Question

我有一个文件路径的字符向量：

> tail(paths)
[1] "/home/username/data/dir/GCZ98/GCZ98_1998_12_16.asc.gz"
[2] "/home/username/data/dir/GCZ98/GCZ98_1998_12_20.asc.gz"
[3] "/home/username/data/dir/GCZ99/GCZ99_1999_12_21.asc.gz"
[4] "/home/username/data/dir/GCZ99/GCZ99_1999_12_23.asc.gz"
[5] "/home/username/data/dir/GCZ99/GCZ99_1999_12_27.asc.gz"
[6] "/home/username/data/dir/GCZ99/GCZ99_1999_12_28.asc.gz"

我想通过父文件夹名称将其拆分为矢量列表，即：

> tail(desired)
$ "/home/username/data/dir/GCZ98"
[1] "/home/username/data/dir/GCZ98/GCZ98_1998_12_16.asc.gz"
[2] "/home/username/data/dir/GCZ98/GCZ98_1998_12_20.asc.gz"
$ "/home/username/data/dir/GCZ98"
[1] "/home/username/data/dir/GCZ99/GCZ99_1999_12_21.asc.gz"
[2] "/home/username/data/dir/GCZ99/GCZ99_1999_12_23.asc.gz"
[3] "/home/username/data/dir/GCZ99/GCZ99_1999_12_27.asc.gz"
[4] "/home/username/data/dir/GCZ99/GCZ99_1999_12_28.asc.gz"

我尝试使用split和strsplit几乎没有成功，但我正在努力寻找能够满足我需求的正则表达式。

感谢您的帮助

Answer 1

您可以合并split和dirname：

path <- c("/home/username/data/dir/GCZ98/GCZ98_1998_12_16.asc.gz",
          "/home/username/data/dir/GCZ98/GCZ98_1998_12_20.asc.gz",
          "/home/username/data/dir/GCZ99/GCZ99_1999_12_21.asc.gz",
          "/home/username/data/dir/GCZ99/GCZ99_1999_12_23.asc.gz",
          "/home/username/data/dir/GCZ99/GCZ99_1999_12_27.asc.gz",
          "/home/username/data/dir/GCZ99/GCZ99_1999_12_28.asc.gz")

## split by basedir
split(path, dirname(path))

# $`/home/username/data/dir/GCZ98`
# [1] "/home/username/data/dir/GCZ98/GCZ98_1998_12_16.asc.gz" "/home/username/data/dir/GCZ98/GCZ98_1998_12_20.asc.gz"
# 
# $`/home/username/data/dir/GCZ99`
# [1] "/home/username/data/dir/GCZ99/GCZ99_1999_12_21.asc.gz" "/home/username/data/dir/GCZ99/GCZ99_1999_12_23.asc.gz" "/home/username/data/dir/GCZ99/GCZ99_1999_12_27.asc.gz"
# [4] "/home/username/data/dir/GCZ99/GCZ99_1999_12_28.asc.gz"

Answer 2

正则表达式方法：

> split(paths, gsub("(.*)/[^/]+$", "\\1", paths))
$`/home/username/data/dir/GCZ98`
[1] "/home/username/data/dir/GCZ98/GCZ98_1998_12_16.asc.gz"
[2] "/home/username/data/dir/GCZ98/GCZ98_1998_12_20.asc.gz"

$`/home/username/data/dir/GCZ99`
[1] "/home/username/data/dir/GCZ99/GCZ99_1999_12_21.asc.gz"
[2] "/home/username/data/dir/GCZ99/GCZ99_1999_12_23.asc.gz"
[3] "/home/username/data/dir/GCZ99/GCZ99_1999_12_27.asc.gz"
[4] "/home/username/data/dir/GCZ99/GCZ99_1999_12_28.asc.gz"

R将文件路径的字符向量拆分为父目录列表？

2 个答案: