R help - Dec 2010 - rpart - how to estimate the “meaningful” predictors for an outcome (in classification trees)

Hi dear R-help memebers,

When building a CART model (specifically classification tree) using rpart,
it is sometimes obvious that there are variables (X's) that are meaningful
for predicting some of the outcome (y) variables - while other predictors
are relevant for other outcome variables (y's only).

*How can it be estimated, which explanatory variable is "used" for
which of
the predicted value in the outcome variable?*

Here is an example code in which x2 is the only important variable for
predicting "b" (one of the y outcomes). There is no predicting
variable for
"c", and x1 is a predictor for "a", assuming that x2 permits
it.

How can this situation be shown using the an rpart fitted model?

N <- 200
set.seed(5123)

x1 <- runif(N)

x2 <- runif(N)

x3 <- runif(N)

y <- sample(letters[1:3], N, T)

y[x1 <.5] <- "a"

y[x2 <.1] <- "b"

fit <- rpart(y ~ x1+x2)

fit2 <- prune(fit, cp= 0.07)

plot(fit2)

text(fit2, use.n=TRUE)

Thanks,

Tal



----------------Contact
Details:-------------------------------------------------------
Contact me: Tal.Galili@gmail.com |  972-52-7275845
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Hi, Tal,

Here is a quick way of getting around. First create two responses via
dummy variables

y1 <- ifelse(y=="a", 1, 0)
y2 <- ifelse(y=="b", 1, 0)

and then built two separate tree models for y1 and y2 separately.

Hope it helps.
Xiaogang


On Tue, Dec 14, 2010 at 8:33 AM, Tal Galili <tal.galili at gmail.com>
wrote:
> Hi dear R-help memebers,
>
> When building a CART model (specifically classification tree) using rpart,
> it is sometimes obvious that there are variables (X's) that are
meaningful
> for predicting some of the outcome (y) variables - while other predictors
> are relevant for other outcome variables (y's only).
>
> *How can it be estimated, which explanatory variable is "used"
for which of
> the predicted value in the outcome variable?*
>
> Here is an example code in which x2 is the only important variable for
> predicting "b" (one of the y outcomes). There is no predicting
variable for
> "c", and x1 is a predictor for "a", assuming that x2
permits it.
>
> How can this situation be shown using the an rpart fitted model?
>
> N <- 200
> set.seed(5123)
>
> x1 <- runif(N)
>
> x2 <- runif(N)
>
> x3 <- runif(N)
>
> y <- sample(letters[1:3], N, T)
>
> y[x1 <.5] <- "a"
>
> y[x2 <.1] <- "b"
>
> fit <- rpart(y ~ x1+x2)
>
> fit2 <- prune(fit, cp= 0.07)
>
> plot(fit2)
>
> text(fit2, use.n=TRUE)
>
> Thanks,
>
> Tal
>
>
>
> ----------------Contact
> Details:-------------------------------------------------------
> Contact me: Tal.Galili at gmail.com | ?972-52-7275845
> Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
> www.r-statistics.com (English)
>
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>
> ? ? ? ?[[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
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> PLEASE do read the posting guide
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> and provide commented, minimal, self-contained, reproducible code.
>


-- 
=============================Xiaogang Su, Ph.D.
Associate Professor, Statistician
School of Nursing, University of Alabama
Birmingham, AL 35294-1210
(205) 934-2355 [Office]
xgsu at uab.edu
xiaogangsu at gmail.com
http://homepage.uab.edu/xgsu/