On Nov 25, 2009, at 1:48 AM, Andreas Wittmann wrote:
Sorry for my bad description, i don't want get a constructed
algorithm without own work. i only hoped to get some advice how to
do this. i don't want to predict any sort of data, i reference only
to newdata which variables are the same as in the model data. But if
factors in the data than i can by possibly that the newdata has a
level which doesn't exist in the original data.
So i have to compare each factor in the data and in the newdata and
if the newdata has a levels which is not in the original data and
drop this variable and do compute the model and prediction again.
I thought this problem is quite common and i can use an algorithm
somebody has already implemented.
best regards
Andreas
If you use str to look at the lm1 object you will find at the bottom a
list called "x":
lm1$x will show you the factors that were present in variables at the
time of the model creation
> lm1$x
$z
[1] "A" "B" "C"
New testing scenario good level and bad level:
test <- data.frame(x=t<-rnorm(2), y=t+rnorm(2), z=c("B", "D") )
lm1 <- lm(x ~ ., data=training)
predict(lm1, subset(test, z %in% lm1$x$z) ) # get prediction for
good level only
1
0.4225204
-------- Original-Nachricht --------
Datum: Wed, 25 Nov 2009 00:48:59 -0500
Von: David Winsemius <dwinsem...@comcast.net>
An: Andreas Wittmann <andreas_wittm...@gmx.de>
CC: r-help@r-project.org
Betreff: Re: [R] predict: remove columns with new levels
automatically
On Nov 24, 2009, at 2:24 PM, Andreas Wittmann wrote:
Dear R-users,
in the follwing thread
http://tolstoy.newcastle.edu.au/R/help/03b/3322.html
the problem how to remove rows for predict that contain levels which
are not in the model.
now i try to do this the other way round and want to remove columns
(variables) in the model which will be later problematic with new
levels for prediction.
## example:
set.seed(0)
x <- rnorm(9)
y <- x + rnorm(9)
training <- data.frame(x=x, y=y, z=c(rep("A", 3), rep("B", 3),
rep("C", 3)))
test <- data.frame(x=t<-rnorm(1), y=t+rnorm(1), z="D")
lm1 <- lm(x ~ ., data=training)
## prediction does not work because the variable z has the new level
"D"
predict(lm1, test)
## solution: the variable z is removed from the model
## the prediction happens without using the information of
variable z
lm2 <- lm(x ~ y, data=training)
predict(lm2, test)
How can i autmatically recognice this and calculate according to
this?
Let me get this straight. You want us to predict in advance (or more
accurately design an algorithm that can see into the future and work
around) any sort of newdata you might later construct????
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
--
Preisknaller: GMX DSL Flatrate für nur 16,99 Euro/mtl.!
http://portal.gmx.net/de/go/dsl02
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
______________________________________________
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.
