Find common directory path

Create a routine that, given a set of strings representing directory paths and a single character directory separator, will 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)

Clojure

<lang clojure>(defn common-prefix [sep paths]

 (let [parts-per-path (map #(.split (re-pattern sep) %) paths)
       common-parts (for [part-list (apply map vector parts-per-path)
                          :when (apply = part-list)] (first part-list))]
       (apply str (interpose sep common-parts))))

(println

(common-prefix "/"
               ["/home/user1/tmp/coverage/test"
                "/home/user1/tmp/covert/operator"
                "/home/user1/tmp/coven/members"]))</lang>

Factor

<lang factor>: take-shorter ( seq1 seq2 -- shorter )

   [ shorter? ] 2keep ? ;

common-head ( seq1 seq2 -- head )

   2dup mismatch [ nip head ] [ take-shorter ] if* ;

common-prefix-1 ( file1 file2 separator -- prefix )

   [ common-head ] dip '[ _ = not ] trim-tail ;

common-prefix ( seq separator -- prefix )

   [ ] swap '[ _ common-prefix-1 ] map-reduce ;</lang>

( scratchpad ) {
                   "/home/user1/tmp/coverage/test"
                   "/home/user1/tmp/covert/operator"
                   "/home/user1/tmp/coven/members" 
               } CHAR: / common-prefix .
"/home/user1/tmp/"

J

Solution: <lang j>parseDirs =: = <;.2 ] 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>

Note: This alternative formulation of parseDirs provides cross-platform support, without the need to specify the path separator. <lang j>parseDirs =: (PATHSEP_j_&= <;.2 ])@jhostpath</lang>

OCaml

<lang ocaml>let rec aux acc paths =

 if List.mem [] paths
 then (List.rev acc) else
 let heads = List.map List.hd paths in
 let item = List.hd heads in
 let all_the_same =
   List.for_all ((=) item) (List.tl heads)
 in
 if all_the_same
 then aux (item::acc) (List.map List.tl paths)
 else (List.rev acc)

let common_prefix sep = function

 | [] -> invalid_arg "common_prefix"
 | dirs ->
     let paths = List.map (Str.split (Str.regexp_string sep)) dirs in
     let res = aux [] paths in
     (sep ^ (String.concat sep res))

let () =

 let dirs = [
   "/home/user1/tmp/coverage/test";
   "/home/user1/tmp/covert/operator";
   "/home/user1/tmp/coven/members";
 ] in
 print_endline (common_prefix "/" dirs);

</lang>

(uses the module Str, str.cma)

Oz

With a few helper functions, we can express the solution like this in Oz: <lang oz>declare

 fun {CommonPrefix Sep Paths}
    fun {GetParts P} {String.tokens P Sep} end
    Parts = {ZipN {Map Paths GetParts}}
    EqualParts = {List.takeWhile Parts fun {$ X|Xr} {All Xr {Equals X}} end}
 in
    {Join Sep {Map EqualParts Head}}
 end

 fun {ZipN Xs}
    if {Some Xs {Equals nil}} then nil
    else
       {Map Xs Head} | {ZipN {Map Xs Tail}}
    end
 end

 fun {Join Sep Xs}
    {FoldR Xs fun {$ X Z} {Append X Sep|Z} end nil}
 end

 fun {Equals C}
    fun {$ X} X == C end
 end

 fun {Head X|_} X end

 fun {Tail _|Xr} Xr end

in

 {System.showInfo {CommonPrefix &/
                   ["/home/user1/tmp/coverage/test"
                    "/home/user1/tmp/covert/operator"
                    "/home/user1/tmp/coven/members"]}}</lang>

PicoLisp

<lang PicoLisp>(de commonPath (Lst Chr)

  (glue Chr
     (make
        (apply find
           (mapcar '((L) (split (chop L) Chr)) Lst)
           '(@ (or (pass <>) (nil (link (next))))) ) ) ) )</lang>

Output:

(commonPath
   (quote
      "/home/user1/tmp/coverage/test"
      "/home/user1/tmp/covert/operator"
      "/home/user1/tmp/coven/members" )
   "/" )

-> "/home/user1/tmp"

PureBasic

PureBasic don't have a path comparator directly but instead have powerful string tools.

Simply by checking the catalog names until they mismatch and add up the correct parts, the task is accomplished. <lang PureBasic>Procedure.s CommonPath(Array InPaths.s(1),separator.s="/")

 Protected SOut$=""
 Protected i, j, toggle
 
 If ArraySize(InPaths())=0
   ProcedureReturn InPaths(0)  ; Special case, only one path
 EndIf
 
 Repeat
   i+1
   toggle=#False
   For j=1 To ArraySize(InPaths())
     If (StringField(InPaths(j-1),i,separator)=StringField(InPaths(j),i,separator))
       If Not toggle
         SOut$+StringField(InPaths(j-1),i,separator)+separator
         toggle=#True
       EndIf
     Else
       ProcedureReturn SOut$
     EndIf      
   Next
 ForEver

EndProcedure</lang>

Example of implementation <lang PureBasic>Dim t.s(2) t(0)="/home/user1/tmp/coverage/test" t(1)="/home/user1/tmp/covert/operator" t(2)="/home/user1/tmp/coven/members"

Debug CommonPath(t(),"/"))</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>>>> from itertools import takewhile >>> def allnamesequal(name): return all(n==name[0] for n in name[1:])

>>> def commonprefix(paths, sep='/'): 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>

Ruby

Uses the standard library abbrev module: Given a set of strings, calculate the set of unambiguous abbreviations for those strings, and return a hash where the keys are all the possible abbreviations and the values are the full strings.

<lang ruby>require 'abbrev'

dirs = %w( /home/user1/tmp/coverage/test /home/user1/tmp/covert/operator /home/user1/tmp/coven/members )

common_prefix = dirs.abbrev.keys.min_by {|key| key.length}.chop # => "/home/user1/tmp/cove" common_directory = common_prefix.sub(%r{/[^/]*$}, ) # => "/home/user1/tmp"</lang>

Implementing without that module: <lang ruby>separator = '/' paths = dirs.collect {|dir| dir.split(separator)} uncommon_idx = paths.transpose.each_with_index.find {|dirnames, idx| dirnames.uniq.length > 1}.last common_directory = paths[0][0 ... uncommon_idx].join(separator) # => "/home/user1/tmp"</lang>

Tcl

<lang tcl>package require Tcl 8.5 proc pop {varname} {

   upvar 1 $varname var
   set var [lassign $var head]
   return $head

}

proc common_prefix {dirs {separator "/"}} {

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

}</lang>

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