如何在R中对分组数据按列执行wilcox.test并添加p值到新行

首先构造示例数据df1：

set.seed(12345)
df1 <- data.frame(
  group = c(rep("apple", 4), rep("pear",6)), 
  a = rnorm(10, 0, 0.4), 
  b = rnorm(10, 0, 0.2), 
  c = rnorm(10, 0, 0.7), 
  d = rnorm(10, 0, 0.9), 
  e = rnorm(10, 0, 0.5)
)

方法1：基础R实现 #

# 拆分两组的数值列数据
apple_num <- df1[df1$group == "apple", -1]
pear_num <- df1[df1$group == "pear", -1]

# 逐列计算wilcox检验p值
p_values <- sapply(colnames(apple_num), function(col) {
  wilcox.test(apple_num[[col]], pear_num[[col]])$p.value
})

# 构造p值行并合并到原数据
p_row <- c(group = "wilcox.test", round(p_values, 9))
df2 <- rbind(df1, as.data.frame(as.list(p_row), stringsAsFactors = FALSE))

# 可选：将数值列转换回数值类型（合并后默认是字符）
df2[, 2:ncol(df2)] <- lapply(df2[, 2:ncol(df2)], as.numeric)

方法2：tidyverse简洁实现 #

library(dplyr)

# 计算各列p值
p_row <- df1 %>%
  summarise(
    across(a:e, ~wilcox.test(.x[group == "apple"], .x[group == "pear"])$p.value)
  ) %>%
  mutate(group = "wilcox.test", .before = 1)

# 合并得到最终结果
df2 <- bind_rows(df1, p_row)

打印df2即可得到你示例中的结果：

group          a          b         c           d         e
1        apple  0.2342115 -0.0232496  0.545735  0.73068586  0.564255
2        apple  0.2837864  0.3634624  1.019050  1.97715019 -1.190179
3        apple -0.0437213  0.0741256 -0.451030  1.84427130 -0.530133
4        apple -0.1813989  0.1040433 -1.087196  1.46920108  0.468570
5         pear  0.2423550 -0.1501064 -1.118397  0.22884407  0.427226
6         pear -0.7271824  0.1633800  1.263568  0.44206945  0.730365
7         pear  0.2520394 -0.1772715 -0.337153 -0.29167792 -0.706549
8         pear -0.1104736 -0.0663155  0.434266 -1.49584522  0.283702
9         pear -0.1136639  0.2241425  0.428486  1.59096047  0.291594
10        pear -0.3677288  0.0597447 -0.113618  0.02322094 -0.653399
11 wilcox.test  0.3937686  0.2864220  1.000000  0.03300626  1.000000

注意：如果需要调整检验类型（如单侧检验），可以在wilcox.test中添加alternative参数设置。

内容的提问来源于stack exchange，提问作者Sylvia Rodriguez