如何根据列名模式计算data.frame中每组三列的均值？

针对你拥有的21列data.frame（第一列为geneName，剩余列每3个对应一个时间点组），这里提供几种R语言的实现方案：

先构造示例数据（方便测试） #

df <- data.frame(
  geneName = c("gene1", "gene2"),
  t11 = c(3296, 4210),
  t12 = c(5133, 5505),
  t13 = c(3466, 4173),
  t21 = c(2166, 2736),
  t22 = c(1759, 2748),
  t23 = c(2099, 3052),
  t31 = c(1916, 2409),
  t32 = c(1379, 1944),
  t33 = c(1570, 2237),
  t41 = c(2533, 1158),
  t42 = c(1794, 3475),
  t43 = c(1016, 1488),
  t51 = c(800, 4023),
  t52 = c(79, 102),
  t53 = c(648, 940),
  t61 = c(99, 265),
  t62 = c(60, 365),
  t63 = c(152, 124)
)

方法1：基础R实现 #

无需额外安装包，直接用原生函数处理：

# 提取数值列的列名
cols <- colnames(df)[-1]
# 生成分组标识（去掉列名最后一位数字，得到t1/t2...t6）
groups <- gsub("\\d$", "", cols)
# 对每个分组计算行均值
mean_cols <- sapply(unique(groups), function(g) {
  rowMeans(df[cols[groups == g]])
})
# 合并geneName与均值结果，并重命名列
result <- cbind(df["geneName"], as.data.frame(mean_cols))
colnames(result)[-1] <- paste0("t", 1:6, "_mean")

方法2：tidyverse（dplyr+tidyr）实现 #

适合习惯tidy风格的用户，逻辑更直观：

library(dplyr)
library(tidyr)

result <- df %>%
  # 宽表转长表，统一处理分组
  pivot_longer(-geneName, names_to = "time_group", values_to = "value") %>%
  # 提取时间组前缀
  mutate(time_group = gsub("\\d$", "", time_group)) %>%
  # 分组计算均值
  group_by(geneName, time_group) %>%
  summarise(mean_value = mean(value), .groups = "drop") %>%
  # 转回宽表格式
  pivot_wider(names_from = time_group, values_from = mean_value, names_prefix = "mean_")

方法3：data.table实现 #

大数据量场景下效率更高：

library(data.table)

dt <- as.data.table(df)
# 生成分组标识
groups <- gsub("\\d$", "", colnames(dt)[-1])
# 计算各分组均值
mean_dt <- dt[, c(
  list(geneName = geneName),
  lapply(unique(groups), function(g) {
    rowMeans(.SD[, grep(g, colnames(.SD)), with = FALSE])
  })
)]
# 重命名列
colnames(mean_dt)[-1] <- paste0("t", 1:6, "_mean")

内容的提问来源于stack exchange，提问作者Assa Yeroslaviz