R语言如何按na列分组提取每组rn列最小值对应的行记录

原始数据准备 #

首先构建原始数据集，注意直接用data.frame()存储可以保留列的原始类型，避免cbind()生成字符矩阵导致数值排序出错：

rn <- c(3,4,5,2,1,5,6,8,10,3,4,5,6,8,9,7)
na <- c("A","A","A","A","A","B","B","B","B","B","CD","CD","CD","CD","CD","CD")
mo <- c("ram","okd","mlu","lom","mpl","mpl","cdd","jjh","yyt","uu","tt","rre","llm","mm","mlp","lok")
dat <- data.frame(rn, na, mo, stringsAsFactors = FALSE)

需求说明 #

按照na列的不同取值分组，筛选每组中rn列数值最小的完整记录，同时保留对应行的mo列取值，预期输出如下：

rn na  mo
   1  A mpl
   3  B  uu
   4  CD tt

实现方案 #

方案1：基础R实现（无第三方包依赖） #

不需要额外安装加载包，直接用R内置函数即可实现：

# 按na列分组，提取每组rn值最小的行
result <- do.call(
  rbind,
  lapply(split(dat, dat$na), function(group_df) group_df[which.min(group_df$rn), ])
)
# 调整行顺序匹配预期输出
result <- result[match(c("A","B","CD"), result$na), ]
# 打印结果
print(result, row.names = FALSE)

运行输出：

rn na  mo
  1  A mpl
  3  B  uu
  4  CD tt

方案2：dplyr实现（简洁流式语法） #

如果日常使用tidyverse生态工具，可以用更简洁的语法实现：

library(dplyr)

result <- dat %>%
  group_by(na) %>%
  # 取每组rn最小的1行，遇到并列最小值默认取第一个
  slice_min(rn, n = 1, with_ties = FALSE) %>%
  ungroup() %>%
  # 调整分组顺序匹配预期
  arrange(match(na, c("A","B","CD")))

print(result, row.names = FALSE)

运行后得到的结果和基础R方案完全一致。

内容的提问来源于stack exchange，提问作者Tpellirn