r - 如何在列表列上应用一个函数并在 dplyr 和 purrr 中返回另一个函数？

Question

我想获取列表列中每个向量中的前 5 个值，并将其作为保存为列表的数据框中的新列返回。

structure(list(sample_num = 1:6, vector = list(c(0, 1, 1, 0, 
1, 2, 0, 0, 3, 0), c(0, 0, 1, 2, 0, 0, 4, 10, 12, 1), c(1, 33, 
4, 4, 2, 2, 6, 9, 14, 2), c(0, 0, 1, 0, 1, 0, 1, 5, 3, 0), c(0, 
1, 1, 0, 0, 0, 1, 4, 3, 0), c(0, 0, 1, 0, 0, 0, 1, 1, 1, 0))), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -6L), .Names = c("sample_num", 
"vector"))

> test
# A tibble: 6 × 2
  sample_num     vector
       <int>     <list>
1          1 <dbl [10]>
2          2 <dbl [10]>
3          3 <dbl [10]>
4          4 <dbl [10]>
5          5 <dbl [10]>
6          6 <dbl [10]>

我尝试使用 lmap 但我收到错误消息

> test  %>% lmap(.$vector,.f = function(x) x[1:5])
Error in .f(.x[i], ...) : 
  unused argument (list(c(0, 1, 1, 0, 1, 2, 0, 0, 3, 0), c(0, 0, 1, 2, 0, 0, 4, 10, 12, 1), c(1, 33, 4, 4, 2, 2, 6, 9, 14, 2), c(0, 0, 1, 0, 1, 0, 1, 5, 3, 0), c(0, 1, 1, 0, 0, 0, 1, 4, 3, 0), c(0, 0, 1, 0, 0, 0, 1, 1, 1, 0)))

谢谢!

Answer 1

这是你想要做的吗？

structure(list(sample_num = 1:6, vector = list(c(0, 1, 1, 0, 
1, 2, 0, 0, 3, 0), c(0, 0, 1, 2, 0, 0, 4, 10, 12, 1), c(1, 33, 
4, 4, 2, 2, 6, 9, 14, 2), c(0, 0, 1, 0, 1, 0, 1, 5, 3, 0), c(0, 
1, 1, 0, 0, 0, 1, 4, 3, 0), c(0, 0, 1, 0, 0, 0, 1, 1, 1, 0))), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -6L), .Names = c("sample_num", 
"vector"))

test$new = lapply(test$vector, function(x) {x[1:5]})
test

# A tibble: 6 × 3
  sample_num     vector       new
       <int>     <list>    <list>
1          1 <dbl [10]> <dbl [5]>
2          2 <dbl [10]> <dbl [5]>
3          3 <dbl [10]> <dbl [5]>
4          4 <dbl [10]> <dbl [5]>
5          5 <dbl [10]> <dbl [5]>
6          6 <dbl [10]> <dbl [5]>


test$vector[3]
[[1]]
 [1]  1 33  4  4  2  2  6  9 14  2

test$new[3]
[[1]]
[1]  1 33  4  4  2

如果要使用dplyr语法，先定义一个函数:

f = function(x) {
    return(list(x[1:5]))
    }

然后，将它逐行应用到列 vector

test = test %>%
    rowwise() %>%
    mutate(new_dplyr = f(vector))

test
# A tibble: 6 × 3
  sample_num     vector new_dplyr
       <int>     <list>    <list>
1          1 <dbl [10]> <dbl [5]>
2          2 <dbl [10]> <dbl [5]>
3          3 <dbl [10]> <dbl [5]>
4          4 <dbl [10]> <dbl [5]>
5          5 <dbl [10]> <dbl [5]>
6          6 <dbl [10]> <dbl [5]>

test$vector[3]
[[1]]
 [1]  1 33  4  4  2  2  6  9 14  2

test$new_dplyr[3]
[[1]]
[1]  1 33  4  4  2