how to get all iterations if I meet NaN?

Uwe Ligges wrote:
Please read the question more carefully, the sin() example was used 
as a method that does not give an error but works as expected (just 
with the warning), but the question is how not to break the loop,
 and so my answer was "see ?try".

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!

Nash wrote:

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)

Please read the question more carefully, the sin() example was used 
as a method that does not give an error but works as expected (just 
with the warning), but the question is how not to break the loop,
 and so my answer was "see ?try".

Uwe Ligges

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