如何在R语言中将单列元素拆分并生成多列存在标记列？

可以使用tidyverse工具包高效处理这个需求，步骤清晰且代码简洁：

方法一：使用tidyverse工具链 #

library(tidyverse)

# 原始数据集
df <- data.frame(identifier = c("a","b","b","c"), 
                 disease = c("heart, lung","lung, heart,,","lung, heart, heart, liver", "kidney, brain "))

# 数据处理流程
processed_df <- df %>%
  # 清洗疾病列：替换多逗号为单逗号、去除首尾空格、拆分疾病为列表
  mutate(disease = str_replace_all(disease, ",+", ","),
         disease = str_trim(disease),
         disease = str_split(disease, ", ")) %>%
  # 将列表形式的疾病展开为多行
  unnest(disease) %>%
  # 标记存在的疾病为Yes
  mutate(status = "Yes") %>%
  # 转换为宽格式，缺失值填充为No
  pivot_wider(names_from = disease, values_from = status, values_fill = "No")

# 输出结果
print(processed_df)

执行后得到的结果：

# A tibble: 4 × 6
  identifier heart lung  liver kidney brain
  <chr>      <chr> <chr> <chr> <chr>  <chr>
1 a          Yes   Yes   No    No     No   
2 b          Yes   Yes   No    No     No   
3 b          Yes   Yes   Yes   No     No   
4 c          No    No    No    Yes    Yes  

关键步骤说明 #

• str_replace_all(disease, ",+", ",")：把连续多个逗号替换为单个，解决类似lung, heart,,的格式问题
• str_trim(disease)：去除字符串首尾空格，避免拆分出空值或带空格的疾病名称
• str_split(disease, ", ")：将每个疾病字符串按, 拆分为列表，方便后续展开
• unnest(disease)：把列表形式的疾病列展开为多行，让每个疾病对应一行数据
• pivot_wider：将长格式数据转换为宽格式，自动按identifier分组，缺失的疾病标记填充为No

方法二：使用基础R实现 #

如果不想依赖第三方包，也可以用基础R代码完成：

# 原始数据
df <- data.frame(identifier = c("a","b","b","c"), 
                 disease = c("heart, lung","lung, heart,,","lung, heart, heart, liver", "kidney, brain "))

# 清洗疾病列
df$disease <- gsub(",+", ",", df$disease)
df$disease <- trimws(df$disease)

# 拆分疾病并提取所有唯一疾病名称
disease_list <- strsplit(df$disease, ", ")
all_diseases <- unique(unlist(disease_list))

# 循环生成每个疾病的标记列
for(d in all_diseases){
  df[[d]] <- sapply(disease_list, function(x) ifelse(d %in% x, "Yes", "No"))
}

# 移除原disease列
df <- df[, !(names(df) %in% "disease")]

# 输出结果
print(df)

这个方法通过循环遍历所有唯一疾病，逐行检查是否包含该疾病，从而生成对应的Yes/No标记列。

内容的提问来源于stack exchange，提问作者Ahmed