如何对R语言中以group_日期命名的list元素按分组、日期先后排序

你之前直接使用order未得到预期结果，是因为直接对元素名称按字符串整体排序，会优先比较分组字段，同分组下才会比较日期，和你给出的预期输出逻辑不符。按以下步骤实现即可：

完整代码 #

# 替换为你自己的原始list变量名
my_list <- list(`groupA_2024-02-01` = structure(list(name = "groupA"), class = "colDef"), 
     `groupB_2024-02-01` = structure(list(name = "groupB"), class = "colDef"), 
     `groupA_2022-04-01` = structure(list(name = "groupA"), class = "colDef"), 
     `groupB_2022-04-01` = structure(list(name = "groupB"), class = "colDef"), 
     `groupA_2021-09-01` = structure(list(name = "groupA"), class = "colDef"), 
     `groupB_2021-09-01` = structure(list(name = "groupB"), class = "colDef"), 
     `groupA_2024-04-01` = structure(list(name = "groupA"), class = "colDef"), 
     `groupB_2024-04-01` = structure(list(name = "groupB"), class = "colDef"), 
     `groupA_2023-02-01` = structure(list(name = "groupA"), class = "colDef"), 
     `groupB_2023-02-01` = structure(list(name = "groupB"), class = "colDef"))

# 1. 拆分元素名称提取排序依据字段
name_parts <- strsplit(names(my_list), "_")
group <- sapply(name_parts, `[`, 1)
date <- as.Date(sapply(name_parts, `[`, 2))

# 2. 生成排序索引
# 适配你给出的预期输出：先按日期升序，同日期下groupA排在groupB前
sort_idx <- order(date, group)
# 如果你需要第一优先级为分组、第二优先级为日期，替换为下行代码即可
# sort_idx <- order(group, date)

# 3. 重排得到最终结果
sorted_list <- my_list[sort_idx]

结果验证 #

执行后打印names(sorted_list)即可看到排序后的元素名称，和你给出的预期输出完全匹配。

内容的提问来源于stack exchange，提问作者Priit Mets