#
# Unit 7 - Vectorization
#
# Practice 
#

gender <- c("M",        "F",        "F",     "F",     "M")
educ   <- c(12,          12,         15,      10,      13)
salary <- c(11912.44, 10753.82, 5113.94, 8883.05, 6532.11)
 
# (1) 
# How many female employees are there? 
gender <- c("M", "F", "F", "F", "M")
print(length(gender[gender == "F"]))


# (2) 
# What is the average years 
# of education of males? 
gender <- c("M", "F", "F", "F", "M")
educ <- c(12, 12, 15, 10, 13)
maleEduc <- educ[gender == "M"]
sumEduc <- 0 
for (years in maleEduc) {
  sumEduc <- sumEduc + years 
}
print(sumEduc / length(maleEduc))


# (3) 
# Increase all salaries by 2000, 
# then find how many salaries are larger than 10000. 
salary <- c(11912.44, 10753.82, 5113.94, 8883.05, 6532.11)
bonusSalary <- salary + 2000
print(length(bonusSalary[bonusSalary > 10000]))




# (4)
# Produce a new vector, bonus: for each employee
# it includes 1000 * number of education years. 
# Then replace with the calculated bonuses
# each salary smaller than the bonus. 
educ <- c(12, 12, 15, 10, 13)
salary <- c(15912.44, 10753.82, 5113.94, 8883.05, 6532.11)
bonusSal <- 1000 * educ 
for (ind in 1:length(salary)) {
  if (salary[ind] < bonusSal[ind]) {
    salary[ind] <- bonusSal[ind]
  }
}
print(salary)


# (5)
# Produce a new vector: salCat
# It includes: "A" - for each salary < 6000,
#              "B" - for each 6000 <= salary < 10000
#              "C" - for each salary >= 10000
# Next produce a vector which includes the 
# number of salaries for each category "A", "B" and "C", 
# in this order. 
salary <- c(11912.44, 10753.82, 5113.94, 8883.05, 6532.11)
#salCat <- salary
salCat <- character(length(salary))
salCat[salary < 6000] <- "A"
salCat[salary >= 6000 & salary < 10000] <- "B"
salCat[salary > 10000] <- "C"
freqs <- c()
for (cat in sort(unique(salCat))) {
  freqs <- c(freqs, length(salCat[salCat == cat]))
}
print(freqs)
print(salCat)






# (6)
# Produce a new vector: salCat
# It includes: "A" - for each salary < 6000,
#              "B" - for reach salary >= 6000
# Then find whether among those employees whose 
# salaries are of category "B" there are 
# both men and women. 
gender <- c("F", "F", "F", "F", "M")
salary <- c(11912.44, 10753.82, 5113.94, 8883.05, 1532.11)
salCat <- ifelse(salary < 6000, "A", "B")
if (length(unique(gender[salCat == "B"])) == 2) {
  print("yes")
} else {
  print("no")
}


# (7) Write the function Q9. 
# (a) 
# Add 2000 to all salaries of men. 
gender <- c("M", "F", "F", "F", "M")
salary <- c(11912.44, 10753.82, 5113.94, 8883.05, 6532.11)
salary[gender == "M"] <- salary[gender == "M"] + 2000

# (b)
# The function has four parameters: 
# - vec1 - a non-empty numeric vector 
# - vec2 - a non-empty nominal vector
#           whose length is equal to vec1's 
#           length and which represents categories
#           of the values in vec1
# - cat - a label in vec2
# - num - a numerical value 
# The function adds num to all vec1's values 
# if they are of category cat in vec2. 
# Next Use the function Q9 to increase the salaries 
# of all men by 2000. 
Q7 <- function(vec1, vec2, cat, num) {
  vec1[vec2 == cat] <- vec1[vec2 == cat] + num
  return(vec1)
}

gender <- c("M", "F", "F", "F", "M")
salary <- c(11912.44, 10753.82, 5113.94, 8883.05, 6532.11)
educ   <- c(12,          12,         15,      10,      13)
print(Q7(salary, gender, "M", -1000))