不均匀的嵌套列表到数据框答案

【问题标题】：uneven nested list to dataframe不均匀的嵌套列表到数据框
【发布时间】：2021-03-18 14:42:51
【问题描述】：

我正在运行一项模拟研究，我的结果存储在一个嵌套列表结构中。列表的第一级代表模型生成的不同超参数。第二层是同一模型的复制次数（改变种子）。

在下面的示例中，我列出了由两个超参数（hyperpar1 和 hyperpar2）控制的模型的输出，其中两个超参数都可以采用 2 个不同的值，从而导致生成的模型有 4 种不同的组合。此外，4 种可能的组合中的每一种都运行了两次（不同的种子），产生了 8 种可能的组合。最后，从模型的每次可能迭代中恢复了两个性能指标（metric1 和 metric2）以及模型的两个参数的值 beta = list(b1 = value, b2 = value)。

我想将此信息放入data.frame，同时保留两件事。

我想保留对象的类（特别是与time_iter 相关，它以给定单位测量迭代时间）
我希望列表betas 在每次迭代中的每个组件都有一个单独的列，比如说b1 和b2。

样本数据：

res <-list(
  list(list(modeltype = "tree", time_iter = structure(0.7099, class = "difftime", units = "secs"),seed = 1, nobs = 75, hyperpar1 = 0.5, hyperpar2 = 0.5, metric1 = 0.4847, metric2 = 0.2576, beta = list(b1 = 0.575, b2 =0.745)),     
       list(modeltype = "tree", time_iter = structure(0.058 , class = "difftime", units = "secs"),seed = 2, nobs = 75, hyperpar1 = 0.5, hyperpar2 = 0.5, metric1 = 0.4013, metric2 = 0.2569, beta = list(b1 = 0.535, b2 =0.775))), 
  list(list(modeltype = "tree", time_iter = structure(0.046 , class = "difftime", units = "secs"),seed = 1, nobs = 75, hyperpar1 = 0.8, hyperpar2 = 0.5, metric1 = 0.4755, metric2 = 0.2988, beta = list(b1 = 0.541, b2 =0.702) ), 
       list(modeltype = "tree", time_iter = structure(0.0474, class = "difftime", units = "secs"),seed = 2, nobs = 75, hyperpar1 = 0.8, hyperpar2 = 0.5, metric1 = 0.2413, metric2 = 0.2147, beta = list(b1 = 0.545, b2 =0.793) )), 
  list(list(modeltype = "tree", time_iter = structure(0.0502, class = "difftime", units = "secs"),seed = 1, nobs = 75, hyperpar1 = 0.5, hyperpar2 = 1  , metric1 = 0.7131, metric2 = 0.5024, beta = list(b1 = 0.500, b2 =0.722) ), 
       list(modeltype = "tree", time_iter = structure(2.9419, class = "difftime", units = "secs"),seed = 2, nobs = 75, hyperpar1 = 0.5, hyperpar2 = 1  , metric1 = 0.4254, metric2 = 0.2824, beta = list(b1 = 0.555, b2 =0.712) )), 
  list(list(modeltype = "tree", time_iter = structure(0.041 , class = "difftime", units = "secs"),seed = 1, nobs = 75, hyperpar1 = 0.8, hyperpar2 = 1  , metric1 = 0.6709, metric2 = 0.4092, beta = list(b1 = 0.578, b2 =0.701) ), 
       list(modeltype = "tree", time_iter = structure(0.0396, class = "difftime", units = "secs"),seed = 2, nobs = 75, hyperpar1 = 0.8, hyperpar2 = 1  , metric1 = 0.4585, metric2 = 0.4115, beta = list(b1 = 0.501, b2 =0.777) )))

这是我的尝试。首先我未列出列表，然后 cbind 每个组件。如您所见，它失败得很惨。

un <- do.call(c, unlist(res, recursive=FALSE))
do.call(rbind.data.frame, un)

              b1     b2
modeltype    tree   tree
time_iter  0.7099 0.7099
seed            1      1
nobs           75     75
hyperpar1     0.5    0.5
hyperpar2     0.5    0.5
metric1    0.4847 0.4847
metric2    0.2576 0.2576
beta        0.575  0.745
modeltype1   tree   tree
time_iter1  0.058  0.058
seed1           2      2
nobs1          75     75
hyperpar11    0.5    0.5
hyperpar21    0.5    0.5
metric11   0.4013 0.4013
metric21   0.2569 0.2569
beta1       0.535  0.775
modeltype2   tree   tree
time_iter2  0.046  0.046

