【发布时间】：2021-03-25 11:15:48
【问题描述】：
我正在进行一项模拟研究，我的结果存储在嵌套列表结构中。列表的第一级代表模型生成的不同超参数。第二个层次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（如下图str(res, max = 2) 所示）。最后，从模型的每个可能迭代中恢复了两个性能指标（metric1 和 metric2）。
我的问题是，在我的真实数据中，迭代次数（列表的第二级）是巨大的（高达 10000 次），并且在某些情况下超参数数量的全阶乘高达 2000 次。因此，退市过程变得相当缓慢。
下面我列出了我当前的过程以及我想要的输出，但同样，它相对较慢。特别是，有一个部分是我将所有内容都取消列出，然后将它们放在一个大数据框架中，这需要很长时间，但我还没有以更快的方式解决这个问题。
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 ),
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 )),
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 ),
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 )),
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 ),
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 )),
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 ),
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 )))
hyperpar1 <- c(0.5 , 0.8 )
hyperpar2 <- c(0.5 , 1 )
expand.grid(hyperpar1 = hyperpar1, hyperpar2 = hyperpar2)
# hyperpar1 hyperpar2
# 1 0.5 0.5
# 2 0.8 0.5
# 3 0.5 1.0
# 4 0.8 1.0
#List structure:
#The 4 elements represent the 4 combinations of the hyperparams
#Inside each of the 4 combinations of the hyperparams, 2 lists represent the 2 simulations (with different seeds)
str(res, max = 1)
#Finally, inside each of the final level (level=3) there is a list of 10 objects that are the results of each simulation
str(res, max = 2)
# List of 4
# $ :List of 2
# ..$ :List of 8
# ..$ :List of 8
# $ :List of 2
# ..$ :List of 8
# ..$ :List of 8
# $ :List of 2
# ..$ :List of 8
# ..$ :List of 8
# $ :List of 2
# ..$ :List of 8
# ..$ :List of 8
#e.g Fist iteration of first model
t(res[[1]][[1]])
# modeltype time_iter seed nobs hyperpar1 hyperpar2 metric1 metric2
# [1,] "tree" 0.7099 1 75 0.5 0.5 0.4847 0.2576
解析列表
在以下代码中，我将列表取消嵌套并将所有内容放入 data.frame。
#Unlist the nested structure of the list `res`
all_in_list <- lapply(1:length(res), function(i) {
unlisting <- unlist(res[i],recursive = FALSE)
to_df <- do.call(rbind, lapply(unlisting, as.data.frame))
return(to_df)})
#Here is where averything get really really slow when the list is huge
all_in_df <- do.call(rbind, lapply(all_in_list, as.data.frame))
# modeltype time_iter seed nobs hyperpar1 hyperpar2 metric1 metric2
# 1 tree 0.7099 secs 1 75 0.5 0.5 0.4847 0.2576
# 2 tree 0.0580 secs 2 75 0.5 0.5 0.4013 0.2569
# 3 tree 0.0460 secs 1 75 0.8 0.5 0.4755 0.2988
# 4 tree 0.0474 secs 2 75 0.8 0.5 0.2413 0.2147
# 5 tree 0.0502 secs 1 75 0.5 1.0 0.7131 0.5024
# 6 tree 2.9419 secs 2 75 0.5 1.0 0.4254 0.2824
# 7 tree 0.0410 secs 1 75 0.8 1.0 0.6709 0.4092
# 8 tree 0.0396 secs 2 75 0.8 1.0 0.4585 0.4115
输出的汇总统计
在下文中，我恢复了性能指标的均值和标准差，将它们的下标添加到每个 colname 以便以后识别（用于绘图）。
#auxiliar function to compute the metrics at aggregate level.
foo_summary <- function(df ,
metrics =c("time_iter","metric1", "metric2") ,
by = c("nobs","hyperpar1", "hyperpar2", "modeltype"),
summary_function = mean)
{
#compute the aggregate metrics
out <- as.data.frame(aggregate(
x = df[metrics],
by = df[by],
FUN = summary_function,
na.rm = TRUE))
#rename conviniently the metric computed
oldnames <- colnames(out[metrics])
names(out)[match(oldnames,names(out))] <- paste(colnames(out[metrics]),
as.character(substitute(summary_function)),
sep = "_")
return(out)
}
df_mean <- foo_summary(df = all_in_df,
metrics =c("time_iter","metric1", "metric2"),
by = c("nobs","hyperpar1", "hyperpar2", "modeltype"),
summary_function = mean)
df_sd <- foo_summary(df = all_in_df,
metrics =c("time_iter","metric1", "metric2"),
by = c("nobs","hyperpar1", "hyperpar2", "modeltype"),
summary_function = sd)
final_df <- merge(df_mean,df_sd )
最后是想要的输出。
# nobs hyperpar1 hyperpar2 modeltype time_iter_mean metric1_mean metric2_mean time_iter_sd metric1_sd metric2_sd
# 1 75 0.5 0.5 tree 0.38395 secs 0.44300 0.25725 0.4609629107 0.05897271 0.0004949747
# 2 75 0.5 1.0 tree 1.49605 secs 0.56925 0.39240 2.0447406792 0.20343462 0.1555634919
# 3 75 0.8 0.5 tree 0.04670 secs 0.35840 0.25675 0.0009899495 0.16560441 0.0594676803
# 4 75 0.8 1.0 tree 0.04030 secs 0.56470 0.41035 0.0009899495 0.15018948 0.0016263456
【问题讨论】：
-
如果数据真的很大，我会说来自
base R的aggregate会很慢。使用data.table或tidyverse（如果包没问题）
标签： r dataframe data.table aggregate lapply