Re: [R] 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.
So,
Do you have any solution about his problem ?
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!
On Sat, 28 Mar 2009 16:35:24 +0100, Uwe Ligges wrote
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]]
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Re: [R] how to get all iterations if I meet NaN?
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 Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - (p.dalga...@biostat.ku.dk) FAX: (+45) 35327907
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Re: [R] how to get all iterations if I meet NaN?
Okay!
Thank you!
On Sun, 29 Mar 2009 10:01:29 +0200, Peter Dalgaard wrote
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 喒ter Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - (p.dalga...@biostat.ku.dk) FAX: (+45) 35327907
--
Nash - morri...@ibms.sinica.edu.tw
__
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Re: [R] 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!
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Nash - morri...@ibms.sinica.edu.tw
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to get all iterations if I meet NaN?
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]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Nash - morri...@ibms.sinica.edu.tw
__
R-help@r-project.org mailing list
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and provide commented, minimal, self-contained, reproducible code.
__
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[R] 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]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to get all iterations if I meet NaN?
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?
See ?try
Uwe Ligges
Thanks!
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to get all iterations if I meet NaN?
2009/3/27 huiming song huimi...@gmail.com:
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?
not sure I properly understood. Consider:
x = seq(-pi,pi,length=1001)
z = sin(1/x)
# Warning message:
# In sin(1/x) : NaNs produced
x[500:502]; z[500:502]
# [1] -0.006283185 0.0 0.006283185
# [1] -0.8754095NaN 0.8754095
one NaN and one warning have been created, the remaining 1000
calculations has been executed.
lim(x-0)sin(1/x) not exists so sin(1/0) is not a number nan
z1 = z[!is.nan(z)]
x1 = x[!is.nan(z)]
# x and z without the z's nan position
hope that help
Patrizio
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