如何在R中移除名称前缀Gain数字不同的命名列表元素？

在下方的My_list中，需要移除所有名称里前缀"Gain"后附带数字不同的元素：

• 例如元素名称Gain1(conventional) - Gain2(conventional)，因为Gain1和Gain2的数字不同，需要移除；
• 而元素名称Gain1(conventional) - Gain1(framework notes)，因为前后数字都是1，需要保留。
  请问在R中如何实现？

My_list <- 
list(`Gain1(conventional) - Gain2(conventional)` = c(5L, -1L, 
-9L, 1L), `Gain1(conventional) - Gain1(framework notes)` = c(5L, 
-1L, -6L, 2L), `Gain1(conventional) - Gain2(framework notes)` = c(5L, 
-1L, -10L, 2L), `Gain1(conventional) - Gain1(note-taking instruction)` = c(5L, 
-1L, -7L, 3L), `Gain1(conventional) - Gain2(note-taking instruction)` = c(5L, 
-1L, -11L, 3L), `Gain1(conventional) - Gain1(vocabulary notebook)` = c(5L, 
-1L, -8L, 4L), `Gain1(conventional) - Gain2(vocabulary notebook)` = c(5L, 
-1L, -12L, 4L), `Gain2(conventional) - Gain1(framework notes)` = c(9L, 
-1L, -6L, 2L), `Gain2(conventional) - Gain2(framework notes)` = c(9L, 
-1L, -10L, 2L), `Gain2(conventional) - Gain1(note-taking instruction)` = c(9L, 
-1L, -7L, 3L), `Gain2(conventional) - Gain2(note-taking instruction)` = c(9L, 
-1L, -11L, 3L), `Gain2(conventional) - Gain1(vocabulary notebook)` = c(9L, 
-1L, -8L, 4L), `Gain2(conventional) - Gain2(vocabulary notebook)` = c(9L, 
-1L, -12L, 4L), `Gain1(framework notes) - Gain2(framework notes)` = c(6L, 
-2L, -10L, 2L), `Gain1(framework notes) - Gain1(note-taking instruction)` = c(6L, 
-2L, -7L, 3L), `Gain1(framework notes) - Gain2(note-taking instruction)` = c(6L, 
-2L, -11L, 3L), `Gain1(framework notes) - Gain1(vocabulary notebook)` = c(6L, 
-2L, -8L, 4L), `Gain1(framework notes) - Gain2(vocabulary notebook)` = c(6L, 
-2L, -12L, 4L), `Gain2(framework notes) - Gain1(note-taking instruction)` = c(10L, 
-2L, -7L, 3L), `Gain2(framework notes) - Gain2(note-taking instruction)` = c(10L, 
-2L, -11L, 3L), `Gain2(framework notes) - Gain1(vocabulary notebook)` = c(10L, 
-2L, -8L, 4L), `Gain2(framework notes) - Gain2(vocabulary notebook)` = c(10L, 
-2L, -12L, 4L), `Gain1(note-taking instruction) - Gain2(note-taking instruction)` = c(7L, 
-3L, -11L, 3L), `Gain1(note-taking instruction) - Gain1(vocabulary notebook)` = c(7L, 
-3L, -8L, 4L), `Gain1(note-taking instruction) - Gain2(vocabulary notebook)` = c(7L, 
-3L, -12L, 4L), `Gain2(note-taking instruction) - Gain1(vocabulary notebook)` = c(11L, 
-3L, -8L, 4L), `Gain2(note-taking instruction) - Gain2(vocabulary notebook)` = c(11L, 
-3L, -12L, 4L), `Gain1(vocabulary notebook) - Gain2(vocabulary notebook)` = c(8L, 
-4L, -12L, 4L))

可以通过提取列表名称中Gain后的数字，筛选出前后数字一致的元素，提供两种实现方式：

方式一：使用stringr包（简洁直观） #

library(stringr)

# 提取每个列表名称中的两个Gain数字
gain_nums <- str_extract_all(names(My_list), "(?<=Gain)\\d+", simplify = TRUE)

# 筛选出前后数字相等的元素
filtered_list <- My_list[gain_nums[, 1] == gain_nums[, 2]]

方式二：基础R实现（无需额外包） #

# 提取每个列表名称中的两个Gain数字
gain_nums <- do.call(rbind, regmatches(names(My_list), gregexpr("(?<=Gain)\\d+", names(My_list), perl = TRUE)))

# 筛选出前后数字相等的元素
filtered_list <- My_list[gain_nums[, 1] == gain_nums[, 2]]

代码说明 #

• 正则表达式(?<=Gain)\\d+用于精准提取紧跟在Gain后的数字，(?<=Gain)是正向先行断言，确保只匹配Gain之后的数字部分；
• 提取结果为二维数组，每行对应一个列表名称的两个Gain数字；
• 通过比较每行的两个数字是否相等，保留符合条件的列表元素。

内容的提问来源于Stack Exchange，提问作者Simon Harmel