通过新增列转换t检验结果的长格式dataframe

可以使用tidyverse工具包完成这个转换，步骤简洁高效：

步骤1：加载工具包并还原原始数据 #

library(tidyverse)

# 原始输入数据
df <- structure(list(region = c("region 1", "region 1", "region 1", "region 2", "region 2", "region 2", "region 3", "region 3", "region 3", "region 4", "region 4", "region 4", "region 5", "region 5", "region 5", "region 6", "region 6", "region 6"), group = c("Male", "Female", "p-value", "Male", "Female", "p-value", "Male", "Female", "p-value", "Male", "Female", "p-value", "Male", "Female", "p-value", "Male", "Female", "p-value"), value = c(1.1, 0.9, 0.001, 0.5, 1.2, 0.612, 0.9, 0.1, 0.001, 0.9, 0.8, 0.7, 1.3, 0.8, 0.04, 2.3, 1.5, 0.561)), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, -18L))

步骤2：数据转换 #

result_df <- df %>%
  # 按区域分组，确保每个区域的p-value仅对应本组性别数据
  group_by(region) %>%
  # 提取当前组的p-value值，生成新列并在组内重复填充
  mutate(`p-value` = value[group == "p-value"]) %>%
  # 过滤掉p-value行，仅保留Male和Female的数据
  filter(group != "p-value") %>%
  # 取消分组状态
  ungroup() %>%
  # 调整列顺序为需求格式
  select(region, group, value, `p-value`)

验证结果 #

运行代码后，result_df的结构与期望完全匹配：

# 查看转换后的数据
result_df

输出示例：

# A tibble: 12 × 4
   region   group  value `p-value`
   <chr>    <chr>  <dbl>     <dbl>
 1 region 1 Male    1.1      0.001
 2 region 1 Female  0.9      0.001
 3 region 2 Male    0.5      0.612
 4 region 2 Female  1.2      0.612
 5 region 3 Male    0.9      0.001
 6 region 3 Female  0.1      0.001
 7 region 4 Male    0.9      0.7  
 8 region 4 Female  0.8      0.7  
 9 region 5 Male    1.3      0.04 
10 region 5 Female  0.8      0.04 
11 region 6 Male    2.3      0.561
12 region 6 Female  1.5      0.561

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