Question

我有两个data.frames firstdf和secondf，（下面的示例数据。）我正在尝试创建一个输出结果的函数，如下面的ExampleList数据。我想创建一个列表列表，它从firstdf中获取第一行条目，并将它们的值放在exampleList中的thing和test字段中，然后从seconddf中的其他字段中获取前三个值，将它们连接在一起，然后保存它们在exampleList的otherthing字段中，然后移到firstdf中的下一行和seconddf中的下一行。循环对我来说有点棘手，所以提示绝对值得赞赏。

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"

Answer 1

您可以使用Map，lapply的多变量版本（split为otherthing）。第一个参数是一个应用于多个参数的函数，这些参数将并行迭代，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"

从两个data.frames

1 个答案: