Question

我有一个嵌套列表，需要计算frt和srt的相关性

   $`bs. bs`
    fapp  frt sapp  srt
1   bs 2280   bs 0.25
2   bs 2287   bs 0.25
3   bs 2288   bs 0.25
4   bs 2289   bs 0.25

$`bs. lhc`
    fapp  frt sapp      srt
5   bs 2320  lhc 0.250000
6   bs 2333  lhc 0.250214
7   bs 2524  lhc 0.316449

源：

    structure(list(`bs. bs` = structure(list(fapp = structure(c(1L,
    1L, 1L, 1L), .Label = "bs", class = "factor"), frt = c(2280L,
    2287L, 2288L, 2289L), sapp = structure(c(1L, 1L, 1L, 1L), .Label = c(" bs",
    " lhc"), class = "factor"), srt = c(0.25, 0.25, 0.25, 0.25)), .Names = c("fapp",
    "frt", "sapp", "srt"), row.names = c(NA, 4L), class = "data.frame"),
   `bs. lhc` = structure(list(fapp = structure(c(1L, 1L, 1L), .Label = "bs", class=         "factor"),
    frt = c(2320L, 2333L, 2524L), sapp = structure(c(2L,
    2L, 2L), .Label = c(" bs", " lhc"), class = "factor"),
    srt = c(0.25, 0.250214, 0.316449)), .Names = c("fapp",
"frt", "sapp", "srt"), row.names = 5:7, class = "data.frame")), .Names = c("bs. bs",
   "bs. lhc"))

类似

    ddply(y,.(fapp + sapp),cor)

或

    ddply(y,.(fapp,sapp),cor)

不起作用

Answer 1

> ldply(y, function(x) { x$corr <- cor(x$frt, x$srt); x })
      .id fapp  frt sapp      srt      corr
1  bs. bs   bs 2280   bs 0.250000        NA
2  bs. bs   bs 2287   bs 0.250000        NA
3  bs. bs   bs 2288   bs 0.250000        NA
4  bs. bs   bs 2289   bs 0.250000        NA
5 bs. lhc   bs 2320  lhc 0.250000 0.9985343
6 bs. lhc   bs 2333  lhc 0.250214 0.9985343
7 bs. lhc   bs 2524  lhc 0.316449 0.9985343
Warning message:
In cor(x$frt, x$srt) : the standard deviation is zero
Calls: ldply -> llply -> structure -> lapply -> FUN -> cor

或将结果保留为列表

> llply(y, function(x) { x$corr <- cor(x$frt, x$srt); x })
$`bs. bs`
  fapp  frt sapp  srt corr
1   bs 2280   bs 0.25   NA
2   bs 2287   bs 0.25   NA
3   bs 2288   bs 0.25   NA
4   bs 2289   bs 0.25   NA

$`bs. lhc`
  fapp  frt sapp      srt      corr
5   bs 2320  lhc 0.250000 0.9985343
6   bs 2333  lhc 0.250214 0.9985343
7   bs 2524  lhc 0.316449 0.9985343

Warning message:
In cor(x$frt, x$srt) : the standard deviation is zero
Calls: llply -> structure -> lapply -> FUN -> cor

Answer 2

你想使用ldply，它将你的功能应用于列表的元素，然后像你的ddply调用的第二个版本应该有效。没有示例数据，我无法证明它。

Answer 3

ddply明确地设计用于处理data.frame。所以，我首先将列表的所有元素放在data.frame：

中

dat = do.call("rbind", nested_list)

然后使用ddply：

ddply(dat, .(fapp, sapp), corr)

如何ddply嵌套列表

3 个答案: