R语言单行DataFrame按相近列名分组计算均值的实现问题

核心思路 #

你的列名遵循「公共前缀+末尾数字」的命名规则，无需使用模糊匹配，直接提取列名的非数字后缀作为分组依据即可实现精准分组，完全规避agrep的误匹配问题，同时自动适配每组不定数量的列。 #

可运行实现代码 #

方案1：Base R 原生实现（无需安装第三方包） #

# 读入测试数据
df <- structure(list(poms_tat1 = 0.370882118644872, poms_tat2 = 0.491592168116328, 
    poms_tat3 = 0.395819547420188, tens1 = -0.113960576459638, 
    tens2 = -0.146254484825426, tens3 = -0.44114940169153, bat_ratio1 = 1, 
    isi1 = 0.0944068640061701, isi2 = 0.597785124823513, isi3 = 0.676617801589949, 
    isi4 = 0.143940321201716, sleepqual = 0.378902118888194, 
    se1 = 0.393610946830482, se2 = 0.0991899501072693, se3 = 0.501745206004254, 
    challenge1 = 0.417855447018672, challenge2 = 0.393610946830482, 
    challenge3 = 0.417855447018672, threat1 = -0.13014390184863, 
    threat2 = -0.34027852368936, threat3 = -0.269679944985297, 
    reat1 = 0.565825152115738, reat2 = 0.571605347479646, reat3 = 0.497468338163091, 
    reat4 = 0.484881137876427, reat5 = 0.494727444918154, selfman1 = 0.389249472080761, 
    selfman2 = 0.40609787800914, selfman3 = 0.418121005003545, 
    selfman4 = 0.467099366496914, selfman5 = 0.205356548067582, 
    selfman6 = 0.464385939554693, selfman7 = 0.379071252751718, 
    eli1 = 0.250872603002127, eli2 = 0, eli3 = 0.265908011739155), row.names = 1L, class = "data.frame")

# 提取列名分组：去掉末尾所有数字得到公共前缀
col_groups <- sub("\\d+$", "", names(df))
# 按分组拆分列并计算每组均值
result <- as.data.frame(sapply(split.default(df, col_groups), function(x) rowMeans(x, na.rm = TRUE)))

运行后result的输出完全符合预期，例如poms_tat列均值为0.4194313，tens列均值为-0.2337882，无数字后缀的sleepqual列会保留原值。 #

方案2：tidyverse 风格实现（可读性更高） #

library(tidyverse)

result <- df %>%
  # 宽表转长表
  pivot_longer(everything(), names_to = "col_name") %>%
  # 提取分组前缀
  mutate(group = str_remove(col_name, "\\d+$")) %>%
  # 分组求均值
  group_by(group) %>%
  summarise(mean_val = mean(value, na.rm = TRUE)) %>%
  # 转回宽表
  pivot_wider(names_from = group, values_from = mean_val)

方案优势 #

• 分组精准：基于列名规则硬匹配，不会出现threat和reat误归为一类的问题
• 适配性强：自动支持每组任意数量的列，无论1列、3列还是7列都能正确计算
• 代码简洁：无需嵌套循环，维护成本低

内容的提问来源于stack exchange，提问作者Leo