下面是问题代码的更正版本。

bso <- function(effort){
  ifelse(effort == 0, 4500,
         ifelse(effort == 1, 4300,
                ifelse(effort == 2, 3700,
                       ifelse(effort == 3, 4700, 5300)
                )
         )
  )
}

但是下面的版本要快得多。它使用输入变量effort作为索引向量。

bso2 <- function(effort){
  c(4500, 4300, 3700, 5700, 5300)[effort + 1L]
}

使用microbenchmark封装进行性能测试。

library(microbenchmark)

n <- 1e4
X <- sample(0:4, n, TRUE)
mb <- microbenchmark(
  bso = bso(X),
  bso2 = bso2(X)
)
print(mb, order = "median")
#Unit: microseconds
# expr      min        lq      mean    median       uq       max neval cld
# bso2   78.660  119.9295  127.4464  129.7135  139.198   193.971   100  a 
#  bso 1469.328 1892.0490 2781.1861 2115.7405 2156.303 17435.984   100   b

对于case_when也可能有用

library(dplyr)
bso <- function(effort){
  case_when(effort == 0 ~ 4500,
            effort == 1 ~ 4300,
            effort == 2 ~ 3700,
            effort == 3 ~ 4700, 
            TRUE ~ 5300)
  
}