对于data.frames对象的单个列表，我通常可以轻松地进行转换：
list1 list2 list3
1 x i
1 x j
1 x l
1 y i
1 y j
1 y l
2 x i
2 x j
2 x l
2 y i
2 y j
2 y l
但是，我当前拥有的列表对象是嵌套的。它是data.frames列表的列表。需要注意的几点：
这是我正在处理的事情的简化示例：
my_df <- do.call("rbind", lapply(my_list, data.frame))
上面的List of 3
$ 1 :List of 2
..$ : NULL
..$ : NULL
$ 2 :List of 2
..$ :'data.frame': 3 obs. of 2 variables:
.. ..$ name : chr [1:3] "jack" "jim" "joe" "jon"
.. ..$ value : chr [1:3] "10" "12" "13" "14"
..$ :'data.frame': 4 obs. of 2 variables:
.. ..$ name : chr [1:4] "jacky" "jane" "juanita" "julia"
.. ..$ value : chr [1:4] "11" "9" "10" "14"
$ 3 :List of 1
..$ :'data.frame': 5 obs. of 2 variables:
.. ..$ name : chr [1:5] "adam" "ashley" "arnold" "avery" "arthur"
.. ..$ value : chr [1:5] "6" "7" "11" "12" "11"
方法报告了一个错误do.call，看来我的具有不同行号的data.frames列表导致了此问题？
我尝试了this post中描述的一些策略，但是每次尝试都有其自己的独特错误。
arguments imply differing number of rows“ rbindlist”方法和data.table“ bind_rows”方法均报告：
dplyr感谢有关如何处理这种情况的任何提示。
答案 0 :(得分：0)
考虑两个select l.*
from (select l.*,
row_number() over (partition by name
order by (feature_code = 'RGN') desc, id
) as seqnum
from locations l
) l
where seqnum = 1;
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var m = Math.floor(d % 3600 / 60);
var s = Math.floor(d % 3600 % 60);
var hDisplay = h > 0 ? h + (h == 1 ? " hour, " : " hours, ") : "";
var mDisplay = m > 0 ? m + (m == 1 ? " minute, " : " minutes, ") : "";
var sDisplay = s > 0 ? s + (s == 1 ? " second" : " seconds") : "";
return hDisplay + mDisplay + sDisplay;
}
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element.innerHTML = "Time Remaining "+h;
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之间运行：
do.call答案 1 :(得分：0)
首先，NULL的值无关紧要，因为rbind(NULL, NULL, ..., A, B, C, ...)与rbind(A, B, C, ...)相同。
第二，列表的结构很重要。如果嵌套列表与示例一样简单，那么答案也很简单。下面的代码可以解决此问题：
# This list is the same as your example
test <- list(
list(NULL, NULL),
list(data.frame(name = c("jack", "jim", "joe", "jon"),
value = c("10", "12", "13", "14")),
data.frame(name = c("jacky", "jane", "juanita", "julia"),
value = c("11", "9", "10", "14"))),
list(data.frame(name = c("adam", "ashley", "arnold", "avery", "arthur"),
value = c("6", "7", "11", "12", "11")))
)
# This function rbinds the dataframes inside a list
ls_rbind <- function(ls_of_df) do.call(rbind, ls_of_df)
# Applying "ls_rbind" to each of your child lists creates a list, each of whose elements is either a NULL or a dataframe
# Applying "ls_rbind" again to the result list gives you a dataframe you want
result_df <- ls_rbind(lapply(test, ls_rbind))
但是，如果嵌套列表实际上更复杂，那么您可能需要一种更通用的方式来处理它。例如，每个子列表可以是以下项之一：
NULL NULL的列表在这种情况下，递归可能会有所帮助。考虑以下代码：
# These two lines complicate the list structure
test[[4]] <- test
test[[1]][[3]] <- test
recr_ls_rbind <- function(ls_of_ls) {
are_lists <- lapply(ls_of_ls, class) == "list"
# Any child lists will be recursively evaluated. "part1" is always a list
part1 <- lapply(ls_of_ls[are_lists], recr_ls_rbind)
# Applying the above function "ls_rbind" to all non-list child items and then coerce the result into a list.
part2 <- list(ls_rbind(ls_of_ls[!are_lists]))
# Put part1 and part2 into the same list and apply "ls_rbind" to it.
ls_rbind(c(part1, part2))
}
result_df <- recr_ls_rbind(test)