如何在保留名称的同时将多个文件导入列表？

使用basename和tools::file_path_sans_ext：

filenames <- head(list.files("~/StackOverflow", pattern = "^[^#].*\\.R", recursive = TRUE, full.names = TRUE))
filenames
# [1] "C:\\Users\\r2/StackOverflow/1000343/61469332.R"  "C:\\Users\\r2/StackOverflow/10087004/61857346.R"
# [3] "C:\\Users\\r2/StackOverflow/10097832/60589834.R" "C:\\Users\\r2/StackOverflow/10214507/60837843.R"
# [5] "C:\\Users\\r2/StackOverflow/10215127/61720149.R" "C:\\Users\\r2/StackOverflow/10226369/60778116.R"
basename(filenames)
# [1] "61469332.R" "61857346.R" "60589834.R" "60837843.R" "61720149.R" "60778116.R"
tools::file_path_sans_ext(basename(filenames))
# [1] "61469332" "61857346" "60589834" "60837843" "61720149" "60778116"

somedat <- setNames(lapply(filenames, readLines, n=2), 
                    tools::file_path_sans_ext(basename(filenames)))
names(somedat)
# [1] "61469332" "61857346" "60589834" "60837843" "61720149" "60778116"

str(somedat)
# List of 6
#  $ 61469332: chr [1:2] "# https://stackoverflow.com/questions/61469332/determine-function-name-within-that-function/61469380" ""
#  $ 61857346: chr [1:2] "# https://stackoverflow.com/questions/61857346/how-to-use-apply-family-instead-of-nested-for-loop-for-my-problem?noredirect=1" ""
#  $ 60589834: chr [1:2] "# https://stackoverflow.com/questions/60589834/add-columns-to-data-frame-based-on-function-argument" ""
#  $ 60837843: chr [1:2] "# https://stackoverflow.com/questions/60837843/how-to-remove-all-parentheses-from-a-vector-of-string-except-whe"| __truncated__ ""
#  $ 61720149: chr [1:2] "# https://stackoverflow.com/questions/61720149/extracting-the-original-data-based-on-filtering-criteria" ""
#  $ 60778116: chr [1:2] "# https://stackoverflow.com/questions/60778116/how-to-shift-data-by-a-factor-of-two-months-in-r" ""

每个“名称”都是（在这种情况下）stackoverflow问题编号的字符表示形式，已".R"删除。（由于我通常将普通URL作为第一行，然后在我用来测试/播放和回答SO问题的文件中包含一个空行，因此所有这些文件在前两行看起来都相似。）