以R为基数：

df[] <- lapply(df, function(x) gsub("var","new", x))

输出：

#   feature1 feature2 feature3 feature4
# 1    new_a    new_c    new_b    new_a
# 2    new_b    new_a    new_a    new_c
# 3    new_c    new_b    new_c    new_b

数据

df <- read.table(text = "feature1   feature2    feature3    feature4
var_a   var_c   var_b   var_a
var_b   var_a   var_a   var_c
var_c   var_b   var_c   var_b", header = TRUE)
df <- data.table::data.table(df)

这是一个接受 Dataframe 、旧字符串及其替换的函数。

library(tidyverse)

replace_func <- function(df, var, new_var) {
  df %>%
    mutate(across(everything(), ~ .x %>%
                    str_replace_all(var, new_var)))
}

replace_func(df, "var", "new")

# A tibble: 3 × 4
  feature1 feature2 feature3 feature4
  <chr>    <chr>    <chr>    <chr>   
1 new_a    new_c    new_b    new_a   
2 new_b    new_a    new_a    new_c   
3 new_c    new_b    new_c    new_b

df[, lapply(.SD, sub, pattern = "var", replacement = "new", fixed = TRUE)]
#    feature1 feature2 feature3 feature4
# 1:    new_a    new_c    new_b    new_a
# 2:    new_b    new_a    new_a    new_c
# 3:    new_c    new_b    new_c    new_b

使用此示例数据：

library(data.table)
df = fread(text = 'feature1     feature2    feature3    feature4
var_a   var_c   var_b   var_a
var_b   var_a   var_a   var_c
var_c   var_b   var_c   var_b')

使用dplyr和across

library(dplyr)

df %>% 
  mutate(across(1:4, ~ sub(".*(_)", "new\\1", .x)))
  feature1 feature2 feature3 feature4
1    new_a    new_c    new_b    new_a
2    new_b    new_a    new_a    new_c
3    new_c    new_b    new_c    new_b