更新：感谢Gregor托马斯：

library(dplyr)

left_join(df1,df2, by=c("reference"="data")) %>% 
  count(reference)

输出：

reference n
1       cat 1
2       dog 2

我们可以使用semi_join和count：

library(dplyr)

semi_join(df2,df1, by=c("data"="reference")) %>% 
  count(data)

data n
1  cat 1
2  dog 2

基于tidyverse的可能解决方案：

library(tidyverse)

df1 <- data.frame(reference=c("cat","dog"))
df2 <- data.frame(data=c("cat","car","catt","cart","dog","dog","pitbull"))

df1 %>% 
  group_by(animal = reference) %>% 
  summarise(n = sum(reference == df2$data), .groups = "drop")

#> # A tibble: 2 × 2
#>   animal     n
#>   <chr>  <int>
#> 1 cat        1
#> 2 dog        2

使用连接可能会更快

library(data.table)
setDT(df2)[, .(animals = data)][df1, .(n = .N), 
     on = .(animals = reference), by = .EACHI]
   animals n
1:     cat 1
2:     dog 2

或者在subset执行base R中的数据后使用table

table(subset(df2, data %in% df1$reference, select = data))

这是第三种选择：

library(tidyverse)

df1 <- tibble(reference=c("cat","dog"))
df2 <- tibble(data=c("cat","car","catt","cart","dog","dog","pitbull"))

df2 |>
  count(data) |>
  filter(data %in% df1$reference) |>
  rename(animal = data)
#> # A tibble: 2 x 2
#>   animal     n
#>   <chr>  <int>
#> 1 cat        1
#> 2 dog        2

我们可以使用str_count将第二个df中的列折叠成一个字符串。

library(tidyverse)

df1 %>%
  transmute(animals = reference, n = str_c(df2$data, collapse = " ") %>%
    str_count(str_c("\\b", reference, "\\b")) )
#>   animals n
#> 1     cat 1
#> 2     dog 2

创建于2021-12-29由reprex package（v2.0.1）

df1$n <- colSums(outer(df2$data, df1$reference, '=='))

df1
#>   reference n
#> 1       cat 1
#> 2       dog 2