如何在列表列上应用函数并在 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。

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structure(list(sample_num = 1:6, vector = list(c(0, 1, 1, 0, \n1, 2, 0, 0, 3, 0), c(0, 0, 1, 2, 0, 0, 4, 10, 12, 1), c(1, 33, \n4, 4, 2, 2, 6, 9, 14, 2), c(0, 0, 1, 0, 1, 0, 1, 5, 3, 0), c(0, \n1, 1, 0, 0, 0, 1, 4, 3, 0), c(0, 0, 1, 0, 0, 0, 1, 1, 1, 0))), class = c("tbl_df", \n"tbl", "data.frame"), row.names = c(NA, -6L), .Names = c("sample_num", \n"vector"))\n\n> test\n# A tibble: 6 \xc3\x97 2\n  sample_num     vector\n       <int>     <list>\n1          1 <dbl [10]>\n2          2 <dbl [10]>\n3          3 <dbl [10]>\n4          4 <dbl [10]>\n5          5 <dbl [10]>\n6          6 <dbl [10]>\n

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我尝试使用 lmap 但收到错误消息

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> test  %>% lmap(.$vector,.f = function(x) x[1:5])\nError in .f(.x[i], ...) : \n  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)))\n

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谢谢你！

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Answer 1

这就是你想要做的吗？

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structure(list(sample_num = 1:6, vector = list(c(0, 1, 1, 0, \n1, 2, 0, 0, 3, 0), c(0, 0, 1, 2, 0, 0, 4, 10, 12, 1), c(1, 33, \n4, 4, 2, 2, 6, 9, 14, 2), c(0, 0, 1, 0, 1, 0, 1, 5, 3, 0), c(0, \n1, 1, 0, 0, 0, 1, 4, 3, 0), c(0, 0, 1, 0, 0, 0, 1, 1, 1, 0))), class = c("tbl_df", \n"tbl", "data.frame"), row.names = c(NA, -6L), .Names = c("sample_num", \n"vector"))\n\ntest$new = lapply(test$vector, function(x) {x[1:5]})\ntest\n\n# A tibble: 6 \xc3\x97 3\n  sample_num     vector       new\n       <int>     <list>    <list>\n1          1 <dbl [10]> <dbl [5]>\n2          2 <dbl [10]> <dbl [5]>\n3          3 <dbl [10]> <dbl [5]>\n4          4 <dbl [10]> <dbl [5]>\n5          5 <dbl [10]> <dbl [5]>\n6          6 <dbl [10]> <dbl [5]>\n\n\ntest$vector[3]\n[[1]]\n [1]  1 33  4  4  2  2  6  9 14  2\n\ntest$new[3]\n[[1]]\n[1]  1 33  4  4  2\n

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如果要使用 dplyr 语法，请先定义一个函数：

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

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然后，将其按行应用到列vector

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test = test %>%\n    rowwise() %>%\n    mutate(new_dplyr = f(vector))\n\ntest\n# A tibble: 6 \xc3\x97 3\n  sample_num     vector new_dplyr\n       <int>     <list>    <list>\n1          1 <dbl [10]> <dbl [5]>\n2          2 <dbl [10]> <dbl [5]>\n3          3 <dbl [10]> <dbl [5]>\n4          4 <dbl [10]> <dbl [5]>\n5          5 <dbl [10]> <dbl [5]>\n6          6 <dbl [10]> <dbl [5]>\n\ntest$vector[3]\n[[1]]\n [1]  1 33  4  4  2  2  6  9 14  2\n\ntest$new_dplyr[3]\n[[1]]\n[1]  1 33  4  4  2\n

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