WPA8

Question 1

drunk <- c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

jacks.equation <- function(squaremeters = 0,
                           trees = 0,
                           drunk = 0) {
  output <- squaremeters * trees - drunk * 324 + log(squaremeters)
  return(output)}
                           

jacks.equation(squaremeters = 1000,
               trees = 30,
               drunk = 7)

## [1] 27738.91

Question 2

standardize.me <- function(x) {
  output <- (x - mean(x)) / sd(x)
  return(output)}

data <- c(6, 3, 8, 6, 3, 2, 3, 2, 100)

standardize.me(data)

## [1] -0.2740789 -0.3677514 -0.2116305 -0.2740789 -0.3677514 -0.3989756
## [7] -0.3677514 -0.3989756  2.6609937

Question 3

how.many <- function(data, value){
  result <- sum(data == value)
  return(result)
}

how.many(data = c(1, 1, 9, 3, 2, 1, 1), value = 1)

## [1] 4

how.many(data = c(1, 1, 9, 3, 2, 1, 1), value = -100)

## [1] 0

Question 4

recode.numeric <- function(x, lb, ub) {
  outliers <- x < lb | x > ub 
  x[outliers] <- NA

  return(x)}

recode.numeric(x = c(5, 6, -10, 2, 1000, 2), lb = 0, ub = 100)

## [1]  5  6 NA  2 NA  2

survey.fixed <- data.frame(
                     id = 1:6,
                     q1 = c(6, 2, 5, -1, 11, 100),
                     q2 = c(-5, 4, 65, 3, 7, 6),
                     q3 = c(2, 1, 2, 45, 5, -5)
                     )

survey.fixed$id <- recode.numeric(survey.fixed$id, lb = 1, ub = 10)
survey.fixed$q1 <- recode.numeric(survey.fixed$q1, lb = 1, ub = 10)
survey.fixed$q2 <- recode.numeric(survey.fixed$q2, lb = 1, ub = 10)
survey.fixed$q3 <- recode.numeric(survey.fixed$q3, lb = 1, ub = 10)

survey.fixed 

##   id q1 q2 q3
## 1  1  6 NA  2
## 2  2  2  4  1
## 3  3  5 NA  2
## 4  4 NA  3 NA
## 5  5 NA  7  5
## 6  6 NA  6 NA

Question 5

orders <- c("coke light", 
            "coke", 
            "pepsi", 
            "coke", 
            "coke light", 
            "water", 
            "pepsi", 
            "pepsi light", 
            "water", 
            "water")

recode.factor <- function(x, old, new) {
  x[x == old[1]] <- new[1]
  x[x == old[2]] <- new[2]
  x[x == old[3]] <- new[3]
  
  return(x)
  
}

recode.factor(orders, old = c("coke", "coke light", "water"), new = c("pepsi", "pepsi light", "pepsi max"))

##  [1] "pepsi light" "pepsi"       "pepsi"       "pepsi"       "pepsi light"
##  [6] "pepsi max"   "pepsi"       "pepsi light" "pepsi max"   "pepsi max"

Question 6

madlib <- function(adjective, name, plural.noun) {
  output <- paste("if you talk to an ", adjective, " pirate like ", name, ", you may find that he/she spends more time talking about ", plural.noun, " than the pirate arts.", sep = "")
  return(output)}

madlib("hipster", "Bruce", "kale")

## [1] "if you talk to an hipster pirate like Bruce, you may find that he/she spends more time talking about kale than the pirate arts."

Question 7

remove.outliers <- function(x) {
  out.log <- x > (mean(x) + 2 * sd(x)) | x < (mean(x) - 2 * sd (x))
  return(x[out.log == FALSE])}

data <- c(rep(1, 50), -529484903)
remove.outliers(data)

##  [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## [36] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Question 8 (started)

ttest.apa <- function(x, null, p.critical) {
  test.result <- t.test(x, mu = null)
  test.statistic <- round(test.result$statistic, 2)
  df <- round(test.result$parameter, 2)
  p.value <- round(test.result$.p.value, 2) }