predict: remove columns with new levels automatically
Thank you all for the good advice.
Now i did a fast hack, which does want i was looking for, maybe anyone
else finds this usefull
set.seed(0)
x <- rnorm(9)
y <- x + rnorm(9)
training <- data.frame(x=x, y=y,
z1=c(rep("A", 3), rep("B", 3), rep("C", 3)),
z2=c(rep("F", 4), rep("G", 5)))
test <- data.frame(x=t<-rnorm(1), y=t+rnorm(1), z1="D", z2="F")
`predict.drop` <- function(f, dat, newdat)
{
datlev <- vector("list", ncol(dat))
newdatlev <- vector("list", ncol(newdat))
`filllevs` <- function(dat, veclev)
{
for (j in 1:ncol(dat))
{
if (is.factor(dat[,j]))
veclev[[j]] <- levels(dat[,j])
else
veclev[[j]] <- NULL
}
return(veclev)
}
datlev <- filllevs(dat, datlev)
newdatlev <- filllevs(newdat, newdatlev)
if (ncol(dat) == ncol(newdat))
{
drop <- logical(ncol(dat))
names(drop) <- colnames(dat)
for (j in 1:ncol(dat))
{
if (!is.null(datlev[[j]]))
{
if (!(newdatlev[[j]] %in% datlev[[j]]))
drop[j] <- TRUE
}
}
}
else
stop("dat and newdat must have the same column length!")
m <- lm(formula(f), data=dat[,(1:ncol(dat))[!drop]])
p <- predict(m, newdat)
return(list(drop=drop, p=p))
}
predict.drop(x ~ ., training, test)
best regards
Andreas
David Winsemius wrote:
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 <dwinsemius at comcast.net> An: Andreas Wittmann <andreas_wittmann at gmx.de> CC: r-help at 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 f?r nur 16,99 Euro/mtl.! http://portal.gmx.net/de/go/dsl02
David Winsemius, MD Heritage Laboratories West Hartford, CT