Question

我有以下数据表：

>

 dt
    folder  type             value
 1:      1  prob 0.683541877259838
 2:      1 label                 1
 3:      1  prob 0.495066106474617
 4:      1 label                 1
 5:      1  prob 0.640406598486647
 6:      1 label                 1
 7:      1  prob 0.482671032167111
 8:      1 label                 1
 9:      1  prob 0.683541877259838
10:      1 label                 1
11:      1  prob 0.640406598486647
12:      1 label                 0
13:      1  prob 0.726221221435974
14:      1 label                 0
15:      1  prob 0.565276604111103
16:      1 label                 1
17:      1  prob                 0
18:      1 label                 0
19:      1  prob 0.565276604111103
20:      1 label                 1
21:      1  prob 0.726221221435974
22:      1 label                 1
23:      1  prob 0.738455509755984
24:      1 label                 1
25:      1  prob                 1
26:      1 label                 0
27:      1  prob 0.640406598486647
28:      1 label                 0
29:      1  prob 0.743529095250444
30:      1 label                 0
31:      1  prob 0.640406598486647
32:      1 label                 0
33:      1  prob 0.726221221435974
34:      1 label                 1
35:      1  prob 0.683541877259838
36:      1 label                 1
    folder  type             value

我想实现的是一个嵌套的列表列表，如下所示（第一级是“类型”数据表列中包含的变量，第二级是“文件夹”，然后是所有值）：

labels[[1]]
  [1]  1 1 1 1 1 0
  [6]  0 0 1

labels[[2]]
  [1]  1 1 1 1 1 0
  [6]  0 0 1

prob [[1]]
  [1]  0.1 0.21 0.1 0.1 0.43 0.0
  [6] 0.0 0 1

prob[[2]]
  [1]  0.56 0.64 0.3 0.21 0.81 0.5
  [6]  0.33 0.4

我尝试过类似的事情：

dt2 = dt[, {
    l1 = list()
    l1[[type]][[folder]]= as.list(value)
    list(l1)
  }, by = list(type,folder)]


  dt3 = dt2[, list({
    l2 = list()
    l2[[type]][[folder]] = unlist(V1, recursive = FALSE)
    l2
  }), by = list(type, folder) ]  


  out = dt3[['V1']]
  names(out) = dt3[['type']]

但是结果不是预期的

$prob
$prob$`1`
$prob$`1`$`1`
$prob$`1`$`1`[[1]]
[1] "0.683541877259838"

$prob$`1`$`1`[[2]]
[1] "0.495066106474617"

$prob$`1`$`1`[[3]]
[1] "0.640406598486647"

$prob$`1`$`1`[[4]]
[1] "0.482671032167111"

$prob$`1`$`1`[[5]]
[1] "0.683541877259838"

$prob$`1`$`1`[[6]]
[1] "0.640406598486647"

$prob$`1`$`1`[[7]]
[1] "0.726221221435974"

$prob$`1`$`1`[[8]]
[1] "0.565276604111103"

$prob$`1`$`1`[[9]]
[1] "0"

$prob$`1`$`1`[[10]]
[1] "0.565276604111103"

$prob$`1`$`1`[[11]]
[1] "0.726221221435974"

$prob$`1`$`1`[[12]]
[1] "0.738455509755984"

$prob$`1`$`1`[[13]]
[1] "1"

$prob$`1`$`1`[[14]]
[1] "0.640406598486647"

$prob$`1`$`1`[[15]]
[1] "0.743529095250444"

$prob$`1`$`1`[[16]]
[1] "0.640406598486647"

$prob$`1`$`1`[[17]]
[1] "0.726221221435974"

$prob$`1`$`1`[[18]]
[1] "0.683541877259838"




$label
$label$`1`
$label$`1`$`1`
$label$`1`$`1`[[1]]
[1] "1"

$label$`1`$`1`[[2]]
[1] "1"

$label$`1`$`1`[[3]]
[1] "1"

$label$`1`$`1`[[4]]
[1] "1"

$label$`1`$`1`[[5]]
[1] "1"

$label$`1`$`1`[[6]]
[1] "0"

$label$`1`$`1`[[7]]
[1] "0"

$label$`1`$`1`[[8]]
[1] "1"

$label$`1`$`1`[[9]]
[1] "0"

$label$`1`$`1`[[10]]
[1] "1"

$label$`1`$`1`[[11]]
[1] "1"

$label$`1`$`1`[[12]]
[1] "1"

$label$`1`$`1`[[13]]
[1] "0"

$label$`1`$`1`[[14]]
[1] "0"

$label$`1`$`1`[[15]]
[1] "0"

$label$`1`$`1`[[16]]
[1] "0"

$label$`1`$`1`[[17]]
[1] "1"

$label$`1`$`1`[[18]]
[1] "1"

关于如何实现此列表格式的任何建议？

Answer 1

解决了以下问题：

> #create support lists
> f.list = vector('list', 2)    #list for variables
> o.list=vector('list',n_folds) #list for folds
> 
> #list the levels of type (variables)
> t<-as.data.frame(table(dt$type))
> 
> for (i in 1:2) {
+   for (j in 1:n_folds) {
+     o.list[[j]] = as.numeric(filter(dt, folder==j & type==t[i,1])$value)
+   }
+   f.list[[i]] = o.list
+ }
> 
> #rename list names
> names(f.list)[[1]]="label"
> names(f.list)[[2]]="prob"
> 
> str(f.list)
List of 2
 $ label:List of 10
  ..$ : num [1:18] 0 1 1 0 1 0 0 1 1 0 ...
  ..$ : num [1:19] 0 1 1 1 1 1 0 0 1 1 ...
  ..$ : num [1:18] 1 1 1 0 1 1 1 0 1 1 ...
  ..$ : num [1:19] 1 1 1 1 1 1 1 1 0 0 ...
  ..$ : num [1:20] 0 1 1 1 0 1 0 0 1 1 ...
  ..$ : num [1:19] 1 1 1 1 1 1 0 0 1 0 ...
  ..$ : num [1:19] 1 1 1 0 1 0 1 0 1 1 ...
  ..$ : num [1:18] 1 1 0 0 1 0 0 0 1 1 ...
  ..$ : num [1:19] 1 1 1 0 0 1 1 1 1 0 ...
  ..$ : num [1:18] 0 1 1 1 0 0 1 0 1 1 ...
 $ prob :List of 10
  ..$ : num [1:18] 0 0.906 0.691 0.731 0.691 ...
  ..$ : num [1:19] 0.711 0.673 0.673 0.711 0.673 ...
  ..$ : num [1:18] 1 0.619 0.668 0.668 0.788 ...
  ..$ : num [1:19] 0.688 0.723 0.723 0.723 0.688 ...
  ..$ : num [1:20] 0.696 0.696 0.696 0.696 0.696 ...
  ..$ : num [1:19] 0.645 0.696 0.696 0.645 0.696 ...
  ..$ : num [1:19] 0.645 0.691 0.645 0.645 0.691 ...
  ..$ : num [1:18] 0.66 0.66 0.59 0.617 0.704 ...
  ..$ : num [1:19] 0.836 0.73 0.689 0.689 0.836 ...
  ..$ : num [1:18] 0.688 0.651 0.804 0.688 0.579 ...
> f.list
$label
$label[[1]]
 [1] 0 1 1 0 1 0 0 1 1 0 1 1 1 1 1 0 0 1

$label[[2]]
 [1] 0 1 1 1 1 1 0 0 1 1 0 0 0 1 1 0 1 1 0

根据数据表创建列表列表

1 个答案: