## R Assignment Solution on Employee Shift Management

• 27th Oct, 2021
• 22:07 PM

library(infix)
options(warn=-1)

#Exercise 1:

```Enter_Hours <- as.numeric(readline(prompt="Enter Hours: "))

pay_computation_1 <- function(Enter_Hours, Enter_Rate){
return(Enter_Hours*Enter_Rate)
}

pay_computation_1(Enter_Hours, Enter_Rate)```

#Exercise 2:

```
pay_computation_2 <- function(){

if(is.numeric(Enter_Hours)==T & is.numeric(Enter_Rate)==T){
if(is.na(Enter_Hours*Enter_Rate) == T){
}
return((Enter_Hours)*(Enter_Rate))
}
}

# Now we will test this function :```

pay_computation_2()
20
nine

pay_computation_2()

forty
9

#Exercise 3 :

```Grade_allocation <- function(){
if(is.na(score*1) == T ) return("Bad score")

if(score > 1 || score < 0>     return("Bad Score")
}
if(score >= 0.9) return("A")
if(score >= 0.8) return("B")
if(score >= 0.7) return("C")
if(score >= 0.6) return("D")
if(score < 0> }

#Example```

0.95

perfect

10

0.75

#Exercise 4

```summary_1 <- function(){
i = 1
number = c()
while(1) {
number[i] <- (readline(prompt="Enter a number: "))
if(number[i] == "done"){
break;
}
if(is.na(as.numeric(number[i])*1) == T){return("Invalid input")}
i = i + 1
}
number = as.numeric(number)[-length(number)]
return(
cat(paste("count is:", length(number), "\n" , "total is:", sum(number), "\n", "average is:", mean(number),"\n" ,sep = " "))

)

}```

summary_1()
4
5

summary_1()
7
done

#Exercise 5:

```
summary_2 <- function(){
i = 1
number = c()
while(1) {
number[i] <- (readline(prompt="Enter a number: "))
if(number[i] == "done"){
break;
}
if(is.na(as.numeric(number[i])*1) == T){return("Invalid input")}
i = i + 1
}
number = as.numeric(number)[1:(length(number) - 1)]
return(
cat(paste("Maximum is:", max(number), "\n" , "Minimum is:", min(number), "\n" ,sep = " "))

)

}```

summary_2()
1
1
1
2
3
done