Exercises in this document can be completed after a basic R training. Your code’s output should be very similar to the output given below questions. Some commands have set.seed() function. Actions requiring randomness can be synced with randomness seeds.
In a loop from 1 to 15, if a value can be exactly divided by 3, print “Industrial”; if by 5, print “Engineering”; if by 15, print “Industrial Engineering”. (p.s. You shouldn’t print Industrial or Engineering when the value can be exactly divided by 15). (Tip: You can use modulo operator (%%). For example 4%%3 is 1 and 4%%2 is 0.)
for(i in 1:15){
print(i)
#Check modulo with 15
if(i%%15==0){
print("Industrial Engineering")
#If not check modulo 5
}else if(i%%5 == 0){
print("Engineering")
#If not check modulo 3
}else if(i%%3 == 0){
print("Industrial")
}
}## [1] 1
## [1] 2
## [1] 3
## [1] "Industrial"
## [1] 4
## [1] 5
## [1] "Engineering"
## [1] 6
## [1] "Industrial"
## [1] 7
## [1] 8
## [1] 9
## [1] "Industrial"
## [1] 10
## [1] "Engineering"
## [1] 11
## [1] 12
## [1] "Industrial"
## [1] 13
## [1] 14
## [1] 15
## [1] "Industrial Engineering"my_vec<-rep("",15)
my_vec[(1:15)%%3 == 0] <- "Industrial"
my_vec[(1:15)%%5 == 0] <- "Engineering"
my_vec[(1:15)%%15 == 0] <- "Industrial Engineering"
names(my_vec)<-1:15
print(my_vec)## 1 2 3
## "" "" "Industrial"
## 4 5 6
## "" "Engineering" "Industrial"
## 7 8 9
## "" "" "Industrial"
## 10 11 12
## "Engineering" "" "Industrial"
## 13 14 15
## "" "" "Industrial Engineering"You can see distances between 10 cities on the data set below.
#Copy paste the following lines to get the distance matrix
set.seed(2017)
distance_matrix<-matrix(sample(10:50,100,replace=TRUE),ncol=10)
distance_matrix[lower.tri(distance_matrix)]<-0
distance_matrix <- distance_matrix + t(distance_matrix)
diag(distance_matrix)<-0
distance_matrix## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
## [1,] 0 42 49 28 27 22 36 50 36 35
## [2,] 42 0 31 17 16 18 19 26 19 38
## [3,] 49 31 0 14 21 12 29 26 50 48
## [4,] 28 17 14 0 28 33 32 23 30 25
## [5,] 27 16 21 28 0 28 49 28 50 36
## [6,] 22 18 12 33 28 0 18 15 10 15
## [7,] 36 19 29 32 49 18 0 41 48 27
## [8,] 50 26 26 23 28 15 41 0 22 25
## [9,] 36 19 50 30 50 10 48 22 0 37
## [10,] 35 38 48 25 36 15 27 25 37 0Consider a salesperson operating in these cities. For a given route create a function which will return the total distance that the salesperson need to cover. Use distance matrix as input.
calculate_total_distance<-function(distance_input,route_info){
total_distance<-0
for(i in 1:(length(route_info)-1)){
total_distance <- total_distance + distance_input[route_info[i],route_info[i+1]]
}
return(paste0("Total distance: ",total_distance))
}Calculate the total distance of the route 1-5-2-6-7-2-4-1.
calculate_total_distance(distance_input=distance_matrix,route_info=c(1,5,2,6,7,2,4,1))## [1] "Total distance: 143"Have the function to choose a number between 1 and 100 and create a game where the user is asked to guess that number in at most 10 trials. At each trial report whether the guessed number is higher or lower than the actual number. It is either a win or a game over after 10 trials.
guess_number <-function(){
my_number<-sample(1:100,1)
print("I have a number on my mind. It is between 1 and 100. Can you guess it in 10 trials?")
for(i in 10:1){
player_guess<-readline("Enter a number between 1 and 100: ")
if(player_guess == my_number){
return("Congratulations! Correct answer.")
