有没有办法列出 dplyr 中 na_if 函数中的值？

Question

我想列出小标题中所有出现的情况，我想使用na_ifdplyr 包中的函数将其转换为缺失，但我似乎没有得到正确的结果。有线索吗？

library(dplyr)

set.seed(123)

df <- tibble(
  a1 = c("one", "three", "97", "twenty", "98"),
  a2 = c("R", "Python", "99", "Java", "97"),
  a3 = c("statistics", "Data", "Programming", "99", "Science"),
  a4 = floor(rnorm(5, 80, 2))
)

#--- The long route

df1 <- df %>%
  mutate(across(where(is.character), ~na_if(., "97")),
         across(where(is.character), ~na_if(., "98")),
         across(where(is.character), ~na_if(., "99")))

#---- Trial

df2 <- df %>%
  mutate(across(where(is.character),
                ~na_if(., c("97", "98", "99"))))

Answer 1

您可以使用：

\n

df %>%\n  mutate(\n      across(\n          where(is.character),\n          ~if_else(. %in% c("97", "98", "99"), NA_character_, .)\n      )\n  )\n

\n

# A tibble: 5 \xc3\x97 4\n  a1     a2     a3             a4\n  <chr>  <chr>  <chr>       <dbl>\n1 one    R      statistics     80\n2 three  Python Data           80\n3 NA     NA     Programming    76\n4 twenty Java   NA             83\n5 NA     NA     Science        78\n

\n\n

原因na_if在这里不起作用是因为~na_if(., c("97", "98", "99"))基本上相当于if_else(. == c("97", "98", "99"), NA_character_, .). 换句话说，它仅以成对方式比较向量。您可以看到为什么这是一个问题：

\n

# A tibble: 5 \xc3\x97 4\n  a1     a2     a3             a4\n  <chr>  <chr>  <chr>       <dbl>\n1 one    R      statistics     80\n2 three  Python Data           80\n3 NA     NA     Programming    76\n4 twenty Java   NA             83\n5 NA     NA     Science        78\n

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