使用across()尝试以下解决方案：

library(dplyr)
library(magrittr)
#Data
a = c("aa", "bb", "cc", NA, "aa") 
b = c("aa", "bb", "cc", "bb", "bb") 
c = c(NA, "aa", "aa", NA, "bb") 
d = c(1, 1, 2, 2, 3)
df = data.frame(a,b,c,d,stringsAsFactors = F)
#Vector
cols <- c("a","b","c")
#Code
df %>%
  mutate(across(cols, ~ ifelse(.=='aa' | is.na(.), 'Missing', .)))

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输出量：

a       b       c d
1 Missing Missing Missing 1
2      bb      bb Missing 1
3      cc      cc Missing 2
4 Missing      bb Missing 2
5 Missing      bb      bb 3

型

你可以像这样使用匿名函数

library(tidyverse)

a <-  c("aa", "bb", "cc", NA, "aa") 
b <-  c("aa", "bb", "cc", "bb", "bb") 
c <-  c(NA, "aa", "aa", NA, "bb") 
d <-  c(1, 1, 2, 2, 3)
df <- data.frame(a,b,c,d)

cols <- c("a","b","c")

df %>%
  mutate(across(.cols = where(is.character),
                .fns = ~ case_when(.x %>% is.na ~ "Missing",
                .x == "aa" ~ "Missing",
                TRUE ~ .x)
  )
)

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另一种选择是使用常规函数

test_function <- function(col) {
  case_when(col %>% is.na() ~ "Missing",
            col == "aa" ~ "Missing",
            TRUE ~ col)
}

df %>% 
  mutate(across(.cols = where(is.character),
                .fns = test_function)
  )

型

您可以使用na_if()将"aa"替换为NA，并使用replace_na()将NA替换为"missing"。

library(dplyr)
library(tidyr)

cols <- c("a", "b", "c")

df %>%
  mutate(across(cols, ~ replace_na(na_if(.x, "aa"), "missing")))

#         a       b       c d
# 1 missing missing missing 1
# 2      bb      bb missing 1
# 3      cc      cc missing 2
# 4 missing      bb missing 2
# 5 missing      bb      bb 3

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你也可以从forcats使用fct_explicit_na()。

library(forcats)

df %>%
  mutate(across(cols, ~ fct_explicit_na(na_if(.x, "aa"))))

#           a         b         c d
# 1 (Missing) (Missing) (Missing) 1
# 2        bb        bb (Missing) 1
# 3        cc        cc (Missing) 2
# 4 (Missing)        bb (Missing) 2
# 5 (Missing)        bb        bb 3

型

Base R解决方案如下：

df[df=="aa" | is.na(df)]="missing"

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数据：

a <-  c("aa", "bb", "cc", NA, "aa") 
b <-  c("aa", "bb", "cc", "bb", "bb") 
c <-  c(NA, "aa", "aa", NA, "bb") 
d <-  c(1, 1, 2, 2, 3)
df <- data.frame(a,b,c,d)
cols <- c("a","b","c")

型

输出：

a       b       c d
1 missing missing missing 1
2      bb      bb missing 1
3      cc      cc missing 2
4 missing      bb missing 2
5 missing      bb      bb 3

型