R 巨型嵌套ifelse语句的备选方案

我有一个如下所示的数据框：

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

我想添加一个名为“Period”的列，其值由Period No.和Frequency列中的内容决定。因此：

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

现在，我正在使用嵌套的if语句来实现这一点。例如：

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'
)
)
)

data$Period如果条件本身是嵌套的，那么实际上只需要嵌套
ifelse
语句。在这种情况下，条件是互斥的，因此您可以单独指定给每个集合。我们还可以巧妙地使用
粘贴和一些数学来创建结果，而不是列出所有的可能性：

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

如果您喜欢
dplyr
，我建议使用
case\u when
功能：

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])
))


使用此示例数据：

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)

如果条件本身是嵌套的，那么实际上只需要嵌套
ifelse
语句。在这种情况下，条件是互斥的，因此您可以单独指定给每个集合。我们还可以巧妙地使用
粘贴和一些数学来创建结果，而不是列出所有的可能性：

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

如果您喜欢
dplyr
，我建议使用
case\u when
功能：

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])
))


使用此示例数据：

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)