I have constructed a decision tree using rpart for a dataset.
I have then divided the data into 2 parts - a training dataset and a test dataset. A tree has been constructed for the dataset using the training data. I want to calculate the accuracy of the predictions based on the model that was created.
My code is shown below:
library(rpart)
#reading the data
data = read.table("source")
names(data) <- c("a", "b", "c", "d", "class")
#generating test and train data - Data selected randomly with a 80/20 split
trainIndex <- sample(1:nrow(x), 0.8 * nrow(x))
train <- data[trainIndex,]
test <- data[-trainIndex,]
#tree construction based on information gain
tree = rpart(class ~ a + b + c + d, data = train, method = 'class', parms = list(split = "information"))
I now want to calculate the accuracy of the predictions generated by the model by comparing the results with the actual values train and test data however I am facing an error while doing so.
My code is shown below:
t_pred = predict(tree,test,type="class")
t = test['class']
accuracy = sum(t_pred == t)/length(t)
print(accuracy)
I get an error message that states -
Error in t_pred == t : comparison of these types is not implemented In
addition: Warning message: Incompatible methods ("Ops.factor",
"Ops.data.frame") for "=="
On checking the type of t_pred, I found out that it is of type integer however the documentation
(https://stat.ethz.ch/R-manual/R-devel/library/rpart/html/predict.rpart.html)
states that the predict() method must return a vector.
I am unable to understand why is the type of the variable is an integer and not a list. Where have I made the mistake and how can I fix it?
Try calculating the confusion matrix first:
confMat <- table(test$class,t_pred)
Now you can calculate the accuracy by dividing the sum diagonal of the matrix - which are the correct predictions - by the total sum of the matrix:
accuracy <- sum(diag(confMat))/sum(confMat)
My response is very similar to #mtoto's one but a bit more simply... I hope it also helps.
mean(test$class == t_pred)
Related
I have written some code in R. This code takes some data and splits it into a training set and a test set. Then, I fit a "survival random forest" model on the training set. After, I use the model to predict observations within the test set.
Due to the type of problem I am dealing with ("survival analysis"), a confusion matrix has to be made for each "unique time" (inside the file "unique.death.time"). For each confusion matrix made for each unique time, I am interested in the corresponding "sensitivity" value (e.g. sensitivity_1001, sensitivity_2005, etc.). I am trying to get all these sensitivity values : I would like to make a plot with them (vs unique death times) and determine the average sensitivity value.
In order to do this, I need to repeatedly calculate the sensitivity for each time point in "unique.death.times". I tried doing this manually and it is taking a long time.
Could someone please show me how to do this with a "loop"?
I have posted my code below:
#load libraries
library(survival)
library(data.table)
library(pec)
library(ranger)
library(caret)
#load data
data(cost)
#split data into train and test
ind <- sample(1:nrow(cost),round(nrow(cost) * 0.7,0))
cost_train <- cost[ind,]
cost_test <- cost[-ind,]
#fit survival random forest model
ranger_fit <- ranger(Surv(time, status) ~ .,
data = cost_train,
mtry = 3,
verbose = TRUE,
write.forest=TRUE,
num.trees= 1000,
importance = 'permutation')
#optional: plot training results
plot(ranger_fit$unique.death.times, ranger_fit$survival[1,], type = 'l', col = 'red') # for first observation
lines(ranger_fit$unique.death.times, ranger_fit$survival[21,], type = 'l', col = 'blue') # for twenty first observation
#predict observations test set using the survival random forest model
ranger_preds <- predict(ranger_fit, cost_test, type = 'response')$survival
ranger_preds <- data.table(ranger_preds)
colnames(ranger_preds) <- as.character(ranger_fit$unique.death.times)
From here, another user (Justin Singh) from a previous post (R: how to repeatedly "loop" the results from a function?) suggested how to create a loop:
sensitivity <- list()
for (time in names(ranger_preds)) {
prediction <- ranger_preds[which(names(ranger_preds) == time)] > 0.5
real <- cost_test$time >= as.numeric(time)
confusion <- confusionMatrix(as.factor(prediction), as.factor(real), positive = 'TRUE')
sensitivity[as.character(i)] <- confusion$byclass[1]
}
But due to some of the observations used in this loop, I get the following error:
Error in confusionMatrix.default(as.factor(prediction), as.factor(real), :
The data must contain some levels that overlap the reference.
Does anyone know how to fix this?
Thanks
Certain values in prediction and/or real have only 1 unique value in them. Make sure the levels of the factors are the same.
sapply(names(ranger_preds), function(x) {
prediction <- factor(ranger_preds[[x]] > 0.5, levels = c(TRUE, FALSE))
real <- factor(cost_test$time >= as.numeric(x), levels = c(TRUE, FALSE))
confusion <- caret::confusionMatrix(prediction, real, positive = 'TRUE')
confusion$byClass[1]
}, USE.NAMES = FALSE) -> result
result
I try to use kknn + loop to create a leave-out-one cross validation for a model, and compare that with train.kknn.
