R：如何用中位数替换 NAs？

Question

按列用中位数替换 NA 是一项非常简单的任务。但是如何用行中位数替换 NA 值？我试过matrixStats::rowMedians 还是不行。

示例：

 tibble(
   name = LETTERS[1:5],
   name2 = LETTERS[9:13],
   id = 1:5,
   val1 = rnorm(5, 0.05, 0.5),
   val2 = rnorm(5, 0, 1),
   val3 = c(1, 2, NA, 7, 0.55),
   val4 = c(NA, 2.33, 12, -0.444, 0)
  )


# A tibble: 5 x 7
  name  name2    id    val1   val2  val3   val4
  <chr> <chr> <int>   <dbl>  <dbl> <dbl>  <dbl>
1 A     I         1  0.160  -1.62   1    NA    
2 B     J         2  0.194   0.345  2     2.33 
3 C     K         3  0.681   1.18  NA    12    
4 D     L         4  0.0168 -0.385  7    -0.444
5 E     M         5 -0.509  -1.10   0.55  0   

我尝试了这段代码，但它给了我一个错误：

sample <- sample %>%
       mutate_all(~ifelse(is.na(.), matrixStats::rowMedians(., na.rm = T), .))

Problem with `mutate()` input `val3`.
x Argument 'dim.' must be an integer vector of length two.
i Input `val3` is `(structure(function (..., .x = ..1, .y = ..2, . = ..1) ...`.
Run `rlang::last_error()` to see where the error occurred.

我了解matrixStats::rowMedians 希望我将数据转换为矩阵。但是当我将数据转换为矩阵时，我无法执行mutate 函数。当我尝试立即实施rowMedians 时，我收到错误：

sample <- matrixStats::rowMedians(sample, cols = c("val1", "val2", "val3", "val4"))
Error in matrixStats::rowMedians(sample, cols = c("val1", "val2", "val3",  :
  Argument 'x' must be of type logical, integer or numeric, not 'character'.

as.matrix 将我的数据从numeric 转换为character。但是，在我的原始数据集上，我收到另一个错误：

Error in matrixStats::rowMedians(original_df, cols = c(val1, val2, val3,  :
  object 'val1' was not found

Answer 1

如果你想留在tidyverse，一种方法是重塑数据：

library(dplyr)
library(tidyr)

df %>%
  pivot_longer(cols = starts_with('val'), 
               names_to = 'col') %>%
  group_by(id) %>%
  mutate(value = replace(value, is.na(value), median(value, na.rm = TRUE))) %>%
  pivot_wider(names_from = col, values_from = value) %>%
  ungroup

在base R中，我们可以使用apply：

cols <- grep('val', names(df))
df[cols] <- t(apply(df[cols], 1, function(x) 
              replace(x, is.na(x), median(x, na.rm = TRUE))))

Answer 2

使用dplyr 和purrr 的一个选项可能是：

df %>%
 mutate(across(val1:val4, 
               ~ if_else(is.na(.), pmap_dbl(across(val1:val4), ~ median(c(...), na.rm = TRUE)), .)))
                    
  name  name2    id   val1   val2  val3   val4
  <chr> <chr> <int>  <dbl>  <dbl> <dbl>  <dbl>
1 A     I         1 -0.660  1.68   1     1    
2 B     J         2  0.145  1.04   2     2.33 
3 C     K         3 -1.26   2.54   2.54 12    
4 D     L         4 -0.788 -0.562  7    -0.444
5 E     M         5  0.821  1.74   0.55  0