告诉ifelse忽略NA的直接方法 [英] Direct way of telling ifelse to ignore NA

问题描述

如此处所述，当ifelse(test, yes, no)中的测试条件为NA时，评估结果也为NA .因此，以下返回...

As explained here when the test condition in ifelse(test, yes, no) is NA, the evaluation is also NA. Hence the following returns...

df <- data.frame(a = c(1, 1, NA, NA, NA ,NA),
                 b = c(NA, NA, 1, 1, NA, NA),
                 c = c(rep(NA, 4), 1, 1))
ifelse(df$a==1, "a==1", 
    ifelse(df$b==1, "b==1", 
        ifelse(df$c==1, "c==1", NA)))
#[1] "a==1" "a==1" NA     NA     NA     NA    

...而不是所需的

#[1] "a==1" "a==1" "b==1" "b==1"  "c==1" "c==1" 

正如Cath所建议的，我可以通过正式指定测试条件不应包含NA来规避此问题:

As suggested by Cath, I can circumvent this problem by formally specifying that the test condition should not include NA:

ifelse(df$a==1 &  !is.na(df$a), "a==1", 
    ifelse(df$b==1 & !is.na(df$b), "b==1", 
        ifelse(df$c==1 & !is.na(df$c), "c==1", NA)))

但是，正如akrun还指出的那样，随着列数的增加，该解决方案变得相当冗长.

However, as akrun also noted, this solution becomes rather lengthy with increasing number of columns.

一种解决方法是先将所有NA替换为data.frame中不存在的值(例如，在本例中为2):

A workaround would be to first replace all NAs with a value not present in the data.frame (e.g, 2 in this case):

df_noNA <- data.frame(a = c(1, 1, 2, 2, 2 ,2),
                 b = c(2, 2, 1, 1, 2, 2),
                 c = c(rep(2, 4), 1, 1))

ifelse(df_noNA$a==1, "a==1", 
    ifelse(df_noNA$b==1, "b==1", 
        ifelse(df_noNA$c==1, "c==1", NA)))
#[1] "a==1" "a==1" "b==1" "b==1"  "c==1" "c==1" 

但是，我想知道是否有一种更直接的方式告诉ifelse忽略NAs ?还是为& !is.na编写函数最直接的方法?

However, I was wondering if there was a more direct way to tell ifelse to ignore NAs? Or is writing a function for & !is.na the most direct way?

ignorena <- function(column) {
        column ==1 & !is.na(column)
}
ifelse(ignorena(df$a), "a==1", 
    ifelse(ignorena(df$b), "b==1", 
        ifelse(ignorena(df$c), "c==1", NA)))
#[1] "a==1" "a==1" "b==1" "b==1"  "c==1" "c==1" 

推荐答案

您可以使用%in%而不是==来排序忽略NA s.

You can use %in% instead of == to sort-of ignore NAs.

ifelse(df$a %in% 1, "a==1", 
       ifelse(df$b %in% 1, "b==1", 
              ifelse(df$c %in% 1, "c==1", NA)))

不幸的是，与原始版本相比，这没有任何性能提升，而@arkun的解决方案快了约3倍.

Unfortunately, this does not give any performance gain compared to the original while @arkun's solution is about 3 times faster.

solution_original <- function(){
  ifelse(df$a==1 &  !is.na(df$a), "a==1", 
         ifelse(df$b==1 & !is.na(df$b), "b==1", 
                ifelse(df$c==1 & !is.na(df$c), "c==1", NA)))
}

solution_akrun <- function(){
  v1 <- names(df)[max.col(!is.na(df)) * NA^!rowSums(!is.na(df))]
  i1 <- !is.na(v1)
  v1[i1] <- paste0(v1[i1], "==1")
}

solution_mine <- function(x){
  ifelse(df$a %in% 1, "a==1", 
         ifelse(df$b %in% 1, "b==1", 
                ifelse(df$c %in% 1, "c==1", NA)))
}
set.seed(1)
df <- data.frame(a = sample(c(1, rep(NA, 4)), 1e6, T),
                 b = sample(c(1, rep(NA, 4)), 1e6, T),
                 c = sample(c(1, rep(NA, 4)), 1e6, T))
microbenchmark::microbenchmark(
  solution_original(),
  solution_akrun(),
  solution_mine()
)
## Unit: milliseconds
##                expr      min       lq     mean   median       uq       max neval
## solution_original() 701.9413 839.3715 845.0720 853.1960 875.6151 1051.6659   100
##    solution_akrun() 217.4129 242.5113 293.2987 253.2144 387.1598  564.3981   100
##     solution_mine() 698.7628 845.0822 848.6717 858.7892 877.9676 1006.2872   100

受此启发: R:处理TRUE，FALSE， NA和NaN

修改

在@arkun发表评论之后，我重述了基准并修改了声明.

Following the comment by @arkun, I redid the benchmark and revised the statement.

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