我们可以捕获字符（(...)）直到'Word'作为一个组，然后在replacement（str_replace）中使用捕获组的反向引用（\\1）。.*表示我们丢弃的其余字符。str_replace也被向量化以进行替换，因此我们不需要任何循环

library(dplyr)
library(stringr)
df1 %>%
   mutate(After = str_replace(String, sprintf("(.*%s).*", Word), "\\1"))

• 输出

String   Word                  After
1         tasty red apple number 1  apple        tasty red apple
2       tasty red apple and banana  apple        tasty red apple
3 tasty banana and apple and peach  apple tasty banana and apple
4           tasty banana and peach banana           tasty banana
5            tasty peach and apple  peach            tasty peach

数据

df1 <- structure(list(String = c("tasty red apple number 1",
 "tasty red apple and banana", 
"tasty banana and apple and peach", "tasty banana and peach", 
"tasty peach and apple"), Word = c("apple", "apple", "apple", 
"banana", "peach")), class = "data.frame", row.names = c(NA, 
-5L))

在mapply中对gsub使用lookbehind以删除字符串中不需要的部分。

transform(dat, After=mapply(\(x, y) gsub(sprintf('(?<=%s).*',  x), '', y, perl=TRUE), Word, String))
#                             String   Word                  After
# 1         tasty red apple number 1  apple        tasty red apple
# 2       tasty red apple and banana  apple        tasty red apple
# 3 tasty banana and apple and peach  apple tasty banana and apple
# 4           tasty banana and peach banana           tasty banana
# 5            tasty peach and apple  peach            tasty peach

• 数据：*

dat <- structure(list(String = c("tasty red apple number 1", "tasty red apple and banana", 
"tasty banana and apple and peach", "tasty banana and peach", 
"tasty peach and apple"), Word = c("apple", "apple", "apple", 
"banana", "peach")), class = "data.frame", row.names = c(NA, 
-5L))

试试这个：

df1 %>%
  mutate(After = str_replace(String, str_c("(.*\\b", Word, "\\b).*"), "\\1"))
                            String   Word                  After
1         tasty red apple number 1  apple        tasty red apple
2       tasty red apple and banana  apple        tasty red apple
3 tasty banana and apple and peach  apple tasty banana and apple
4           tasty banana and peach banana           tasty banana
5            tasty peach and apple  peach            tasty peach

在这里我们（i）将Word Package 到字边界\\b中以防止包含Word值的较大字（例如，“dapple”和“apple”）被匹配。（ii）我们将该子串用括号括起来，以将其强制到捕获组中，然后我们（iii）在str_replace替换自变量中引用，而捕获组（.*）之后的任何内容都被省略。

您也可以使用str_extract而不是str_replace来获得稍微简单的语法：

df1 |>
  mutate(After = str_extract(String, str_c(".*\\b", Word, "\\b")))