how to get all iterations if I meet NaN?

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!

	[[alternative HTML version deleted]]

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?

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!

	[[alternative HTML version deleted]]

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!

	[[alternative HTML version deleted]]

guide.html

--
Nash - morrison at ibms.sinica.edu.tw

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!

	[[alternative HTML version deleted]]

guide.html

--
Nash - morrison at ibms.sinica.edu.tw

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".

So,
Do you have any solution about his problem ?

Nash wrote:

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".

So,
Do you have any solution about his problem ?

Yes he does: Use try().

This may not be very specific, but neither was the original 
question. The poster didn't give an example that we could fix, only 
the sin() example that he wanted his own code to work _like_.

So to answer the question with working code, one first needs to 
invent a broken example. Perhaps something like

for(i in 2:-2) {
    x <- try(chol(as.matrix(i)))
    if(inherits(x,"try-error")) x <- NaN
    print(x)
}

-- 
    O__  ---- Peter Dalgaard             ?ter Farimagsgade 5, Entr.B
   c/ /'_ --- Dept. of Biostatistics     PO Box 2099, 1014 Cph. K

  (*) \(*) -- University of Copenhagen   Denmark      Ph:  (+45) 35327918
~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk)              FAX: (+45) 35327907