根据 R 中另一列中的值替换一列中第二次出现的字符串

Question

这是一个示例数据框：

a <- c("cat", "dog", "mouse")
b <- c("my cat is a tabby cat and is a friendly cat", "walk the dog", "the mouse is scared of the other mouse")
df <- data.frame(a,b)

我希望能够删除 b 列中 a 列中第二次出现的值。

这是我想要的输出：

a      b
cat    my cat is a tabby and is a friendly cat
dog    walk the dog
mouse  the mouse is scared of the other

我已经尝试了 gsub 和一些 stringr 函数的不同组合，但我什至还没有接近于能够删除 col a 和 col b 中字符串的第二次（也是仅第二次）出现。我想我问的是类似于 this 的问题，但我不熟悉 Perl，无法将其翻译成 R。

谢谢！

Answer 1

构建正确的正则表达式需要一些工作。

P1 = paste(a, collapse="|")
PAT = paste0("((", P1, ").*?)(\2)")

sub(PAT, "\1", b, perl=TRUE)
[1] "my cat is a tabby  and is a friendly cat"
[2] "walk the dog"                            
[3] "the mouse is scared of the other "

Answer 2

你可以这样做...

library(stringr)
df$b <- str_replace(df$b, 
                    paste0("(.*?",df$a,".*?) ",df$a), 
                    "\1")

df
      a                                       b
1   cat my cat is a tabby and is a friendly cat
2   dog                            walk the dog
3 mouse        the mouse is scared of the other

正则表达式查找第一个字符串中某处带有 df$a 的字符，后跟一个 space 和另一个 df$a。捕获组是直到第二次出现（由 (...) 表示）之前 space 的文本，并且整个文本（包括第二次出现）被捕获组 \1 替换（具有删除第二个 df$a 及其前面的 space 的效果）。第二个 df$a 之后的任何内容都不会受到影响。

Answer 3

我实际上找到了另一个解决方案，虽然更长，但对于其他正则表达式初学者来说可能更清晰：

library(stringr)
# Replace first instance of col a in col b with "INTERIM" 
df$b <- str_replace(b, a, "INTERIM")

# Now that the original first instance of col a is re-labeled to "INTERIM", I can again replace the first instance of col a in col b, this time with an empty string
df$b <- str_replace(df$b, a, "")

# And I can re-replace the re-labeled "INTERIM" to the original string in col a
df$b <- str_replace(df$b, "INTERIM", a)

# Trim "double" whitespace
df$b <- str_replace(gsub("\s+", " ", str_trim(df$b)), "B", "b")


df
a            b
cat          my cat is a tabby and is a friendly cat
dog          walk the dog
mouse        the mouse is scared of the other

Answer 4

基础 R，拆分应用组合解决方案：

# Split-apply-combine: 

data.frame(do.call("rbind", lapply(split(df, df$a), function(x){

        b <- paste(unique(unlist(strsplit(x$b, "\s+"))), collapse = " ")

        return(data.frame(a = x$a, b = b))

      }

    )

  ), 

  stringsAsFactors = FALSE, row.names = NULL

)

数据：

df <- data.frame(a = c("cat", "dog", "mouse"),
                 b = c("my cat is a tabby cat and is a friendly cat", "walk the dog", "the mouse is scared of the other mouse"), 
                 stringsAsFactors = FALSE)

根据 R 中另一列中的值替换一列中第二次出现的字符串

Replace second occurrence of a string in one column based on value in other column in R

regex

string

r

find-occurrences