SVM Plot e1071 plots single color - r

I am trying to plot my SVM classification with usinf e1071 library.However, the classification plot shows only single value for my parameters.Even though I change the selected parameters to create 2d classification plot it is wrong.
require(caTools)
library(caret)
dataset <-read.csv("income_evaluation.csv")
# fnlwgt row remowed since it is not necessary
df_income <- subset(dataset,select=-c(fnlwgt))
# turn binary attribute into 0 and 1
df_income$income <-ifelse(df_income$income==" >50K",1,0)
df_income$native.country
apply(X=df_income,2,FUN=function(x) length(which(x==' ?')))
# handling missing values
#define function to calculate mode
find_mode <- function(x) {
u <- unique(x)
tab <- tabulate(match(x, u))
u[tab == max(tab)]
}
mod_workclass_df = find_mode(df_income$workclass)
mod_occupation_df = find_mode(df_income$occupation)
mod_country_df = find_mode(df_income$native.country)
# replacing the missing values with the mod values
df_income$workclass[df_income$workclass == ' ?'] <- mod_workclass_df
df_income$occupation[df_income$occupation == ' ?'] <- mod_occupation_df
df_income$native.country[df_income$native.country == ' ?'] <- mod_country_df
# one hot encoding for train set
dmy <- dummyVars(" ~ .", data = df_income, fullRank = T)
df_income <- data.frame(predict(dmy, newdata = df_income))
# sampling
set.seed(101)
sample = sample.split(df_income$income, SplitRatio = .75)
trainingSet = subset(df_income, sample == TRUE)
testSet = subset(df_income, sample == TRUE)
# isolaate y cariable
Y_train <- trainingSet$income
Y_test <- testSet$income
#isolate x cariable
X_test <- subset(testSet,select=-c(income))
# evalution of svm
library(e1071)
svm_classifier = svm(formula=income ~ .,data=trainingSet,type="C-classification",kernel="radial",scale=TRUE,cost=10)
Y_pred = predict(svm_classifier,newdata= X_test)
confusionMatrix(table(Y_test,Y_pred))
# cross validation
# in creating the folds we specify the target feature (dependent variable) and # of folds
folds = createFolds(trainingSet$income, k = 10)
# in cv we are going to applying a created function to our 'folds'
cv = lapply(folds, function(x) { # start of function
# in the next two lines we will separate the Training set into it's 10 pieces
training_fold = trainingSet[-x, ] # training fold = training set minus (-) it's sub test fold
test_fold = trainingSet[x, ] # here we describe the test fold individually
# now apply (train) the classifer on the training_fold
classifier = svm_classifier
Y_pred = predict(svm_classifier,newdata= test_fold[-97])
cm = table(test_fold[, 97], Y_pred)
accuracy = (cm[1,1] + cm[2,2]) / (cm[1,1] + cm[2,2] + cm[1,2] + cm[2,1])
return(accuracy)
})
accuracy = mean(as.numeric(cv))
accuracy
trainingSet$income <-as.factor(trainingSet$income)
# Visualising the Training set results
plot(svm_classifier,trainingSet,education.num~age)
library(ggplot2)
svm_classifier
table(predicted=svm_classifier$fitted,actual=trainingSet$income)
Here is my code above and the plot below.I could not find the problem why there is only one color background and why there is any red color in the background.

Related

Why is my model so accurate when using knn(), where k=1?

