将嵌套列表的第一个元素绑定到嵌套列表的后续元素
将嵌套列表的第一个元素绑定到嵌套列表的后续元素
提问于 2019-02-22 02:54:50
我有一个嵌套列表，我希望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)
以下是前两个更高级别的列表。目标是输出一个包含两列的数据帧。列1是第一行，列2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"回答 1
Stack Overflow用户
回答已采纳
发布于 2019-02-22 04:06:46
编写一个执行所需操作的函数，并使用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" simple "：
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，而不是数据帧。如果您想要取出数据帧，您可以使用cbind.data.frame而不只是cbind。您还可以像在示例输出中那样添加列名，例如
foo = function(x) cbind.data.frame(Location = x[1], Bid = x[-1])页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/54814280
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