Is there a way to loop within an R function?

0
votes

I am having trouble with a function I am trying to create. I want to convert numbers into assigned days of the week. For example: 1='Monday', 2='Tuesday', 3='Wednesday', 4='Thursday', 5='Friday', 6='Saturday', 0='Sunday'

Below is my attempt a writing the function, but I get an error, and I also think there must be a way to loop it. I just don't know how.

#experiment
pow <- function(x) {
   
 if (x == 1)

    {

    print("Monday")  

    }

    else if (x == 2)

    {

    print("Tuesday")

    } 

    else if (x == 3)

    {
    print("Wednesday")
    } 
    else if (x == 4)
    {
    print("Thursday")
    } 
    else if (x == 5)
    {
    print("Friday")
    } 
    else if (x == 6)
    {
    print("Saturday")
    } 
    else if (x == 0)
    {
    print("Sunday")
    } 
    else if (is.na(x) ==TRUE)
     {print("Sorry, please enter two numbers.")
    }
}
2
rfunctionloops

2 Answers

1
votes

I would use case_when here from the dplyr package, and instead would just have your function map an input to some string output.

library(dplyr)

pow <- function(x) {
    output = case_when(
        x == 0 ~ "Sunday",
        x == 1 ~ "Monday",
        x == 2 ~ "Tuesday",
        x == 3 ~ "Wednesday",
        x == 4 ~ "Thursday",
        x == 5 ~ "Friday",
        x == 6 ~ "Saturday",
        TRUE ~ "Sorry, please enter two numbers."
    )

    return(output)
}
1
votes

You can create a vector and subset it :

pow <- function(x) {
  days <- c('Sunday','Monday','Tuesday','Wednesday','Thursday','Friday','Saturday')
  days[x + 1]
}

pow(2)
#[1] "Tuesday"
pow(0)
#[1] "Sunday"

#You can also pass more than 1 number to the function
pow(c(1, 5, 6, NA, 3))
#[1] "Monday"    "Friday"    "Saturday"  NA     "Wednesday"