the priority of operators in the for 
statement ( the colon before the minus operator, for example) is a 
deliberate design decision. I have tested Matlab and found that it 
interprets my original statement correctly without an extra 

paranthesis.

2. Faced with a similiar problem in the future, what is a 

smart way of 

debugging in R to locate a problem. With this problem, I checked and 
double checked every single statement in the program, except the for 
statement because I just did not expect any problem there. I 

have seen 

that there is a debug package but I have not used it. Can 

such tools be 

used to locate a problem with greater ease? Can somebody 

give a concrete 

example (for the following program, for example) of a 

debugging routine.

*************************************************************

************'

# Bayesian Data Analysis
## source("M:/programming/Rfolder/Assignments/fortest.txt")

# #Remove all objects from the workspace
rm(list=ls())
# #We will also try to note the time that the program takes
# #We will start the clock at starttime
starttime <- proc.time()[3];

my.function<-function(x) {
s2<-sqrt(2);
if ((x>=0) & (x<s2)) return(x/2)
else 
if ((x>=s2) & (x<1+s2)) return(0.2)
else 
if ((x>=1+s2) & (x<1.5+s2)) return(0.6)
else 
if ((x>1.5+s2) | (x<0)) return(0)
}

alphayx<-function(y,x) {
fy<-my.function(y)
fx<-my.function(x)
fyx<-fy/fx
# to account for 0/0 division
if (is.na(fyx)) fyx<-0
#fyx<-ifelse(is.na(fyx),0,fyx);
alpha<-min(1,fyx)
return(alpha)
}

sigma<-0.5;
#nr is the number of iterations
nr<-20
x<-numeric(nr);
x[1]<-1;
t<-1:nr;

for (i in 1:nr-1) {
xi<-x[i];
yi<-rnorm(1,mean=xi,sd=sigma);
ui<-runif(1,0,1);
ualphai<-alphayx(yi,xi);
xn<-ifelse(ui<=ualphai,yi,xi);
x[i+1]<-xn;
}

plot(t,x,type="p")

endtime<-proc.time()[3];
elapsedTime<-endtime-starttime;
cat("Elapsed time is", elapsedTime, "seconds", "\n")
*************************************************************

****************'

 
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