期望的输出

structure(list(modeltype = c("tree", "tree", "tree", "tree","tree", "tree", "tree", "tree"), 
               time_iter = structure(c(0.7099,0.058, 0.046, 0.0474, 0.0502, 2.9419, 0.041, 0.0396), 
               class = "difftime", units = "secs"),
               seed = c(1, 2, 1, 2, 1, 2, 1, 2), 
               nobs = c(75, 75, 75, 75, 75, 75, 75, 75), 
               hyperpar1 = c(0.5, 0.5, 0.8, 0.8, 0.5, 0.5,0.8, 0.8), 
               hyperpar2 = c(0.5, 0.5, 0.5, 0.5, 1, 1, 1, 1), 
               metric1 = c(0.4847, 0.4013, 0.4755, 0.2413, 0.7131, 0.4254, 0.6709, 0.4585), 
               metric2 = c(0.2576, 0.2569, 0.2988, 0.2147, 0.5024, 0.2824, 0.4092, 0.4115), 
               b1 = c(0.575, 0.535, 0.541, 0.545, 0.5, 0.555, 0.578, 0.501), 
               b2 = c(0.745, 0.775, 0.702,0.793, 0.722, 0.712, 0.701, 0.777)), 
               row.names = c(NA, -8L), class = "data.frame")

  modeltype   time_iter seed nobs hyperpar1 hyperpar2 metric1 metric2    b1    b2
1      tree 0.7099 secs    1   75       0.5       0.5  0.4847  0.2576 0.575 0.745
2      tree 0.0580 secs    2   75       0.5       0.5  0.4013  0.2569 0.535 0.775
3      tree 0.0460 secs    1   75       0.8       0.5  0.4755  0.2988 0.541 0.702
4      tree 0.0474 secs    2   75       0.8       0.5  0.2413  0.2147 0.545 0.793
5      tree 0.0502 secs    1   75       0.5       1.0  0.7131  0.5024 0.500 0.722
6      tree 2.9419 secs    2   75       0.5       1.0  0.4254  0.2824 0.555 0.712
7      tree 0.0410 secs    1   75       0.8       1.0  0.6709  0.4092 0.578 0.701
8      tree 0.0396 secs    2   75       0.8       1.0  0.4585  0.4115 0.501 0.777

最后，我想说我检查了Converting nested list to dataframe、How to convert a list consisting of vector of different lengths to a usable data frame in R?、How to convert a list consisting of vector of different lengths to a usable data frame in R? 和Convert R list to dataframe with missing/NULL elements，这些问题/答案都没有解决我的问题，因为我的示例的嵌套结构不同。

【问题讨论】：

标签： r list dataframe

【解决方案1】：

您可以将其取消列出并将其存储在矩阵中，然后更改为数据框。

# Helper function
conv=function(x) {
  hr=floor(x/3600) 
  mins1=x%%3600
  mins=floor(mins1/60)
  secs=mins1%%60
  return(paste0(hr,":",mins,":",secs))
}

# Mutation
library(dplyr)
library(lubridate)
n=names(unlist(res))[1:10]
f=matrix(unlist(res), ncol=10, byrow=TRUE)
f=data.frame(f, stringsAsFactors = FALSE)
colnames(f)=n
g=rename(f, b1=beta.b1, b2=beta.b2) %>%
  mutate(across(time_iter:b2, as.numeric), 
         time_iter=time_iter*10000, 
         time_iter=conv(time_iter),
         time_iter=as.difftime(time_iter, "%H:%M:%S", "secs"),
         time_iter=time_iter/10000)

输出

  modeltype   time_iter seed nobs hyperpar1 hyperpar2 metric1 metric2    b1    b2
1      tree 0.7099 secs    1   75       0.5       0.5  0.4847  0.2576 0.575 0.745
2      tree 0.0580 secs    2   75       0.5       0.5  0.4013  0.2569 0.535 0.775
3      tree 0.0460 secs    1   75       0.8       0.5  0.4755  0.2988 0.541 0.702
4      tree 0.0474 secs    2   75       0.8       0.5  0.2413  0.2147 0.545 0.793
5      tree 0.0502 secs    1   75       0.5       1.0  0.7131  0.5024 0.500 0.722
6      tree 2.9419 secs    2   75       0.5       1.0  0.4254  0.2824 0.555 0.712
7      tree 0.0410 secs    1   75       0.8       1.0  0.6709  0.4092 0.578 0.701
8      tree 0.0396 secs    2   75       0.8       1.0  0.4585  0.4115 0.501 0.777

【讨论】：