程序代写 ##########################################

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## Homework 2
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## Question 1. Identify Course Combinations
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rm(list=ls())

# Load and clean data.
course.df <- read.csv("./Data/Coursetopics.csv") course.mat <- as.matrix(course.df) head(course.mat, 10) library(arules) # Recast incidence matrix into transcations list. course.trans <- as(course.mat, "??") # Generate rules with the highest lift. options(digits = 2, scipen = 1) rules <- ?? inspect(head(sort(rules, by = "??"), 5)) ########################################## ## Question 2 ########################################## rm(list=ls()) #load the data bank.df <- read.csv("./Data/UniversalBankFull.csv") #consider only the required variables bank.df <- bank.df[ , c(13, 14, 10)] bank.df$Online <- as.factor(bank.df$Online) bank.df$CreditCard <- as.factor(bank.df$CreditCard) bank.df$Personal.Loan <- as.factor(bank.df$Personal.Loan) str(bank.df) #partition the data into history (60%) and future (40%) sets #set the seed for the random number generator for reproducing the partition. set.seed(12345) ntotal <- length(bank.df$Personal.Loan) #Sample row numbers randomly. nhistory.index <- sort(sample(ntotal, round(ntotal * 0.6))) history.df <- bank.df[nhistory.index, ] future.df <- bank.df[-nhistory.index, ] #check if variables in the dataset are correctly identified for their types str(bank.df) str(history.df) # Find P(Personal.Loan = 1|CreditCard=1, Online=1) library(??) loan.nb <- naiveBayes(??, data = history.df) ## predict probabilities loan.pred.prob <- predict(loan.nb, newdata = future.df, type = "raw") loan.combined.df <- data.frame(actual = future.df$Personal.Loan, loan.pred.prob) str(loan.combined.df) head(loan.combined.df[??, ]) # Find P(CreditCard=0|Personal.Loan=1) ########################################## ## Question 3 ########################################## rm(list=ls()) set.seed(100) r40000 <- ??norm(40000) ??(r40000, breaks= 200, probability=T, xlab="value", ylab="density") ########################################## ## Question 4 ########################################## rm(list=ls()) benefit.df <- read.csv('./Data/benefits.csv') # read data head(benefit.df) str(benefit.df) ??(benefit.df, conf.level = 0.95) PME <- 0.02 conf.level <- ?? alpha <- ?? z = qnorm(1 - alpha/2) n = z^2*p*(1-p)/PME^2 程序代写 CS代考 加微信: powcoder QQ: 1823890830 Email: powcoder@163.com