I am currently using genomic expression levels, age, and smoking intensity levels to predict the number of days Lung Cancer Patients have to live. I have a small amount of data; 173 patients and 20,438 variables, including gene expression levels (which make up for 20,436). I have split up my data into test and training, utilizing an 80:20 ratio. There are no missing values in the data.
I am using knn() to train the model. Here is what the code looks like:
prediction <- knn(train = trainData, test = testData, cl = trainAnswers, k=1)
Nothing seems out of the ordinary until you notice that k=1. "Why is k=1?" you may ask. The reason k=1 is because when k=1, the model is the most accurate. This makes no sense to me. There are quite a few concerns:
I am using knn() to predict a continuous variable. I should be using something along the lines of, cox maybe.
The model is waaaaaaay too accurate. Here are a few examples of the test answer and the model's predictions. For the first patient, the number of days to death is 274. The model predicts 268. For the second patient, test: 1147, prediction: 1135. 3rd, test: 354, prediction: 370. 4th, test: 995, prediction 995. How is this possible? Out of the entire test data, the model was only off by and average of 9.0625 days! The median difference was 7 days, and the mode was 6 days. Here is a graph of the results:
Bar Graph.
So I guess my main question is what does knn() do, what does k represent, and how is the model so accurate when k=1? Here is my entire code (I am unable to attach the actual data):
# install.packages(c('caret', 'skimr', 'RANN', 'randomForest', 'fastAdaboost', 'gbm', 'xgboost', 'caretEnsemble', 'C50', 'earth'))
library(caret)
# Gather the data and store it in variables
LUAD <- read.csv('/Users/username/Documents/ClinicalData.csv')
geneData <- read.csv('/Users/username/Documents/GenomicExpressionLevelData.csv')
geneData <- data.frame(geneData)
row.names(geneData) = geneData$X
geneData <- geneData[2:514]
colNamesGeneData <- gsub(".","-",colnames(geneData),fixed = TRUE)
colnames(geneData) = colNamesGeneData
# Organize the data
# Important columns are 148 (smoking), 123 (OS Month, basically how many days old), and the gene data. And column 2 (barcode).
LUAD = data.frame(LUAD$patient, LUAD$TOBACCO_SMOKING_HISTORY_INDICATOR, LUAD$OS_MONTHS, LUAD$days_to_death)[complete.cases(data.frame(LUAD$patient, LUAD$TOBACCO_SMOKING_HISTORY_INDICATOR, LUAD$OS_MONTHS, LUAD$days_to_death)), ]
rownames(LUAD)=LUAD$LUAD.patient
LUAD <- LUAD[2:4]
# intersect(rownames(LUAD),colnames(geneData))
# ind=which(colnames(geneData)=="TCGA-778-7167-01A-11R-2066-07")
gene_expression=geneData[, rownames(LUAD)]
# Merge the two datasets to use the geneomic expression levels in your model
LUAD <- data.frame(LUAD,t(gene_expression))
LUAD.days_to_death <- LUAD[,3]
LUAD <- LUAD[,c(1:2,4:20438)]
LUAD <- data.frame(LUAD.days_to_death,LUAD)
set.seed(401)
# Number of Rows in the training data (createDataPartition(dataSet, percentForTraining, boolReturnAsList))
trainRowNum <- createDataPartition(LUAD$LUAD.days_to_death, p=0.8, list=FALSE)
# Training/Test Dataset
trainData <- LUAD[trainRowNum, ]
testData <- LUAD[-trainRowNum, ]
x = trainData[, c(2:20438)]
y = trainData$LUAD.days_to_death
v = testData[, c(2:20438)]
