R语言：将行数不同的data.frame列表合并为带NA补全的data.frame

首先构造示例数据方便复现：

# 构造示例数据
mylist <- list(
  elem1 = data.frame(elem1 = c("E7", "E7", "E4", "E8", "E9", "E8"),
                     row.names = paste0("sample-", sprintf("%02d", 1:6))),
  elem2 = data.frame(elem2 = c("E1", "E1", "E8", "E3"),
                     row.names = paste0("sample-", sprintf("%02d", 1:4))),
  elem3 = data.frame(elem3 = c("E2", "E2", "E7", "E2"),
                     row.names = paste0("sample-", sprintf("%02d", 5:8)))
)

方法一：Base R 实现 #

通过收集所有行名，逐个合并并保留所有行：

# 获取所有唯一行名
all_samples <- unique(unlist(lapply(mylist, rownames)))

# 初始化结果data.frame
result_base <- data.frame(row.names = all_samples)

# 循环合并每个列表元素
for (df in mylist) {
  result_base <- merge(result_base, df, by = "row.names", all.x = TRUE)
  rownames(result_base) <- result_base$Row.names
  result_base$Row.names <- NULL
}

# 按行名排序（确保顺序为sample-01到sample-08）
result_base <- result_base[order(rownames(result_base)), ]

方法二：tidyverse 实现（更简洁） #

利用dplyr和purrr的函数链，先转长格式再转宽格式：

library(tidyverse)

result_tidy <- mylist %>%
  # 为每个data.frame添加行名列，保留原列表元素名作为列标识
  imap(~ .x %>% rownames_to_column("sample")) %>%
  # 绑定所有行
  bind_rows() %>%
  # 转宽格式，缺失值填充NA
  pivot_wider(
    names_from = .y, 
    values_from = starts_with("elem"), 
    values_fill = NA
  ) %>%
  # 将sample列转回行名
  column_to_rownames("sample") %>%
  # 按行名排序
  arrange(rownames(.))

两种方法最终都会得到目标格式：

elem1 elem2 elem3
sample-01   E7   E1   NA
sample-02   E7   E1   NA
sample-03   E4   E8   NA
sample-04   E8   E3   NA
sample-05   E9   NA   E2
sample-06   E8   NA   E2
sample-07   NA   NA   E7
sample-08   NA   NA   E2

内容的提问来源于stack exchange，提问作者abraham