如何将R语言中的指定小型列表转换为DataFrame？

你的列表里每个元素都是带行名的单列结构化数据，常规的列表转DataFrame方法没法直接生成你要的笛卡尔积样式结果，得先拆分提取每个部分的类别与对应值，再生成全组合并填充数据。

方法一：基础R实现 #

# 提取time的类别和对应值
time_data <- data.frame(
  group = rownames(my_list$time),
  value = as.vector(my_list$time),
  stringsAsFactors = FALSE
)

# 提取treatment的类别和对应值
treatment_data <- data.frame(
  group = rownames(my_list$treatment),
  value = as.vector(my_list$treatment),
  stringsAsFactors = FALSE
)

# 生成所有组合
full_combinations <- expand.grid(
  time = time_data$group,
  treatment = treatment_data$group,
  stringsAsFactors = FALSE
)

# 匹配填充对应数值
full_combinations$time <- time_data$value[match(full_combinations$time, time_data$group)]
full_combinations$treatment <- treatment_data$value[match(full_combinations$treatment, treatment_data$group)]

# 设置行名
rownames(full_combinations) <- paste(
  rownames(my_list$time)[match(full_combinations$time, time_data$value)],
  rownames(my_list$treatment)[match(full_combinations$treatment, treatment_data$value)],
  sep = "-"
)

运行后得到的full_combinations就是目标结构：

time treatment
Pre-Control       0         0
Post-Control      1         0
Pre-Treatment     0         1
Post-Treatment    1         1

方法二：用tidyverse简化 #

如果习惯用tidyverse工具链，可以更简洁：

library(tidyr)
library(dplyr)

result <- crossing(
  time = rownames(my_list$time),
  treatment = rownames(my_list$treatment)
) %>%
  mutate(
    time = as.integer(my_list$time[time, ]),
    treatment = as.integer(my_list$treatment[treatment, ])
  ) %>%
  `rownames<-`(paste(rownames(my_list$time)[match(.$time, time_data$value)], 
                     rownames(my_list$treatment)[match(.$treatment, treatment_data$value)],
                     sep = "-"))

两种方法核心逻辑一致：先把每个列表元素的行名（类别标签）和数值对应绑定，再生成两个类别所有可能的组合，最后匹配填充对应数值，就能得到你要的DataFrame结构。

内容的提问来源于stack exchange，提问作者Andrzej Andrzej