函数来创建从两个data.frames列表的列表

我有两个data.frames firstdf和secondf，（示例数据如下。）我试图创建一个函数，可以输出结果，如下面的ExampleList数据。我想创建一个列表列表，它从firstdf获取第一行条目，并将它们的值放入exampleList中的thing和test字段中，然后从seconddf中的otherthing字段获取前3个值，并将它们连接在一起，并保存他们在exampleList中的其他字段中，然后移到firstdf中的下一行和seconddf中的下3个行中。循环对我来说有点棘手，所以提示是绝对赞赏的。函数来创建从两个data.frames列表的列表

data: 

dput(firstdf) 
structure(list(thing = structure(1:3, .Label = c("thing1", "thing2", 
"thing3"), class = "factor"), test = structure(1:3, .Label = c("test1", 
"test2", "test3"), class = "factor")), .Names = c("thing", "test" 
), row.names = c(NA, -3L), class = "data.frame") 

dput(seconddf) 
    structure(list(otherthing = structure(c(4L, 5L, 6L, 7L, 8L, 9L, 
    1L, 2L, 3L), .Label = c("thing10", "thing11", "thing12", "thing4", 
    "thing5", "thing6", "thing7", "thing8", "thing9"), class = "factor"), 
     other = structure(c(9L, 6L, 7L, 2L, 3L, 1L, 8L, 4L, 5L), .Label = c("fads", 
     "oiu", "qwer", "rewa", "rewq", "sfas", "sfwg", "tre", "xdfs" 
     ), class = "factor")), .Names = c("otherthing", "other"), row.names = c(NA, 
    -9L), class = "data.frame")

输出：

dput(ExampleList) 
list(structure(list(thing = "thing1", test = "test1", otherthing = c("thing4", 
"thing5", "thing6")), .Names = c("thing", "test", "otherthing" 
)), structure(list(thing = "thing2", test = "test2", otherthing = c("thing7", 
"thing8", "thing9")), .Names = c("thing", "test", "otherthing" 
)), structure(list(thing = "thing3", test = "test3", otherthing = c("thing10", 
"thing11", "thing12")), .Names = c("thing", "test", "otherthing" 
))) 
[[1]] 
[[1]]$thing 
[1] "thing1" 

[[1]]$test 
[1] "test1" 

[[1]]$otherthing 
[1] "thing4" "thing5" "thing6" 


[[2]] 
[[2]]$thing 
[1] "thing2" 

[[2]]$test 
[1] "test2" 

[[2]]$otherthing 
[1] "thing7" "thing8" "thing9" 


[[3]] 
[[3]]$thing 
[1] "thing3" 

[[3]]$test 
[1] "test3" 

[[3]]$otherthing 
[1] "thing10" "thing11" "thing12"

来源

2017-01-23 user3476463

的可能的复制http://stackoverflow.com/questions/26177565/converting-nested-list-to-dataframe – akrun

@akrun我愿意把这个作为重复，但我们需要一个更好的目标imho – Jaap

我说尽可能，而不是肯定 – akrun

你可以使用Map，多元版lapply（与split为otherthing）。第一个参数是施加于将被并行迭代上的多个参数的函数，所以

ExampleList <- Map(list, 
    thing = as.character(firstdf$thing), 
    test = as.character(firstdf$test), 
    otherthing = split(as.character(seconddf[[1]]), rep(1:3, each = 3))) 

str(ExampleList) 

## List of 3 
## $ thing1:List of 3 
## ..$ thing  : chr "thing1" 
## ..$ test  : chr "test1" 
## ..$ otherthing: chr [1:3] "thing4" "thing5" "thing6" 
## $ thing2:List of 3 
## ..$ thing  : chr "thing2" 
## ..$ test  : chr "test2" 
## ..$ otherthing: chr [1:3] "thing7" "thing8" "thing9" 
## $ thing3:List of 3 
## ..$ thing  : chr "thing3" 
## ..$ test  : chr "test3" 
## ..$ otherthing: chr [1:3] "thing10" "thing11" "thing12"

来源

2017-01-23 05:11:20 alistaire

谢谢你，你的建议与我的样本数据非常吻合。但是，我的真实数据secondddf数据更长，并且当我尝试您的建议时出现以下错误。有没有办法修改你的代码来处理更长的seconddf？错误：警告信息：在split.default（as.character（seconddf [[1]]），rep（长度（firstdf $ thing），：数据长度不是分割变量的倍数 – user3476463

检出'？split' 。从这个例子中，我假设你想要三分裂，但如果它对于你的真实数据是不同的，你需要构造一个向量来拆分，这个向量与要拆分的向量长度相同，每个拆分的唯一值我想结束。 – alistaire

函数来创建从两个data.frames列表的列表

回答

相关问题