w = testData$LUAD.days_to_death
# Imputing missing values into the data
preProcess_missingdata_model <- preProcess(trainData, method='knnImpute')
library(RANN)
if (anyNA(trainData)) {
trainData <- predict(preProcess_missingdata_model, newdata = trainData)
}
anyNA(trainData)
# Normalizing the data
preProcess_range_model <- preProcess(trainData, method='range')
trainData <- predict(preProcess_range_model, newdata = trainData)
trainData$LUAD.days_to_death <- y
apply(trainData[,1:20438], 2, FUN=function(x){c('min'=min(x), 'max'=max(x))})
preProcess_range_model_Test <- preProcess(testData, method='range')
testData <- predict(preProcess_range_model_Test, newdata = testData)
testData$LUAD.days_to_death <- w
apply(testData[,1:20438], 2, FUN=function(v){c('min'=min(v), 'max'=max(v))})
# To uncomment, select the text and press 'command' + 'shift' + 'c'
# set.seed(401)
# options(warn=-1)
# subsets <- c(1:10)
# ctrl <- rfeControl(functions = rfFuncs,
# method = "repeatedcv",
# repeats = 5,
# verbose = TRUE)
# lmProfile <- rfe(x=trainData[1:20437], y=trainAnswers,
# sizes = subsets,
# rfeControl = ctrl)
# lmProfile
trainAnswers <- trainData[,1]
testAnswers <- testData[,1]
library(class)
prediction <- knn(train = trainData, test = testData, cl = trainAnswers, k=1)
#install.packages("plotly")
library(plotly)
Test_Question_Number <- c(1:32)
prediction2 <- data.frame(prediction[1:32])
prediction2 <- as.numeric(as.vector(prediction2[c(1:32),]))
data <- data.frame(Test_Question_Number, prediction2, testAnswers)
names(data) <- c("Test Question Number","Prediction","Answer")
p <- plot_ly(data, x = ~Test_Question_Number, y = ~prediction2, type = 'bar', name = 'Prediction') %>%
add_trace(y = ~testAnswers, name = 'Answer') %>%
layout(yaxis = list(title = 'Days to Death'), barmode = 'group')
p
merge <- data.frame(prediction2,testAnswers)
difference <- abs((merge[,1])-(merge[,2]))
difference <- sort(difference)
meanDifference <- mean(difference)
medianDifference <- median(difference)
modeDifference <- names(table(difference))[table(difference)==max(table(difference))]
cat("Mean difference:", meanDifference, "\n")
cat("Median difference:", medianDifference, "\n")
cat("Mode difference:", modeDifference,"\n")
Lastly, for clarification purposes, ClinicalData.csv is the age, days to death, and smoking intensity data. The other .csv is the genomic expression data. The data above line 29 doesn't really matter, so you can just skip to the part of the code where it says "set.seed(401)".
Edit: Some samples of the data:
days_to_death OS_MONTHS
121 3.98
NACC1 2001.5708 2363.8063 1419.879
NACC2 58.2948 61.8157 43.4386
NADK 706.868 1053.4424 732.1562
NADSYN1 1628.7634 912.1034 638.6471
NAE1 832.8825 793.3014 689.7123
NAF1 140.3264 165.4858 186.355
NAGA 1523.3441 1524.4619 1858.9074
NAGK 983.6809 899.869 1168.2003
NAGLU 621.3457 510.9453 1172.511
NAGPA 346.9762 257.5654 275.5533
NAGS 460.7732 107.2116 321.9763
NAIF1 217.1219 202.5108 132.3054
NAIP 101.2305 87.8942 77.261
NALCN 13.9628 36.7031 48.0809
NAMPT 3245.6584 1257.8849 5465.6387
Because K = 1 is the most complex knn model. It has the most flexible decision boundary. It creates an overfit. It will perform well within the training data by poorly on a holdout set (but not always).

