我有一个嵌套列表，我想将每个列表的第一个元素绑定到该嵌套列表的其余元素。（对整个集合中的每个列表执行此操作）
这是前 2 个更高级别的列表。目标是输出具有两列的数据帧。第 1 列是第一行，第 2 列是后续行。
简化版：
list(c("Location1",
"Location1_Bid1",
"Location1_Bid2",
"Location1_Bid3"),
c("Location2",
"Location2_Bid1",
"Location2_Bid2",
"Location2_Bid3"),
c("Location3",
"Location3_Bid1",
"Location3_Bid2",
"Location3_Bid3",
"Location3_Bid4")
, c("Location4",
"Location4_Bid1",
"Location4_Bid2"))
例如：
Location | Bid
"Location1" | "Location1_Bid1"
"Location1" | "Location1_Bid2"
"Location1" | "Location1_Bid3"
"Location2" | "Location2_Bid1"
"Location2" | "Location2_Bid2"
"Location2" | "Location2_Bid3"
"Location3" | "Location3_Bid1"
"Location3" | "Location3_Bid2"
"Location3" | "Location3_Bid3"
"Location3" | "Location3_Bid4"
"Location4" | "Location4_Bid1"
"Location4" | "Location4_Bid2"
编写一个函数来执行您想要的操作并用于lapply对每个列表项执行此操作：
foo = function(x) cbind(x[1], x[-1])
result = lapply(your_list, foo)
将您的简单示例称为“简单”：
lapply(simple, foo)
# [[1]]
# [,1] [,2]
# [1,] "Location1" "Location1_Bid1"
# [2,] "Location1" "Location1_Bid2"
# [3,] "Location1" "Location1_Bid3"
#
# [[2]]
# [,1] [,2]
# [1,] "Location2" "Location2_Bid1"
# [2,] "Location2" "Location2_Bid2"
# [3,] "Location2" "Location2_Bid3"
#
# [[3]]
# [,1] [,2]
# [1,] "Location3" "Location3_Bid1"
# [2,] "Location3" "Location3_Bid2"
# [3,] "Location3" "Location3_Bid3"
# [4,] "Location3" "Location3_Bid4"
#
# [[4]]
# [,1] [,2]
# [1,] "Location4" "Location4_Bid1"
# [2,] "Location4" "Location4_Bid2"
这些是矩阵，而不是数据框。如果您想要数据帧，您可以使用cbind.data.frame而不仅仅是cbind。您还可以在示例输出中添加列名称，例如
foo = function(x) cbind.data.frame(Location = x[1], Bid = x[-1])
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