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] 0.837Without 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.test1poorHlth 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]] > > ______________________________________________ > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > 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. >[[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]] >> >> ______________________________________________ >> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see >> 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. >> >[[alternative HTML version deleted]]