巨型嵌套 ifelse 语句的替代方案

Question

I have a data frame that looks like the following:

Period No.  Frequency
1           Month
2           Month
3           Month
3           Quarter
6           Quarter     
9           Quarter
1           YTD
2           YTD
3           YTD

I want to add on a column called "Period" whose values are determined by what is in the Period No. AND Frequency columns. So:

Period No.  Frequency  Period
1           Month      1
2           Month      2
3           Month      3
3           Quarter    Q1
6           Quarter    Q2 
9           Quarter    Q3
1           YTD        YTD-Jan
2           YTD        YTD-Feb
3           YTD        YTD-Mar

Right now, I'm using nested if statements to do this. For example:

data$Period <-
  ifelse(
    (data$`Period No.` == '3') & (data$Frequency == 'Q1'), 'Q1',
    ifelse(
      (data$`Period No.` == '6') & (data$Frequency == 'Q2'), 'Q2',
      ifelse(
        (data$`Period No.` == '9') & (data$Frequency == 'Q3'), 'Q3', 'ERROR'
)
)
)

If I were to do this for every month for each iteration of Frequency, I would have 30 nested ifelse statements. I'm wondering if there's a more concise method to do what I'm trying to achieve?

Answer 1

You only really need to nest ifelse statements if the conditions themselves are nested. In this case, the conditions are mutually exclusive, so you can assign to each set individually. We can also be a little clever about creating your results with paste and some math rather than listing all the possibilities:

df$Period[df$Frequency == "Month"] = as.character(df$Period_No[df$Frequency == "Month"])
df$Period[df$Frequency == "Quarter"] = paste0("Q", (df$Period_No[df$Frequency == "Quarter"] - 1) %/% 3 + 1)
df$Period[df$Frequency == "YTD"] = paste0("YTD-", month.abb[df$Period_No[df$Frequency == "YTD"]])

df
#   Period_No Frequency  Period
# 1         1     Month       1
# 2         2     Month       2
# 3         3     Month       3
# 4         3   Quarter      Q1
# 5         6   Quarter      Q2
# 6         9   Quarter      Q3
# 7         1       YTD YTD-Jan
# 8         2       YTD YTD-Feb
# 9         3       YTD YTD-Mar

If you like dplyr , I'd recommend the case_when function:

df %>% mutate(Period = case_when(
    Frequency == "Month" ~ as.character(Frequency),
    Frequency == "Quarter" ~ paste0("Q", (Period_No - 1) %/% 3 + 1),
    Frequency == "YTD" ~ paste0("YTD-", month.abb[Period_No])
))

---

Using this sample data:

df = read.table(text = "Period_No  Frequency
1           Month
2           Month
3           Month
3           Quarter
6           Quarter     
9           Quarter
1           YTD
2           YTD
3           YTD", header = T)