您应该在grepl中构造一个正则表达式：

df %>% 
  mutate(check = grepl(paste(names$names, collapse = "|"), sentences))

# A tibble: 3 × 2
  sentences                    check
  <chr>                        <lgl>
1 Bob is looking for something TRUE 
2 Adriana has an umbrella      FALSE
3 Michael is looking at...     TRUE

这是一个R基溶液。

inx <- sapply(names$names, \(pat) grepl(pat, df$sentences))
inx
#>        Bob  Mary Michael  John  Etc.
#> [1,]  TRUE FALSE   FALSE FALSE FALSE
#> [2,] FALSE FALSE   FALSE FALSE FALSE
#> [3,] FALSE FALSE    TRUE FALSE FALSE

inx <- rowSums(inx) > 0L
df$check <- inx
df
#> # A tibble: 3 × 2
#>   sentences                    check
#>   <chr>                        <lgl>
#> 1 Bob is looking for something TRUE 
#> 2 Adriana has an umbrella      FALSE
#> 3 Michael is looking at...     TRUE

创建于2023年1月11日，使用reprex v2.0.2

grep和family期望pattern=的长度为1。类似地，str_detect需要相同长度的 * 字符串 *，而不是逻辑向量，因此不能按原样工作。
我们有几个选择：

• sapply（矩阵），并查看每行是否有一个或多个匹配项：

df %>%
  mutate(check = rowSums(sapply(names$names, grepl, sentences)) > 0)
# # A tibble: 3 × 2
#   sentences                    check
#   <chr>                        <lgl>
# 1 Bob is looking for something TRUE 
# 2 Adriana has an umbrella      FALSE
# 3 Michael is looking at...     TRUE

（我现在明白了，这是鲁伊·巴拉达斯的回答。）

• 使用fuzzyjoin对数据执行模糊连接：

df %>%
  fuzzyjoin::regex_left_join(names, by = c(sentences = "names")) %>%
  mutate(check = !is.na(names))
# # A tibble: 3 × 3
#   sentences                    names   check
#   <chr>                        <chr>   <lgl>
# 1 Bob is looking for something Bob     TRUE 
# 2 Adriana has an umbrella      NA      FALSE
# 3 Michael is looking at...     Michael TRUE

这种方法的一个优点是，它可以告诉您哪个模式（在names中）进行了匹配。

也许我们可以尝试adist + colSums，如下所示

df %>%
  mutate(check = colSums(adist(names$names, sentences, fixed = FALSE) == 0) > 0)

它给出了

# A tibble: 3 × 2
  sentences                    check
  <chr>                        <lgl>
1 Bob is looking for something TRUE
2 Adriana has an umbrella      FALSE
3 Michael is looking at...     TRUE