How to predict test data using a GAM with MRF smooth and neighborhood structure?

I am having a problem using the predict() function for a mgcv::gam (training) model on a new (testing) dataset. The problem arises due to a mrf smooth I have integrated to account for the spatial nature of my data.
I use the following call to create my GAM model
## Run GAM with MRF
m <- gam(crime ~ s(district,k=nrow(traindata),
bs ='mrf',xt=list(nb=nbtrain)), #define MRF smooth
data = traindata,
method = 'REML',
family = scat(), #fit scaled t distribution
gamma = 1.4
)
where I predict the dependent variable crime using the neighbourhood structure, parsed into the model in the smooth term argument xt. The neighbourhood structure comes as a nb object that I created using the poly2nb() function.
Now, if I want to use predict() on a new testing dataset, I don't know how to pass the according neighbourhood structure into the call. Providing just the new data
pred <- predict.gam(m,newdata=testdata)
throws the following error:
Error in predict.gam(m, newdata = testdata) :
7, 16, 20, 28, 35, 36, 37, 43 not in original fit
Here's a full reproduction of the error using the Columbus dataset called from within R directly:
#ERROR REPRODUCTION
## Load packages
require(mgcv)
require(spdep)
require(dplyr)
## Load Columbus Ohio crime data (see ?columbus for details and credits)
data(columb.polys) #Columbus district shapes list
columb.polys <- lapply(columb.polys,na.omit) #omit NAs (unfortunate problem with the Columbus sample data)
data(columb) #Columbus data frame
df <- data.frame(district=numeric(0),x=numeric(0),y= numeric(0)) #Create empty df to store x, y and IDs for each polygon
## Extract x and y coordinates from each polygon and assign district ID
for (i in 1:length(columb.polys)) {
district <- i-1
x <- columb.polys[[i]][,1]
y <- columb.polys[[i]][,2]
df <- rbind(df,cbind(district,x,y)) #Save in df data.frame
}
## Convert df into SpatialPolygons
sp <- df %>%
group_by(district) %>%
do(poly=select(., x, y) %>%Polygon()) %>%
rowwise() %>%
do(polys=Polygons(list(.$poly),.$district)) %>%
{SpatialPolygons(.$polys)}
## Merge SpatialPolygons with data
spdf <- SpatialPolygonsDataFrame(sp,columb)
## Split into training and test sample (80/20 ratio)
splt <- sample(1:2,size=nrow(spdf),replace=TRUE,prob=c(0.8,0.2))
train <- spdf[splt==1,]
test <- spdf[splt==2,]
## Prepapre both samples and create NB objects
traindata <- train#data #Extract data from SpatialPolygonsDataFrame
testdata <- test#data
traindata <- droplevels(as(train, 'data.frame')) #Drop levels
testdata <- droplevels(as(test, 'data.frame'))
traindata$district <- as.factor(traindata$district) #Factorize
testdata$district <- as.factor(testdata$district)
nbtrain <- poly2nb(train, row.names=train$Precinct, queen=FALSE) #Create NB objects for training and test sample
nbtest <- poly2nb(test, row.names=test$Precinct, queen=FALSE)
names(nbtrain) <- attr(nbtrain, "region.id") #Set region.id
names(nbtest) <- attr(nbtest, "region.id")
## Run GAM with MRF
m <- gam(crime ~ s(district, k=nrow(traindata), bs = 'mrf',xt = list(nb = nbtrain)), # define MRF smooth
data = traindata,
method = 'REML', # fast version of REML smoothness selection; alternatively 'GCV.Cp'
family = scat(), #fit scaled t distribution
gamma = 1.4
)
## Run prediction using new testing data
pred <- predict.gam(m,newdata=testdata)
SOLUTION:
I finally found the time to update this post with the solution. Thanks to everyone for helping me out. Here is the code for implementing k-fold CV with a random training-testing split:
#Apply k-fold cross validation
mses <- data.frame() #Create empty df to store CV squared error values
scores <- data.frame() #Create empty df to store CV R2 values
set.seed(42) #Set seed for reproducibility
k <- 10 #Define number of folds
for (i in 1:k) {
# Create weighting column
data$weight <- sample(c(0,1),size=nrow(data),replace=TRUE,prob=c(0.2,0.8)) #0 Indicates testing sample, 1 training sample
#Run GAM with MRF
ctrl <- gam.control(nthreads = 6) #Set controls
m <- gam(crime ~ s(disctrict, k=nrow(data), bs = 'mrf',xt = list(nb = nb)), #define MRF smooth
data = data,
weights = data$weight, #Use only weight==1 observations (training)
method = 'REML',
control = ctrl,
family = scat(),
gamma = 1.4
)
#Generate test dataset
testdata <- data[data$weight==0,] #Select test data by weight
#Predict test data
pred <- predict(m,newdata=testdata)
#Extract MSES
mses[i,1] <- mean((data$R_MeanDiff[data$weight==0] - pred)^2)
scores[i,1] <- summary(m)$r.sq
}
av.mse.GMRF <- mean(mses$V1)
av.r2.GMRF <- mean(scores$V1)
I have one question criticism with the current solution, being that the full dataset was used to "train" the model meaning that the predictions are going to be biased since the testdata was used to train it.
This only requires a couple minor tweaks to fix:
#Apply k-fold cross validation
mses <- data.frame() #Create empty df to store CV squared error values
scores <- data.frame() #Create empty df to store CV R2 values
set.seed(42) #Set seed for reproducibility
k <- 10 #Define number of folds
#For loop for each fold
for (i in 1:k) {
# Create weighting column
data$weight <- sample(c(0,1),size=nrow(data),replace=TRUE,prob=c(0.2,0.8)) #0 Indicates testing sample, 1 training sample
#Generate training dataset
trainingdata <- data[data$weight == 1, ] #Select test data by weight
#Generate test dataset
testdata <- data[data$weight == 0, ] #Select test data by weight
#Run GAM with MRF
ctrl <- gam.control(nthreads = 6) #Set controls
m <- gam(crime ~ s(disctrict, k=nrow(data), bs = 'mrf',xt = list(nb = nb)), #define MRF smooth
data = trainingdata,
weights = data$weight, #Use only weight==1 observations (training)
method = 'REML',
control = ctrl,
family = scat(),
gamma = 1.4
)
#Predict test data
pred <- predict(m,newdata = testdata)
#Extract MSES
mses[i,1] <- mean((data$R_MeanDiff[data$weight==0] - pred)^2)
scores[i,1] <- summary(m)$r.sq
}
#Get average scores from each k-fold test
av.mse.GMRF <- mean(mses$V1)
av.r2.GMRF <- mean(scores$V1)

