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Accepted
Wael
Asked: 2024-11-07 20:52:28 +0800 CST

在 dplyr 管道内使用 which.min 时遇到问题

  • 772

which.min我在使用dplyr 管道内部函数时遇到了一些麻烦solution (*)，我正在寻找一种更紧凑、更优雅的方式来实现这一点

  1. 可重现的例子
library(dplyr)

data=data.frame(s1=c(10,NA,5,NA,NA),s2=c(8,NA,NA,4,20),s3=c(NA,NA,2,NA,10))
data
#>   s1 s2 s3
#> 1 10  8 NA
#> 2 NA NA NA
#> 3  5 NA  2
#> 4 NA  4 NA
#> 5 NA 20 10
  1. 最小值：

在这里min(x,na.rm=TRUE)我可以提取最小值

data%>%
  rowwise()%>%
  mutate(Min_s=min(c(s1,s2,s3),na.rm=TRUE))
#> Warning: There was 1 warning in `mutate()`.
#> ℹ In argument: `Min_s = min(c(s1, s2, s3), na.rm = TRUE)`.
#> ℹ In row 2.
#> Caused by warning in `min()`:
#> ! no non-missing arguments to min; returning Inf
#> # A tibble: 5 × 4
#> # Rowwise: 
#>      s1    s2    s3 Min_s
#>   <dbl> <dbl> <dbl> <dbl>
#> 1    10     8    NA     8
#> 2    NA    NA    NA   Inf
#> 3     5    NA     2     2
#> 4    NA     4    NA     4
#> 5    NA    20    10    10
  1. 提取包含最小值的变量：

在这里，我无法提取哪个变量包含最小值

data%>%
  rowwise()%>%
  mutate(which_s=which.min(c(s1,s2,s3)))
#> Error in `mutate()`:
#> ℹ In argument: `which_s = which.min(c(s1, s2, s3))`.
#> ℹ In row 2.
#> Caused by error:
#> ! `which_s` must be size 1, not 0.
#> ℹ Did you mean: `which_s = list(which.min(c(s1, s2, s3)))` ?

# Solution (*)
data%>%
  rowwise()%>%
  mutate(which_s=if(!is.na(s1)|!is.na(s2)|!is.na(s3)) {which.min(c(s1,s2,s3))} else NA )
#> # A tibble: 5 × 4
#> # Rowwise: 
#>      s1    s2    s3 which_s
#>   <dbl> <dbl> <dbl>   <int>
#> 1    10     8    NA       2
#> 2    NA    NA    NA      NA
#> 3     5    NA     2       3
#> 4    NA     4    NA       2
#> 5    NA    20    10       3

创建于 2024-11-07，使用reprex v2.1.0

3 个回答

  1. Best Answer

    在第二行中，您将integer(0)在列中获得which_s，这就是您无法无错误运行它的要点。

    相反，您可以先将结果存储在列表中，然后unnest（不要忘记启用keep_empty参数unnest）

    data %>%
    rowwise() %>%
    mutate(which_s = list(which.min(c(s1, s2, s3)))) %>%
    unnest(which_s, keep_empty = TRUE)

    由此得出

    # A tibble: 5 × 4
     s1    s2    s3 which_s
  <dbl> <dbl> <dbl>   <int>
1    10     8    NA       2
2    NA    NA    NA      NA
3     5    NA     2       3
4    NA     4    NA       2
5    NA    20    10       3
    • 7

  2. 如果不使用rowwise()，您可以在基础 R 中或使用单个mutate()步骤执行此操作purrr::pmap_chr()：

    基数R：

    data$min_base <- unlist(apply(data, 1, \(x) ifelse(all(is.na(x)), NA, names(data)[which.min(x)])))

    dplyr/purrr

    library(dplyr)

data <- data %>%
  mutate(min_dplyr = purrr::pmap_chr(select(., s1:s3), \(...) {
    ifelse(all(is.na(c(...))), NA, colnames(data)[which.min(c(...))])
  }))

    输出：

    #   s1 s2 s3 min_base min_dplyr
# 1 10  8 NA       s2        s2
# 2 NA NA NA     <NA>      <NA>
# 3  5 NA  2       s3        s3
# 4 NA  4 NA       s2        s2
# 5 NA 20 10       s3        s3

    请注意，在这些答案中，@friede 的基本 R 自定义函数速度明显更快，其次是这个基本 R 方法：

    bigdata <- data[rep(seq_len(nrow(data)), 1e5),]

microbenchmark::microbenchmark(
  rowwise = bigdata %>%
    rowwise() %>%
    mutate(which_s = list(which.min(c(s1, s2, s3)))) %>%
    tidyr::unnest(which_s, keep_empty = TRUE),
  base = unlist(apply(bigdata, 1, \(x) ifelse(all(is.na(x)), NA, names(bigdata)[which.min(x)]))),
  pmap = bigdata %>%
    mutate(min_dplyr = purrr::pmap_chr(select(., s1:s3), \(...) {
      ifelse(all(is.na(c(...))), NA, colnames(bigdata)[which.min(c(...))])
    })),
  custom_row.which.min = row.which.min(bigdata, names = TRUE, ties="first")
)

#                 expr       min       lq      mean    median        uq       max neval cld
#              rowwise 3730.8131 4512.870 6018.3180 4985.6024 5913.5166 53501.838   100 a  
#                 base 2419.1913 3162.745 4309.7700 3557.7805 4427.4588 32814.209   100  b 
#                 pmap 3837.8870 4593.846 6091.5265 5203.0391 5984.0412 22015.418   100 a  
# custom_row.which.min  108.4075  147.695  221.7602  168.5267  240.6043  1419.106   100   c
    • 2

  3. 我有时会错过一个好row.which.min功能。这个功能远非好用，而且无法与{dplyr}-language 很好地协调工作，但在这里可能会有所帮助。

    v0

    row.which.min = \(.data, .cols, .names = FALSE, tm = "first") {
  if(missing(.cols)) .cols = names(.data)
  x = .data[.cols]
  i = rowSums(is.na(x)) < length(.cols)
  nx = -x[i, ]
  nx[is.na(nx)] = -Inf
  y = rep(NA, nrow(.data))
  y[i] = max.col(nx, tm)
  if(!.names) y else names(.data)[y]
}

    给予

    > df0 = data.frame(s1=c(10,NA,5,NA,NA),s2=c(8,NA,NA,4,20),s3=c(NA,NA,2,NA,10))
> row.which.min(df0, .names = TRUE)
[1] "s2" NA   "s3" "s2" "s3"
    • 2

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