获取R中的目录中的文件号？

Question

in shell ,to make a dir:

mkdir  /home/test

then ,to create a file named ".test" in the "/home/test"

a=list.files(path = "/home/test",include.dirs = FALSE)
a
character(0)
a=list.files(path = "/home/test",include.dirs = TRUE)
a
character(0)
a=list.files(path = "/home/test/",include.dirs = TRUE)
a
character(0)
list.files(path = '/home/test', all.files=TRUE,inclued.dirs=FALSE)
[1] "."     ".."    ".test"
a=list.files(path = '/home/test', all.files=TRUE)
length(a)
[1] 3

how can i get length(a) = 1 using regular expression parameters pattern= in list.files to prune . and ..

Answer 1

Use all.files=TRUE to show all file names including hidden files.

list.files(path = '/home/test', all.files=TRUE)

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To answer your edit, one way would be to use a negative number with tail

tail(list.files(path = '/home/test', all.files=TRUE), -2)

---

Using only the pattern argument:

list.files(path='/home/test', all.files=TRUE, pattern="^[^\\.]|\\.[^\\.]")

The pattern says "anything that starts with something other than a dot or anything that starts with a dot followed by anything other than a dot."

---

Although it breaks your requirement to use the pattern argument of list.files , I would actually probably wrap grep around list.statements in this case.

grep("^\\.*\\.$", list.files(path='/home/test', all.files=TRUE), 
     invert=TRUE, value=TRUE)

The above will find any file names that only contain dots, then return everything else. invert=TRUE means "find the names that do not match", and value=TRUE means "return the names instead of their location."