I have split the data into two parts: training (80% data), and test (20% data). In the training data, I exclude one point in the loop to manually create LOOCV.
I think something gets wrong in predict(knn.fit, data.test). I have tried to find how to predict in kknn through the kknn package instruction and online but all the examples are "summary(model)" and "table(validation...)" rather than the prediction on a separate test data. The code predict(model, dataset) works successfully in train.kknn function, so I thought I could use the similar arguments in kknn.
I am not sure if there is such a prediction function in kknn. If yes, what arguments should I give?
Look forward to your suggestion. Thank you.
library(kknn)
for (i in 1:nrow(data.train)) {
train.data <- data.train[-i,]
validation.data <- data.train[i,]
knn.fit <- kknn(as.factor(R1)~., train.data, validation.data, k = 40,
kernel = "rectangular", scale = TRUE)
# train.data + validation.data is the 80% data I split.
}
pred.knn <- predict(knn.fit, data.test) # data.test is 20% data.
Here is the error message:
Error in switch(type, raw = object$fit, prob = object$prob,
stop("invalid type for prediction")) : EXPR must be a length 1
vector
Actually I try to compare train.kknn and kknn+loop to compare the results of the leave-out-one CV. I have two more questions:
1) in kknn: is it possible to use another set of data as test data to see the knn.fit prediction?
2) in train.kknn: I split the data and use 80% of the whole data and intend to use the rest 20% for prediction. Is it an correct common practice?
2) Or should I just use the original data (the whole data set) for train.kknn, and create a loop: data[-i,] for training, data[i,] for validation in kknn? So they will be the counterparts?
I find that if I use the training data in the train.kknn function and use prediction on test data set, the best k and kernel are selected and directly used in generating the predicted value based on the test dataset.
In contrast, if I use kknn function and build a loop of different k values, the model generates the corresponding prediction results based on
the test data set each time the k value is changed. Finally, in kknn + loop, the best k is selected based on the best actual prediction accuracy rate of test data. In short, the best k train.kknn selected may not work best on test data.
Thank you.
For objects returned by kknn, predict gives the predicted value or the predicted probabilities of R1 for the single row contained in validation.data:
predict(knn.fit)
predict(knn.fit, type="prob")
The predict command also works on objects returned by train.knn.
For example:
train.kknn.fit <- train.kknn(as.factor(R1)~., data.train, ks = 10,
kernel = "rectangular", scale = TRUE)
class(train.kknn.fit)
# [1] "train.kknn" "kknn"
pred.train.kknn <- predict(train.kknn.fit, data.test)
table(pred.train.kknn, as.factor(data.test$R1))
The train.kknn command implements a leave-one-out method very close to the loop developed by #vcai01. See the following example:
set.seed(43210)
n <- 500
data.train <- data.frame(R1=rbinom(n,1,0.5), matrix(rnorm(n*10), ncol=10))
library(kknn)
pred.kknn <- array(0, nrow(data.train))
for (i in 1:nrow(data.train)) {
train.data <- data.train[-i,]
validation.data <- data.train[i,]
knn.fit <- kknn(as.factor(R1)~., train.data, validation.data, k = 40,
kernel = "rectangular", scale = TRUE)
pred.kknn[i] <- predict(knn.fit)
}
knn.fit <- train.kknn(as.factor(R1)~., data.train, ks = 40,
kernel = "rectangular", scale = TRUE)
pred.train.kknn <- predict(knn.fit, data.train)
table(pred.train.kknn, pred.kknn)
# pred.kknn
# pred.train.kknn 1 2
# 0 374 14
# 1 9 103
I am using an accelerated failure time / AFT model with a weibull distribution to predict data. I am doing this using the survival package in R. I am splitting my data in training and test, do training on the training set and afterwards try to predict the values for the test set. To do that I am passing the the test set as the newdata parameter, as stated in the references. I get an error, saying that newdata does not have the same size as the training data (obviously!). Then the function seems to evaluate predict the values for the training set.
How can I predict the values for the new data?
# get data
library(KMsurv)
library(survival)
data("kidtran")
n = nrow(kidtran)
kidtran <- kidtran[sample(n),] # shuffle row-wise
kidtran.train = kidtran[1:(n * 0.8),]
kidtran.test = kidtran[(n * 0.8):n,]
# create model
aftmodel <- survreg(kidtransurv~kidtran.train$gender+kidtran.train$race+kidtran.train$age, dist = "weibull")
predicted <- predict(aftmodel, newdata = kidtran.test)
Edit: As mentioned by Hack-R, there was this line of code missing
kidtransurv <- Surv(kidtran.train$time, kidtran.train$delta)
The problem seems to be in your specification of the dependent variable.
The data and code definition of the dependent was missing from your question, so I can't see what the specific mistake was, but it did not appear to be a proper Surv() survival object (see ?survreg).