}else if(player_guess > my_number){
print("My number is lower than your guess. Try again...")
}else{
print("My number is higher than your guess. Try again...")
}
print(paste0("Remaining guesses: ", i-1))
}
return("Game over :/")
}## [1] "I have a number on my mind. It is between 1 and 100. Can you guess it in 10 trials?"
## [1] "Enter a number between 1 and 100: 5"
## [1] "My number is higher than your guess."
## [1] "Low 6"
## [1] "Remaining guesses: 9"
## [1] "Enter a number between 1 and 100: 35"
## [1] "My number is lower than your guess."
## [1] "High 34"
## [1] "Remaining guesses: 8"
## [1] "Enter a number between 1 and 100: 9"
## [1] "My number is higher than your guess."
## [1] "Low 10"
## [1] "Remaining guesses: 7"
## [1] "Enter a number between 1 and 100: 14"
## [1] "My number is higher than your guess."
## [1] "Low 15"
## [1] "Remaining guesses: 6"
## [1] "Enter a number between 1 and 100: 32"
## [1] "My number is lower than your guess."
## [1] "High 31"
## [1] "Remaining guesses: 5"
## [1] "Enter a number between 1 and 100: 15"
## [1] "My number is higher than your guess."
## [1] "Low 16"
## [1] "Remaining guesses: 4"
## [1] "Enter a number between 1 and 100: 20"
## [1] "My number is higher than your guess."
## [1] "Low 21"
## [1] "Remaining guesses: 3"
## [1] "Enter a number between 1 and 100: 29"
## [1] "My number is lower than your guess."
## [1] "High 28"
## [1] "Remaining guesses: 2"
## [1] "Enter a number between 1 and 100: 21"
## [1] "My number is higher than your guess."
## [1] "Low 22"
## [1] "Remaining guesses: 1"
## [1] "Enter a number between 1 and 100: 22"
## [1] "My number is higher than your guess."
## [1] "Low 23"
## [1] "Remaining guesses: 0"## [1] "Game Over :/"Assume today’s USD/TRY is 7.05. We calculate the next 60 days’ parity forecast as follows.
#Copy paste the two lines below
set.seed(135)
USDTRY<-round(7.05*exp(cumsum((0.02-0.5*(0.2^2))/252 + 0.2*rnorm(60)*sqrt(1/252))),4)
#Check the first 10 days with your data
print(USDTRY[1:10])## [1] 7.0106 6.9695 6.9604 7.0850 6.9685 6.9900 7.0969 7.0576 7.0953 7.1821Calculate the following information below with the USDTRY data.
## Solutions to some questions. Rest of them are next to the questions.
##5
# First let's calculate whether USDTRY falls in any day
falls_tf <- c(3.7,USDTRY[-length(USDTRY)]) > USDTRY
# Then sum all TRUEs.
falls_days<-sum(falls_tf)
##7
falls_day<-0
subseq_days<-0
last_fall <- 0
for(i in 1:length(USDTRY)){
if(falls_tf[i]){
falls_day<-falls_day+1
if(falls_day > subseq_days){
subseq_days <- falls_day
last_fall<-i
}
}else{
falls_day<-0
}
}
## 8
change_days<-which(abs(c(3.7,USDTRY[-length(USDTRY)]) - USDTRY)>=0.12)
## 9
perc_change <- USDTRY/c(3.7,USDTRY[-length(USDTRY)])
opportunity_day <- which.max(perc_change) - 1
perc_profit <- round(max(perc_change)-1,3)*100
##10
my_capital <- 100
for(i in 1:length(USDTRY)){
if(perc_change[i] > 1){
my_capital <- my_capital*perc_change[i]
}
}
net_profit <- my_capital - 100
#10 vektörel alternatif
my_capital <- 100
net_profit <- round(my_capital*prod(perc_change[perc_change>1]) - 100,2)