Predict Logistf

I'm using a R package called logistf to make a Logistc Regression and I saw that there's no predict function for new data in this package and predict package does not work with this, so I found a code that show how making this with new data:
fit<-logistf(Tax ~ L20+L24+L28+L29+L31+L32+L33+L36+S10+S15+S16+S17+S20, data=trainData)
betas <- coef(fit)
X <- model.matrix(fit, data=testData)
probs <- 1 / (1 + exp(-X %*% betas))
I want to make a cross validation version with this using fit$predict and the probabilities that probs generate for me. Has anyone ever done something like this before?
Other thing that I want to know is about fit$predict I'm making a binary logistic regression, and this function returns many values, are these values from class 0 or 1, how can I know this? Thanks
While the code that you wrote works perfectly, there is a concise way of getting the same results seemingly:
brglm_model <- brglm(formula = response ~ predictor , family = "binomial", data = train )
brglm_pred <- predict(object = brglm_model, newdata = test , type = "response")
About the CV, you have to write a few lines of code I guess:
#Setting the number of folds, and number of instances in each fold
n_folds <- 5
fold_size <- nrow(dataset) %/% 5
residual <- nrow(dataset) %% 5
#label the instances based on the number of folds
cv_labels <- c(rep(1,fold_size),rep(2,fold_size), rep(3,fold_size), rep(4,fold_size), rep(5,fold_size), rep(5,residual))
# the error term would differ based on each threshold value
t_seq <- seq(0.1,0.9,by = 0.1)
index_mat <- matrix(ncol = (n_folds+1) , nrow = length(t_seq))
index_mat[,1] <- t_seq
# the main loop for calculation of the CV error on each fold
for (i in 1:5){
train <- dataset %>% filter(cv_labels != i)
test <- dataset %>% filter(cv_labels == i )
brglm_cv_model <- brglm(formula = response_var ~ . , family = "binomial", data = train )
brglm_cv_pred <- predict(object = brglm_model, newdata = test , type = "response")
# error formula that you want, e.g. misclassification
counter <- 0
for (treshold in t_seq ) {
counter <- counter + 1
conf_mat <- table( factor(test$response_var) , factor(brglm_cv_pred>treshold, levels = c("FALSE","TRUE") ))
sen <- conf_mat[2,2]/sum(conf_mat[2,])
# other indices can be computed as follows
#spec <- conf_mat[1,1]/sum(conf_mat[1,])
#prec <- conf_mat[2,2]/sum(conf_mat[,2])
#F1 <- (2*prec * sen)/(prec+sen)
#accuracy <- (conf_mat[1,1]+conf_mat[2,2])/sum(conf_mat)
#here I am only interested in sensitivity
index_mat[counter,(i+1)] <- sen
}
}
# final data.frame would be the mean of sensitivity over each threshold value
final_mat <- matrix(nrow = length(t_seq), ncol = 2 )
final_mat[,1] <- t_seq
final_mat[,2] <- apply(X = index_mat[,-1] , MARGIN = 1 , FUN = mean)
final_mat <- data.frame(final_mat)
colnames(final_mat) <- c("treshold","sensitivity")
#why not having a look at the CV-sensitivity of the model over threshold values?
ggplot(data = final_mat) +
geom_line(aes(x = treshold, y = sensitivity ), color = "blue")

"The format of predictions is incorrect"