This variation on your code fixes that, makes some minor formatting improvements, and runs fine:
require(survival)
pacman::p_load(KMsurv)
library(KMsurv)
library(survival)
data("kidtran")
n = nrow(kidtran)
kidtran <- kidtran[sample(n),]
kidtran.train <- kidtran[1:(n * 0.8),]
kidtran.test <- kidtran[(n * 0.8):n,]
# Whatever kidtransurv was supposed to be is missing from your question,
# so I will replace it with something not-missing
# and I will make it into a proper survival object with Surv()
aftmodel <- survreg(Surv(time, delta) ~ gender + race + age, dist = "weibull", data = kidtran.train)
predicted <- predict(aftmodel, newdata = kidtran.test)
head(predicted)
302 636 727 121 85 612
33190.413 79238.898 111401.546 16792.180 4601.363 17698.895
I am new to R and trying to save my svm model in R and have read the documentation but still do not understand what is wrong.
I am getting the error "object is not a matrix" which would seem to mean that my data is not a matrix, but it is... so something is missing.
My data is defined as:
data = read.table("data.csv")
trainSet = as.data.frame(data[,1:(ncol(data)-1)])
Where the last line is my label
I am trying to define my model as:
svm.model <- svm(type ~ ., data=trainSet, type='C-classification', kernel='polynomial',scale=FALSE)
This seems like it should be correct but I am having trouble finding other examples.
Here is my code so far:
# load libraries
require(e1071)
require(pracma)
require(kernlab)
options(warn=-1)
# load dataset
SVMtimes = 1
KERNEL="polynomial"
DEGREE = 2
data = read.table("head.csv")
results10foldAll=c()
# Cross Fold for training and validation datasets
for(timesRun in 1:SVMtimes) {
cat("Running SVM = ",timesRun," result = ")
trainSet = as.data.frame(data[,1:(ncol(data)-1)])
trainClasses = as.factor(data[,ncol(data)])
model = svm(trainSet, trainClasses, type="C-classification",
kernel = KERNEL, degree = DEGREE, coef0=1, cost=1,
cachesize = 10000, cross = 10)
accAll = model$accuracies
cat(mean(accAll), "/", sd(accAll),"\n")
results10foldAll = rbind(results10foldAll, c(mean(accAll),sd(accAll)))
}
# create model
svm.model <- svm(type ~ ., data = trainSet, type='C-classification', kernel='polynomial',scale=FALSE)
An example of one of my samples would be:
10.135338 7.214543 5.758917 6.361316 0.000000 18.455875 14.082668 31
Here, trainSet is a data frame but in the svm.model function it expects data to be a matrix(where you are assigning trainSet to data). Hence, set data = as.matrix(trainSet). This should work fine.
Indeed as pointed out by #user5196900 you need a matrix to run the svm(). However beware that matrix object means all columns have same datatypes, all numeric or all categorical/factors. If this is true for your data as.matrix() may be fine.
In practice more than often people want to model.matrix() or sparse.model.matrix() (from package Matrix) which gives dummy columns for categorical variables, while having single column for numerical variables. But a matrix indeed.
I am learning how to use various random forest packages and coded up the following from example code:
library(party)
library(randomForest)
set.seed(415)
#I'll try to reproduce this with a public data set; in the mean time here's the existing code
data = read.csv(data_location, sep = ',')
test = data[1:65] #basically data w/o the "answers"
m = sample(1:(nrow(factor)),nrow(factor)/2,replace=FALSE)
o = sample(1:(nrow(data)),nrow(data)/2,replace=FALSE)
train2 = data[m,]
train3 = data[o,]
#random forest implementation
fit.rf <- randomForest(train2[,66] ~., data=train2, importance=TRUE, ntree=10000)
Prediction.rf <- predict(fit.rf, test) #to see if the predictions are accurate -- but it errors out unless I give it all data[1:66]
#cforest implementation
fit.cf <- cforest(train3[,66]~., data=train3, controls=cforest_unbiased(ntree=10000, mtry=10))
Prediction.cf <- predict(fit.cf, test, OOB=TRUE) #to see if the predictions are accurate -- but it errors out unless I give it all data[1:66]
Data[,66] is the is the target factor I'm trying to predict, but it seems that by using "~ ." to solve for it is causing the formula to use the factor in the prediction model itself.
How do I solve for the dimension I want on high-ish dimensionality data, without having to spell out exactly which dimensions to use in the formula (so I don't end up with some sort of cforest(data[,66] ~ data[,1] + data[,2] + data[,3}... etc.?
EDIT:
On a high level, I believe one basically
loads full data
breaks it down to several subsets to prevent overfitting
trains via subset data
generates a fitting formula so one can predict values of target (in my case data[,66]) given data[1:65].
so my PROBLEM is now if I give it a new set of test data, let’s say test = data{1:65], it now says “Error in eval(expr, envir, enclos) :” where it is expecting data[,66]. I want to basically predict data[,66] given the rest of the data!
I think that if the response is in train3 then it will be used as a feature.
I believe this is more like what you want:
crtl <- cforest_unbiased(ntree=1000, mtry=3)
mod <- cforest(iris[,5] ~ ., data = iris[,-5], controls=crtl)