Classification Tree Prediction Error

Dear all R experts,

I have a question about using cross-validation to assess results estimated
from a classification tree model. I annotated what each line does in the R
code chunk below. Basically, I split the data, named usedta, into 70% vs.
30%, with the training set having 70% and the test set 30% of the original
cases. After splitting the data, I first run a classification tree off the
training set, and then use the results for cross-validation using the test
set. It turns out that if I don't have any predictors and make predictions
by simply betting on the majority class of the zero-one coding of the
binary response variable, I can do better than what the results from the
classification tree would deliver in the test set. What would this imply
and what would cause this problem? Does it mean that classification tree is
not an appropriate method for my data; or, it's because I have too few
variables? Thanks a lot!

Jun Xu, PhD
Professor
Department of Sociology
Ball State University
Muncie, IN 47306
USA

Using the estimates, I get the following prediction rate (correct
prediction) using the test set. Or we can say the misclassification error
rate is 1-0.837 = 0.163

(tab[1,1] + tab[2,2]) / sum(tab)[1] 0.837

Without any predictors, I can get the following rate by betting on the
majority class every time, again using data from the test set. In this
case, the misclassification error rate is 1-0.85 = 0.15

table(h2.test)h2.test

1poorHlth 0goodHlth
      101       575 > 571/(571+101)[1] 0.85



R Code Chunk

# set the seed for random number generator for replication
set.seed(47306)
# have the 7/3 split with 70% of the cases allotted to the training set
# AND create the training set identifier
class.train = sample(1:nrow(usedta), nrow(usedta)*0.7)
# create the test set indicator
class.test = (-class.train)
# create a vector for the binary response variable from the test set
# for future cross-tabulation.
h2.test <- usedta$h2[class.test]
# count the train set cases
Ntrain = length(usedta$h2[class.train])
# run the classification tree model using the training set
# h2 is the binary response and other variables are predictors
tree.h2 <- tree(h2 ~ age + educ + female + white + married + happy,
                data = usedta, subset = class.train,
                control = tree.control(nobs=Ntrain, mindev=0.003))
# summary results
summary(tree.h2)
# make predictions of h2 using the test set
tree.h2.pred <- predict(tree.h2, usedta[class.test,], type="class")
# cross tab the predictions using the test set
table(tree.h2.pred, h2.test)
tab = table(tree.h2.pred, h2.test)
# calculate the ratio for the correctly predicted in the test set
(tab[1,1] + tab[2,2]) / sum(tab)
# calculate the ratio for the correctly predicted using the naive approach
# by betting on the majority category.
table(h2.test)[2]/sum(tab)

        [[alternative HTML version deleted]]

Purely statistical questions -- as opposed to R programming queries -- are
generally off topic here.
Here is where they are on topic:  https://stats.stackexchange.com/

Suggestion: when you post, do include the package name where you get
tree() from, as there might be
more than one with this function.

Bert Gunter

"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Mon, Aug 24, 2020 at 8:58 AM Xu Jun <junxu.r at gmail.com> wrote:

Dear all R experts,

I have a question about using cross-validation to assess results estimated
from a classification tree model. I annotated what each line does in the R
code chunk below. Basically, I split the data, named usedta, into 70% vs.
30%, with the training set having 70% and the test set 30% of the original
cases. After splitting the data, I first run a classification tree off the
training set, and then use the results for cross-validation using the test
set. It turns out that if I don't have any predictors and make predictions
by simply betting on the majority class of the zero-one coding of the
binary response variable, I can do better than what the results from the
classification tree would deliver in the test set. What would this imply
and what would cause this problem? Does it mean that classification tree
is
not an appropriate method for my data; or, it's because I have too few
variables? Thanks a lot!

Jun Xu, PhD
Professor
Department of Sociology
Ball State University
Muncie, IN 47306
USA

Using the estimates, I get the following prediction rate (correct
prediction) using the test set. Or we can say the misclassification error
rate is 1-0.837 = 0.163

