比R中的ifelse（）函数更快

Question

I have 3 columns Flag, Score, Stage .

Flag will have values 1 or 0, Score will be any values above 0. We need to calculate stage values.

so our data (stagedata) will look like this:

              Flag Score Stage
               1    35
               1    0
               0    12
               ....

IF Flag == 1 and score >= 30, the we calculate stage as 2,

and if Flag ==0 or Flag == 1 and score < 30, stage = 1.

Any other case stage will be calculated as 0 (ie, due to some error in input or if score or flag is missing).

        stagedata$Stage <- ifelse(stagedata$Flag==1,ifelse((stagedata$Score>=30),2,1),ifelse(stagedata$Flag==0,1,0))
        stagedata$Stage[is.na(stagedata$Stage)] <-0

IS there a more efficient way to do this using any other function like apply? The data that we are dealing with are of the order of ten thounsands

Answer 1

We can convert the logical vector to integer with some arithmetic operation

v1 <- with(stagedata, 2 *(Flag == 1 & score >= 30) + (Flag %in% 0:1 & score <30))
v1
#[1] 2 1 1 2 1 0

If there are NA values, then replace it with 0

v1[is.na(v1)] <- 0

data

stagedata <- data.frame(Flag = c(1, 1, 0, 1, 0, 2), score = c(35, 0, 12, 31, 27, 31))

Answer 2

The original answer and the fixed answer are different by 1.07x - not 1.4x - not a meaningful difference

N <- 10000
set.seed(1)
df <- data.frame(Flag = sample(0:1, N, replace=T), Score = sample(c(12, 35), N, replace=T))
  # Flag Score
# 1    0    12
# 2    0    35
# 3    1    35
# 4    1    12
# 5    0    12
# 6    1    12

ifelse_approach <- function() {
  df$Stage <- ifelse(df$Flag==1,ifelse((df$Score>=30),2,1),ifelse(df$Flag==0,1,0))
}

lgl_approach <- function() {
  df$Stage <- with(df, 2 *(Flag == 1 & Score >= 30) + (Flag %in% 0:1 & Score <30))
}

lgl_fix_approach <- function() {
  df$Stage <- with(df, 2 *(Flag == 1 & Score >= 30) + (Flag == 0 | Score < 30))
}

identical(ifelse_approach(), lgl_approach())
# FALSE
identical(ifelse_approach(), lgl_fix_approach())
# TRUE

library(microbenchmark)
microbenchmark(ifelse_approach(), lgl_approach(), lgl_fix_approach(), unit="relative", times=10L)

# Unit: relative
               # expr      min       lq     mean   median       uq       max neval
  # ifelse_approach() 5.949921 6.048253 5.714637 6.737770 7.186373 3.0478402    10
     # lgl_approach() 1.120431 1.111262 1.059140 1.274285 1.376115 0.5364108    10
 # lgl_fix_approach() 1.000000 1.000000 1.000000 1.000000 1.000000 1.0000000    10