Implementation of ROCR curve, kNN ,K 10 fold cross validation.
I am using Ionosphere dataset.
Here is the attribute information for your reference:
-- All 34 are continuous, as described above
-- The 35th attribute is either "good" or "bad" according to the definition
summarized above. This is a binary classification task.
data1<-read.csv('https://archive.ics.uci.edu/ml/machine-learning-databases/ionosphere/ionosphere.data',header = FALSE)
knn on its own works, kNN with kfold also works. But when I put in the ROCR code it doesnt like it.
I get the error: "The format of predictions is incorrect".
I checked the dataframes pred and Class 1. The dimensions are same. I tried with data.test$V35 instead of Class1 I get the same error with this option.
install.packages("class")
library(class)
nrFolds <- 10
data1[,35]<-as.numeric(data1[,35])
# generate array containing fold-number for each sample (row)
folds <- rep_len(1:nrFolds, nrow(data1))
# actual cross validation
for(k in 1:nrFolds) {
# actual split of the data
fold <- which(folds == k)
data.train <- data1[-fold,]
data.test <- data1[fold,]
Class<-data.train[,35]
Class1<-data.test[,35]
# train and test your model with data.train and data.test
pred<-knn(data.train, data.test, Class, k = 5, l = 0, prob = FALSE, use.all = TRUE)
data<-data.frame('predict'=pred, 'actual'=Class1)
count<-nrow(data[data$predict==data$actual,])
total<-nrow(data.test)
avg = (count*100)/total
avg =format(round(avg, 2), nsmall = 2)
method<-"KNN"
accuracy<-avg
cat("Method = ", method,", accuracy= ", accuracy,"\n")
}
install.packages("ROCR")
library(ROCR)
rocrPred=prediction(pred, Class1, NULL)
rocrPerf=performance(rocrPred, 'tpr', 'fpr')
plot(rocrPerf, colorize=TRUE, text.adj=c(-.2,1.7))
Any help is appreciated.
This worked for me..
install.packages("class")
library(class)
library(ROCR)
nrFolds <- 10
data1[,35]<-as.numeric(data1[,35])
# generate array containing fold-number for each sample (row)
folds <- rep_len(1:nrFolds, nrow(data1))
# actual cross validation
for(k in 1:nrFolds) {
# actual split of the data
fold <- which(folds == k)
data.train <- data1[-fold,]
data.test <- data1[fold,]
Class<-data.train[,35]
Class1<-data.test[,35]
# train and test your model with data.train and data.test
pred<-knn(data.train, data.test, Class, k = 5, l = 0, prob = FALSE, use.all = TRUE)
data<-data.frame('predict'=pred, 'actual'=Class1)
count<-nrow(data[data$predict==data$actual,])
total<-nrow(data.test)
avg = (count*100)/total
avg =format(round(avg, 2), nsmall = 2)
method<-"KNN"
accuracy<-avg
cat("Method = ", method,", accuracy= ", accuracy,"\n")
pred <- prediction(Class1,pred)
perf <- performance(pred, "tpr", "fpr")
plot(perf, colorize=T, add=TRUE)
abline(0,1)
}

