Re: [R] glm output format
Look at help(tryCatch) and help(try).
Hope this helps a little
Allan
On 05/06/10 04:21, Bojuan Zhao wrote:
Hello,
I am running a loop to compare some residual deviances obtained from glm, with codes:
OUT-NULL
for (i in 1:10){
myglm-glm(mat ~X1+X2+X3,family = binomial, data =myDATA)
OUT-c(OUT,myglm$deviance) }
...
In the loop, X1, X2, and X3 chage with i. If X1, X2 and X3 are not highly correlated, OUT is a vector of 10 values.
The problem is: if there there is one set of X1,X2,X3 that are highly correlated, the program stopped, with error message: qr.solve(qr.R(object$qr)[p1, p1]) : singular matrix 'a' in solve.
I wonder if there is a way I can skip the highly correlated case and get a NA as the response, instead of letting the whole loop stop with no OUT output at all. That is, I want to get the OUT vector with 9 values and one NA.
Is there a way I can do so?
Many Thanks,
Barbara
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[R] glm output format
Hello,
I am running a loop to compare some residual deviances obtained from glm, with
codes:
OUT-NULL
for (i in 1:10){
myglm-glm(mat ~X1+X2+X3,family = binomial, data =myDATA)
OUT-c(OUT,myglm$deviance) }
...
In the loop, X1, X2, and X3 chage with i. If X1, X2 and X3 are not highly
correlated, OUT is a vector of 10 values.
The problem is: if there there is one set of X1,X2,X3 that are highly
correlated, the program stopped, with error message:
qr.solve(qr.R(object$qr)[p1, p1]) : singular matrix 'a' in solve.
I wonder if there is a way I can skip the highly correlated case and get a NA
as the response, instead of letting the whole loop stop with no OUT output at
all. That is, I want to get the OUT vector with 9 values and one NA.
Is there a way I can do so?
Many Thanks,
Barbara
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__
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.
