告诉 ifelse 忽略 NA 的直接方法

Direct way of telling ifelse to ignore NA

所述,当ifelse(test, yes, no)中的测试条件为NA时,评估也为NA。因此以下 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" 

按照Cath的建议,我可以通过正式指定测试条件不应该包括NA来规避这个问题:

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,此解决方案变得相当冗长。


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

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 更直接的方法告诉 ifelse 忽略 NA?还是为 & !is.na 编写函数是最直接的方法?

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" 

我们可以在没有嵌套 ifelse 循环的情况下更有效地做到这一点。对于第一个数据集,我们为非 NA 元素创建一个逻辑矩阵 (!is.na(df)),获取最大值的列索引,即每行为 TRUE,使用该索引获取列名和 paste ==1

paste0(names(df)[max.col(!is.na(df))], "==1")
#[1] "a==1" "a==1" "b==1" "b==1" "c==1" "c==1"

如果有只有 NA 的行

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

而对于第二个数据集,因为没有NA,我们可以直接与1比较得到一个逻辑矩阵,和之前的步骤一样

paste0(names(df_noNA)[max.col(df_noNA == 1)], "==1")
#[1] "a==1" "a==1" "b==1" "b==1" "c==1" "c==1"

您可以使用 %in% 而不是 == 来排序忽略 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,与原始解决方案相比,这并没有带来任何性能提升,而@arkun 的解决方案大约快了 3 倍。

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: Dealing with TRUE, FALSE, NA and NaN

编辑

根据@arkun 的评论,我重新做了基准测试并修改了声明。

dplyr::case_when 是级联 ifelse 调用的便捷替代方法:

library(dplyr)

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))

df %>% mutate(equals = case_when(a == 1 ~ 'a==1', 
                                 b == 1 ~ 'b==1', 
                                 c == 1 ~ 'c==1'))
#>    a  b  c equals
#> 1  1 NA NA   a==1
#> 2  1 NA NA   a==1
#> 3 NA  1 NA   b==1
#> 4 NA  1 NA   b==1
#> 5 NA NA  1   c==1
#> 6 NA NA  1   c==1

它像ifelse一样级联,所以如果第一个条件为真,即使第二个和第三个条件也为真,也会返回第一个结果。如果 none 为真,则 returns NA:

set.seed(47)
df <- setNames(as.data.frame(matrix(sample(c(1, NA), 30, replace = TRUE), 10)), letters[1:3])

df %>% mutate(equals = case_when(a == 1 ~ 'a==1', 
                                 b == 1 ~ 'b==1', 
                                 c == 1 ~ 'c==1'))
#>     a  b  c equals
#> 1  NA  1  1   b==1
#> 2   1 NA NA   a==1
#> 3  NA  1 NA   b==1
#> 4  NA NA  1   c==1
#> 5  NA NA NA   <NA>
#> 6  NA NA  1   c==1
#> 7   1  1  1   a==1
#> 8   1  1  1   a==1
#> 9  NA  1 NA   b==1
#> 10 NA  1 NA   b==1

而且速度很快:

set.seed(47)
df <- setNames(as.data.frame(matrix(sample(c(1, NA), 3 * 1e5, replace = TRUE), ncol = 3)), letters[1:3])

microbenchmark::microbenchmark(
    original = {
        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)))},
    akrun = {
        v1 <- names(df)[max.col(!is.na(df)) * NA^!rowSums(!is.na(df))]
        i1 <- !is.na(v1)
        v1[i1] <- paste0(v1[i1], "==1")
    },
    amatsuo_net = {
        ifelse(df$a %in% 1, "a==1", 
               ifelse(df$b %in% 1, "b==1", 
                      ifelse(df$c %in% 1, "c==1", NA)))
    },
    alistaire = {
        df %>% mutate(equals = case_when(a == 1 ~ 'a==1', 
                                         b == 1 ~ 'b==1', 
                                         c == 1 ~ 'c==1'))
    }
)
#> Unit: milliseconds
#>         expr      min       lq      mean    median        uq       max neval
#>     original 81.19896 86.11843 110.93882 123.92463 128.58037 171.11026   100
#>        akrun 27.50351 30.99127  38.98353  32.67991  34.64947  77.98958   100
#>  amatsuo_net 83.75744 88.54095 109.22226 110.40066 129.02168 170.92911   100
#>    alistaire 16.57426 18.91951  21.73293  19.29925  24.30350  33.83180   100