r - 如何在 R 中的数据框中用前缀替换后缀？

Question

我在 R 中有一个如下所示的数据框:

df <- data.frame(apple = c("apple", "apple", "applebad", "apple", "apple"),
                orange = c("orangebad", "orangebad", "orange", "orange", "orange"),
                pear = c("pear", "pear", "pear", "pearbad", "pear"))    

然而，从这个数据框中，我想将“bad”的后缀更改为 rotten 的前缀，因此我得到了这样一个数据框:

df_new <- data.frame(apple = c("apple", "apple", "rottenapple", "apple", "apple"),
                orange = c("rottenorange", "rottenorange", "orange", "orange", "orange"),
                pear = c("pear", "pear", "pear", "rottenpear", "pear")) 

所以我知道我可以像这样使用 str_replace(来自 stringr)或 gsub 逐列完成:

df$apple <- gsub("applebad", paste0("rottenapple"), df$apple)
df$apple <- str_replace(df$apple, "applebad", "rottenapple")

但我不确定该怎么做才能遍历数据框而不必声明“rottenapple”、“rottenpear”，因为我的真实数据集有 100 多列。我尝试了以下但不起作用:

item <- c("apple", "orange", "pear")

df[,1] <- str_replace(df[,1], "bad", paste0("rotten", item[1]))

非常感谢任何帮助。

Answer 1

我们可以将其余字符捕获为一组 ((.*))，后跟 'bad' 并替换为 rotten 子字符串和反向引用(\\1) 捕获的组

library(dplyr)
library(stringr)
df <- df %>% 
    mutate(across(everything(),  ~str_replace(.x, "(.*)bad", "rotten\\1")))

-检查

all.equal(df, df_new)
[1] TRUE

---

或者使用base R

 df[] <- lapply(df, gsub, pattern = "(.*)bad", replacement = "rotten\\1")
> df
        apple       orange       pear
1       apple rottenorange       pear
2       apple rottenorange       pear
3 rottenapple       orange       pear
4       apple       orange rottenpear
5       apple       orange       pear