可以使用sapply：

sapply(grep("x", names(dat)), function(y)
       write.csv(dat[, c(1, 2, y)], 
                 paste0(names(dat[y]), ".csv"),
                 row.names = FALSE)
)

grep查找具有x的列，sapply循环通过它们。这将为csv文件命名所选的列名，并将其保存在工作目录中。
注意，您也可以用其他方式指定列。有几种选择：

# using column locations (numbers) directly
sapply(3:5, function(y)
  write.csv(dat[, c(1, 2, y)], 
            paste0(names(dat[y]), ".csv"),
            row.names = FALSE)
)

# using column names
sapply(c("x2006", "x2007", "x2008"), function(y)
       write.csv(dat[, c("country", "industry", y)], 
                 paste0(names(dat[y]), ".csv"),
                 row.names = FALSE))

以下是tidyverse解决方案：

library(tidyverse)

dat %>%
  pivot_longer(cols = starts_with("x"), names_to = "year", values_to = "value") %>%
  split(.$year) %>%
  map(~ select(.x, country, industry, value)) %>%
  map2(names(.), ~ write_csv(.x, file = paste0("dat_", .y, ".csv")))

$x2006
# A tibble: 9 × 3
  country industry value
  <chr>   <chr>    <int>
1 A       X        99954
2 B       Y        27955
3 C       Z        36009
4 A       X         3061
5 B       Y        25612
6 C       Z        67307
7 A       X        96514
8 B       Y        97864
9 C       Z        43014

$x2007
# A tibble: 9 × 3
  country industry value
  <chr>   <chr>    <int>
1 A       X        83954
2 B       Y        96141
3 C       Z        62389
4 A       X        28568
5 B       Y        77503
6 C       Z        70458
7 A       X        34978
8 B       Y        35408
9 C       Z        68731

$x2008
# A tibble: 9 × 3
  country industry value
  <chr>   <chr>    <int>
1 A       X        29498
2 B       Y        62203
3 C       Z        98125
4 A       X        99549
5 B       Y        56839
6 C       Z        21621
7 A       X        84214
8 B       Y        85778
9 C       Z        90275