binning continuous variables by IV value in R

I am building a logistic regression model in R. I want to bin continuous predictors in an optimal way in relationship to the target variable. There are two things that I know of:
the continuous variables are binned such that its IV (information value) is maximized
maximize the chi-square in the two way contingency table -- the target has two values 0 and 1, and the binned continuous variable has the binned buckets
Does anyone know of any functions in R that can perform such binning?
Your help will be greatly appreciated.
For the first point, you could bin using the weight of evidence (woe) with the package woebinning which optimizes the number of bins for the IV
library(woeBinning)
# get the bin cut points from your dataframe
cutpoints <- woe.binning(dataset, "target_name", "Variable_name")
woe.binning.plot(cutpoints)
# apply the cutpoints to your dataframe
dataset_woe <- woe.binning.deploy(dataset, cutpoint, add.woe.or.dum.var = "woe")
It returns your dataset with two extra columns
Variable_name.binned which is the labels
Variable_name.woe.binned which is the replaced values that you can then parse into your regression instead of Variable_name
For the second point, on chi2, the package discretization seems to handle it but I haven't tested it.
The methods used by regression splines to set knot locations might be considered. The rpart package probably has relevant code. You do need to penalize the inferential statistics because this results in an implicit hiding of the degrees of freedom expended in the process of moving the breaks around to get the best fit. Another common method is to specify breaks at equally spaced quantiles (quartiles or quintiles) within the subset with IV=1. Something like this untested code:
cont.var.vec <- # names of all your continuous variables
breaks <- function(var,n) quantiles( dfrm[[var]],
probs=seq(0,1,length.out=n),
na.rm=TRUE)
lapply(dfrm[ dfrm$IV == 1 , cont.var.vec] , breaks, n=5)
s
etwd("D:")
rm(list=ls())
options (scipen = 999)
read.csv("dummy_data.txt") -> dt
head(dt)
summary(dt)
mydata <- dt
head(mydata)
summary(mydata)
##Capping
for(i in 1:ncol(mydata)){
if(is.numeric(mydata[,i])){
val.quant <- unname(quantile(mydata[,i],probs = 0.75))
mydata[,i] = sapply(mydata[,i],function(x){if(x > (1.5*val.quant+1)){1.5*val.quant+1}else{x}})
}
}
library(randomForest)
x <- mydata[,!names(mydata) %in% c("Cust_Key","Y")]
y <- as.factor(mydata$Y)
set.seed(21)
fit <- randomForest(x,y,importance=T,ntree = 70)
mydata2 <- mydata[,!names(mydata) %in% c("Cust_Key")]
mydata2$Y <- as.factor(mydata2$Y)
fit$importance
####var reduction#####
vartoremove <- ncol(mydata2) - 20
library(rminer)
#####
for(i in 1:vartoremove){
rf <- fit(Y~.,data=mydata2,model = "randomForest", mtry = 10 ,ntree = 100)
varImportance <- Importance(rf,mydata2,method="sensg")
Z <- order(varImportance$imp,decreasing = FALSE)
IND <- Z[2]
var_to_remove <- names(mydata2[IND])
mydata2[IND] = NULL
print(i)
}
###########
library(smbinning)
as.data.frame(mydata2) -> inp
summary(inp)
attach(inp)
rm(result)
str(inp)
inp$target <- as.numeric(inp$Y) *1
table(inp$target)
ftable(inp$Y,inp$target)
inp$target <- inp$target -1
result= smbinning(df=inp, y="target", x="X37", p=0.0005)
result$ivtable
smbinning.plot(result,option="badrate",sub="test")
summary(inp)
result$ivtable
boxplot(inp$X2~inp$Y,horizontal=T, frame=F, col="red",main="Distribution")
###Sample
require(caTools)
inp$Y <- NULL
sample = sample.split(inp$target, SplitRatio = .7)
train = subset(inp, sample == TRUE)
test = subset(inp, sample == FALSE)
head(train)
nrow(train)
fit1 <- glm(train$target~.,data=train,family = binomial)
summary(rf)
prediction1 <- data.frame(actual = test$target, predicted = predict(fit1,test ,type="response") )
result= smbinning(df=prediction1, y="actual", x="predicted", p=0.005)
result$ivtable
smbinning.plot(result,option="badrate",sub="test")
tail(prediction1)
write.csv(prediction1 , "test_pred_logistic.csv")
predict_train <- data.frame(actual = train$target, predicted = predict(fit1,train ,type="response") )
write.csv(predict_train , "train_pred_logistic.csv")
result= smbinning(df=predict_train, y="actual", x="predicted", p=0.005)
result$ivtable
smbinning.plot(result,option="badrate",sub="train")
####random forest
rf <- fit(target~.,data=train,model = "randomForest", mtry = 10 ,ntree = 200)
prediction2 <- data.frame(actual = test$target, predicted = predict(rf,train))
result= smbinning(df=prediction2, y="actual", x="predicted", p=0.005)
result$ivtable
smbinning.plot(result,option="badrate",sub="train")
###########IV
library(devtools)
install_github("riv","tomasgreif")
library(woe)
##### K-fold Validation ########
library(caret)
cv_fold_count = 2
folds = createFolds(mydata2$Y,cv_fold_count,list=T);
smpl = folds[[i]];
g_train = mydata2[-smpl,!names(mydata2) %in% c("Y")];
g_test = mydata2[smpl,!names(mydata2) %in% c("Y")];
cost_train = mydata2[-smpl,"Y"];
cost_test = mydata2[smpl,"Y"];
rf <- randomForest(g_train,cost_train)
logit.data <- cbind(cost_train,g_train)
logit.fit <- glm(cost_train~.,data=logit.data,family = binomial)
prediction <- data.f
rame(actual = test$Y, predicted = predict(rf,test))

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