基于tidyverse的统计分析与数据合并代码优化问询

数据集 #

首先加载所需包并导入数据集：

library(tidyverse)

df <- structure(list(L1 = c("Age Class", "Age Class", "Age Class", 
"Age Class", "Gender", "Gender", "Gender", "Age Class", "Age Class", 
"Age Class", "Gender", "Gender", "Age Class", "Age Class", "Age Class", 
"Gender"), L2 = c("Older Youth", "Older Youth", "Younger Youth", 
"Younger Youth", "Female", "Female", "Female", "Younger Youth", 
"Older Youth", "Older Youth", "Male", "Male", "Younger Youth", 
"Older Youth", "Older Youth", "Female"), scr = c(0.78125, 0.90625, 
0.90625, 0.6875, 0.875, 0.78125, 1, 0.65625, 0.75, 0.59375, 0.8125, 
0.75, 0.65625, 0.6875, 0.75, 0.75)), row.names = c(NA, -16L), class = "data.frame")

需求回顾 #

需要完成以下操作：

• 计算整体数据的中位数（median）和标准误（standard error）
• 按L1和L2分组计算中位数和标准误
• 在L1组内针对两个L2水平执行威尔科克森检验（wilcoxon test）
• 合并前两步的统计结果，再与检验的p值关联得到最终输出

简化代码方案 #

方案1：嵌套数据框+链式操作 #

这种写法符合tidy数据理念，避免多层列表嵌套，逻辑清晰：

final_result <- df %>%
  # 复制一份数据标记为整体分组，统一后续统计
  bind_rows(mutate(., L1 = "All", L2 = "All")) %>%
  # 计算所有分组的中位数和标准误
  summarise(
    mdn = median(scr),
    se = sd(scr)/sqrt(n()),
    .by = c(L1, L2)
  ) %>%
  # 关联威尔科克森检验的p值
  left_join(
    df %>%
      # 按L1分组嵌套数据
      nest(data = -L1) %>%
      # 对每个L1组执行检验并提取p值
      mutate(pv = map_dbl(data, ~wilcox.test(scr ~ L2, data = .x)$p.value)) %>%
      select(L1, pv) %>%
      # 给整体分组添加NA的p值（无L2对比）
      add_row(L1 = "All", pv = NA_real_),
    by = "L1"
  )

print(final_result)

方案2：紧凑链式写法 #

如果偏好更简洁的代码结构，可以直接合并整体与分组统计后关联检验结果：

final_result <- df %>%
  {
    bind_rows(
      # 单独计算整体统计
      mutate(., L1 = "All", L2 = "All") %>%
        summarise(mdn = median(scr), se = sd(scr)/sqrt(n()), .by = c(L1, L2)),
      # 计算分组统计
      summarise(., mdn = median(scr), se = sd(scr)/sqrt(n()), .by = c(L1, L2))
    )
  } %>%
  left_join(
    df %>%
      summarise(pv = wilcox.test(scr ~ L2)$p.value, .by = L1) %>%
      add_row(L1 = "All", pv = NA_real_),
    by = "L1"
  )

print(final_result)

代码说明 #

• 两种方案都摒弃了原代码的多层列表处理，用tidyverse原生函数完成合并与统计
• 嵌套数据框的写法（方案1）扩展性更强，后续新增分组检验逻辑更方便
• 整体分组的p值设为NA，符合“无L2水平对比则无检验结果”的逻辑

内容的提问来源于stack exchange，提问作者Salty