how to get all iterations if I meet NaN?

hi, you can try this method.

## if x is a vector
x <- runif(1000)
## if sin(1/0) will appear NaN. we make the situation. 
x[sample(1:length(x),5)] <- 0   
z <- suppressWarnings(ifelse(is.na(sin(1/x)),NA,sin(1/x)))
is.numeric(z)

## if x is a matrix
x=matrix(runif(1000),100,10)
x[sample(1:nrow(x),50),sample(1:ncol(x),5)] <- 0
z <- suppressWarnings(ifelse(is.na(sin(1/x)),NA,sin(1/x)))
is.numeric(z)

On Sat, 28 Mar 2009 01:36:36 +0800, huiming song wrote

hi, everybody, please help me with this question:

If I want to do iteration for 1000 times, however, for the 500th 
iteration, there is NaN appears. Then the iteration will stop. If I 
don't want the stop and want the all the 1000 iterations be done. 
What shall I do?

suppose I have x[1:1000] and z[1:1000],I want to do some calculation 
for all x[1] to x[1000].

z=rep(0,1000)
for (i in 1:1000){
  z[i]=sin(1/x[i])
}

if x[900] is 0, in the above code it will not stop when NaN appears. 
Suppose when sin(1/x[900]) is NaN appears and the iteration will now 
fulfill the rest 100 iterations. How can I write a code to let all 
the 1000 iterations be done?

Thanks!

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Nash - morrison at ibms.sinica.edu.tw