在R中按指定列分组，获取对应列各层级的最小值方法

示例数据集 #

df <- structure(list(ID=c('a','a','a','a','b','b','b','b','c','c','c','c','d','d','d','d','e','e','e','e','f','f','f','f','g','g','g','g','h','h','h','h'),
                     A2=c('d1','d1','d1','d1','d1','d1','d1','d1','d2','d2','d2','d2','d2','d2','d2','d2','d1','d1','d1','d1','d1','d1','d1','d1','d2','d2','d2','d2','d2','d2','d2','d2'),
                     A3=c('g1','g1','g1','g1','g1','g1','g1','g1','g1','g1','g1','g1','g1','g1','g1','g1','g2','g2','g2','g2','g2','g2','g2','g2','g2','g2','g2','g2','g2','g2','g2','g2'),
                     number=c('1','1','2','2','1','1','2','2','1','1','2','2','1','1','2','2','1','1','2','2','1','1','2','2','1','1','2','2','1','1','2','2'),
                     time=c(23,345,123,4,434,76,245,34,135,98,45,678,32,134,76,578,32,145,256,79,311,356,67,12,689,467,98,456,23,45,23,34)), 
                class = "data.frame", row.names = c(NA,-32L))

解决方案 #

方法1：使用dplyr包（推荐，代码简洁易读）

library(dplyr)

df.result <- df %>%
  # 按A2、A3、ID、number分组（需保留ID故加入分组）
  group_by(A2, A3, ID, number) %>%
  # 提取每组time的最小值
  summarize(time = min(time), .groups = "drop") %>%
  # 调整行顺序匹配期望结果
  arrange(A2, A3, ID, number)

方法2：使用base R（无需额外安装包）

# 按指定列分组计算time最小值
df.result <- aggregate(time ~ A2 + A3 + ID + number, data = df, FUN = min)
# 调整行顺序
df.result <- df.result[order(df.result$A2, df.result$A3, df.result$ID, df.result$number), ]
# 重置行名
rownames(df.result) <- NULL

验证结果 #

运行上述代码后，可通过以下代码确认结果与期望一致：

# 期望结果数据集
df.expected <- structure(list(ID=c('a','a','b','b','c','c','d','d','e','e','f','f','g','g','h','h'),
                          A2=c('d1','d1','d1','d1','d2','d2','d2','d2','d1','d1','d1','d1','d2','d2','d2','d2'),
                          A3=c('g1','g1','g1','g1','g1','g1','g1','g1','g2','g2','g2','g2','g2','g2','g2','g2'),
                          number=c('1','2','1','2','1','2','1','2','1','2','1','2','1','2','1','2'),
                          time=c(23,4,76,34,98,45,32,76,32,79,311,12,467,98,23,23)), 
                     class = "data.frame", row.names = c(NA,-16L))

# 检查是否一致
all.equal(df.result, df.expected)

内容的提问来源于stack exchange，提问作者Pegi