如何将按ID匹配的列表转换为R语言DataFrame？

嗨，我来帮你搞定这个合并问题～ 你现在的情况是列表里的每个DataFrame都用行名作为匹配的ID，所以只需要把它们按列合并，再调整列名就可以得到你想要的结果。下面给你几种实用的方法：

方法一：用tidyverse工具链（推荐，和你原代码风格一致） #

因为你已经在用purrr::map，所以用dplyr::bind_cols可以无缝衔接：

首先，先加载必要的包：

library(tidyverse)

然后用你提供的示例列表（或者你自己生成的brca1列表）：

# 先把你给的示例列表赋值（如果是自己生成的brca1可以跳过这步）
mylist <- list(
  structure(list(. = c("815673", "813277", "686858", "601778", "601777", "601776", "564136", "443977", "442989", "442718", "397361", "395470", "155306", "149176", "149167", "146230", "144518", "144379", "58526", "58525")), 
            row.names = c("815673", "813277", "686858", "601778", "601777", "601776", "564136", "443977", "442989", "442718", "397361", "395470", "155306", "149176", "149167", "146230", "144518", "144379", "58526", "58525"), 
            class = "data.frame"),
  structure(list(. = c("copy number gain", "copy number loss", "copy number loss", "copy number gain", "copy number gain", "copy number gain", "copy number loss", "copy number gain", "copy number loss", "copy number gain", "copy number gain", "copy number gain", "copy number gain", "copy number gain", "copy number gain", "copy number gain", "copy number gain", "copy number gain", "copy number gain", "copy number gain")), 
            row.names = c("815673", "813277", "686858", "601778", "601777", "601776", "564136", "443977", "442989", "442718", "397361", "395470", "155306", "149176", "149167", "146230", "144518", "144379", "58526", "58525"), 
            class = "data.frame")
)

接下来合并并调整列名：

# 按行名合并所有子DataFrame
brca1_df <- bind_cols(mylist)

# 给列设置你想要的名字
colnames(brca1_df) <- c("col1", "col2")

# 可选：把行名（也就是匹配ID）转为单独一列，更规范
brca1_df <- brca1_df %>% 
  mutate(match_id = rownames(.)) %>% 
  relocate(match_id)  # 把ID列移到最前面

方法二：用Base R实现（不需要额外包） #

如果你不想加载tidyverse，用原生的do.call+cbind也能搞定：

# 合并列表中的DataFrame
brca1_df_base <- do.call(cbind, mylist)

# 重命名列
colnames(brca1_df_base) <- c("col1", "col2")

# 可选：把行名转为列
brca1_df_base$match_id <- rownames(brca1_df_base)
brca1_df_base <- brca1_df_base[, c("match_id", "col1", "col2")]

优化你的原代码（避免重复改列名） #

其实你可以在生成列表的时候就给每个子DataFrame设置好列名，这样合并后直接能用：

myfunc <- function(a){
  temp_col <- sapply(clinvar_summary, "[[", a) %>% as.data.frame()
  colnames(temp_col) <- a  # 直接用提取的字段名当列名
  return(temp_col)
}

mylist <- c("uid", "obj_type")
brca1 <- map(mylist, myfunc)

# 合并后直接就是带正确列名的DataFrame
brca1_final <- bind_cols(brca1)

这样合并后的结果列名就是uid和obj_type，不用再手动重命名啦～

内容的提问来源于stack exchange，提问作者icedcoffee