Find common directory path

From Rosetta Code
Task
Find common directory path
You are encouraged to solve this task according to the task description, using any language you may know.

Given a set of strings representing directory paths and a single character directory separator; return a string representing that part of the directory tree that is common to all the directories.

Test your routine using the forward slash '/' character as the directory separator and the following three strings as input paths:

 '/home/user1/tmp/coverage/test'
 '/home/user1/tmp/covert/operator'
 '/home/user1/tmp/coven/members'

Note: The resultant path should be the valid directory '/home/user1/tmp' and not the longest common string '/home/user1/tmp/cove'.
If your language has a routine that performs this function (even if it does not have a changeable separator character, then mention it as part of the task)

J

Solution: <lang j>parseDirs =: (PATHSEP_j_&= <;.2 ])@jhostpath getCommonPrefix =: ([: *./\ *./@({. ="1 }.)) ;@# {.

getCommonDirPath=: [: getCommonPrefix parseDirs&></lang>

Example: <lang j> paths=: '/home/user1/tmp/coverage/test';'/home/user1/tmp/covert/operator';'/home/user1/tmp/coven/members'

  getCommonPrefix >paths

/home/user1/tmp/cove

  getCommonDirPath paths

/home/user1/tmp/</lang>

Python

The Python os.path.commonprefix function is broken as it returns common characters that may not form a valid directory path: <lang python>>>> import os >>> os.path.commonprefix(['/home/user1/tmp/coverage/test', '/home/user1/tmp/covert/operator', '/home/user1/tmp/coven/members']) '/home/user1/tmp/cove'</lang>

This result can be fixed: <lang python>>>> def commonprefix(*args, sep='/'): return os.path.commonprefix(*args).rpartition(sep)[0]

>>> commonprefix(['/home/user1/tmp/coverage/test', '/home/user1/tmp/covert/operator', '/home/user1/tmp/coven/members']) '/home/user1/tmp'</lang>

But it may be better to not rely on the faulty implementation at all: <lang python>>>> def allnamesequal(name): return all(n==name[0] for n in name[1:])

>>> def commonprefix(paths, sep='/'): from itertools import takewhile bydirectorylevels = zip(*[p.split(sep) for p in paths]) return sep.join(x[0] for x in takewhile(allnamesequal, bydirectorylevels))

>>> commonprefix(['/home/user1/tmp/coverage/test', '/home/user1/tmp/covert/operator', '/home/user1/tmp/coven/members']) '/home/user1/tmp'</lang>

Tcl

<lang tcl>proc pop {varname {nth 0}} {

   upvar $varname args
   set head [lindex $args $nth]
   set args [lreplace $args $nth $nth]
   return $head

}

proc common_prefix {dirs} {

   set parts [file split [pop dirs]]
   while {[llength $dirs]} {
       set r {}
       foreach cmp $parts elt [file split [pop dirs]] {
           if {$cmp ne $elt} break
           lappend r $cmp
       }
       set parts $r
   }
   return [eval file join $parts]

}</lang>

% common_prefix {/home/user1/tmp/coverage/test /home/user1/tmp/covert/operator /home/user1/tmp/coven/members}
/home/user1/tmp