(tab[1,1] + tab[2,2]) / sum(tab)[1] 0.837

Without any predictors, I can get the following rate by betting on the
majority class every time, again using data from the test set. In this
case, the misclassification error rate is 1-0.85 = 0.15

table(h2.test)h2.test

1poorHlth 0goodHlth
      101       575 > 571/(571+101)[1] 0.85



R Code Chunk

# set the seed for random number generator for replication
set.seed(47306)
# have the 7/3 split with 70% of the cases allotted to the training set
# AND create the training set identifier
class.train = sample(1:nrow(usedta), nrow(usedta)*0.7)
# create the test set indicator
class.test = (-class.train)
# create a vector for the binary response variable from the test set
# for future cross-tabulation.
h2.test <- usedta$h2[class.test]
# count the train set cases
Ntrain = length(usedta$h2[class.train])
# run the classification tree model using the training set
# h2 is the binary response and other variables are predictors
tree.h2 <- tree(h2 ~ age + educ + female + white + married + happy,
                data = usedta, subset = class.train,
                control = tree.control(nobs=Ntrain, mindev=0.003))
# summary results
summary(tree.h2)
# make predictions of h2 using the test set
tree.h2.pred <- predict(tree.h2, usedta[class.test,], type="class")
# cross tab the predictions using the test set
table(tree.h2.pred, h2.test)
tab = table(tree.h2.pred, h2.test)
# calculate the ratio for the correctly predicted in the test set
(tab[1,1] + tab[2,2]) / sum(tab)
# calculate the ratio for the correctly predicted using the naive approach
# by betting on the majority category.
table(h2.test)[2]/sum(tab)

        [[alternative HTML version deleted]]

Thank you for your comment! This tree function is from the tree package.
Although it might be a pure statistical question, it could be related to
how the tree function is used. I will explore the site that you suggested.
But if there is anyone who can figure it out off the top of their head, I'd
very much appreciate it.

Jun

On Mon, Aug 24, 2020 at 1:01 PM Bert Gunter <bgunter.4567 at gmail.com>
wrote:

Purely statistical questions -- as opposed to R programming queries --

are

generally off topic here.
Here is where they are on topic:  https://stats.stackexchange.com/

Suggestion: when you post, do include the package name where you get
tree() from, as there might be
more than one with this function.

Bert Gunter

"The trouble with having an open mind is that people keep coming along

and

sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Mon, Aug 24, 2020 at 8:58 AM Xu Jun <junxu.r at gmail.com> wrote:

Dear all R experts,

I have a question about using cross-validation to assess results

estimated

from a classification tree model. I annotated what each line does in

the R

code chunk below. Basically, I split the data, named usedta, into 70%

vs.

30%, with the training set having 70% and the test set 30% of the

original

cases. After splitting the data, I first run a classification tree off

the

training set, and then use the results for cross-validation using the

test

set. It turns out that if I don't have any predictors and make

predictions

by simply betting on the majority class of the zero-one coding of the
binary response variable, I can do better than what the results from the
classification tree would deliver in the test set. What would this imply
and what would cause this problem? Does it mean that classification tree
is
not an appropriate method for my data; or, it's because I have too few
variables? Thanks a lot!

Jun Xu, PhD
Professor
Department of Sociology
Ball State University
Muncie, IN 47306
USA

Using the estimates, I get the following prediction rate (correct
prediction) using the test set. Or we can say the misclassification

error

rate is 1-0.837 = 0.163

(tab[1,1] + tab[2,2]) / sum(tab)[1] 0.837

Without any predictors, I can get the following rate by betting on the
majority class every time, again using data from the test set. In this
case, the misclassification error rate is 1-0.85 = 0.15

table(h2.test)h2.test

1poorHlth 0goodHlth
      101       575 > 571/(571+101)[1] 0.85



R Code Chunk

# set the seed for random number generator for replication
set.seed(47306)
# have the 7/3 split with 70% of the cases allotted to the training set
# AND create the training set identifier
class.train = sample(1:nrow(usedta), nrow(usedta)*0.7)
# create the test set indicator
class.test = (-class.train)
# create a vector for the binary response variable from the test set
# for future cross-tabulation.
h2.test <- usedta$h2[class.test]
# count the train set cases
Ntrain = length(usedta$h2[class.train])
# run the classification tree model using the training set
# h2 is the binary response and other variables are predictors
tree.h2 <- tree(h2 ~ age + educ + female + white + married + happy,
                data = usedta, subset = class.train,
                control = tree.control(nobs=Ntrain, mindev=0.003))
# summary results
summary(tree.h2)
# make predictions of h2 using the test set
tree.h2.pred <- predict(tree.h2, usedta[class.test,], type="class")
# cross tab the predictions using the test set
table(tree.h2.pred, h2.test)
tab = table(tree.h2.pred, h2.test)
# calculate the ratio for the correctly predicted in the test set
(tab[1,1] + tab[2,2]) / sum(tab)
# calculate the ratio for the correctly predicted using the naive

approach

# by betting on the majority category.
table(h2.test)[2]/sum(tab)

        [[alternative HTML version deleted]]

        [[alternative HTML version deleted]]