Find common directory path
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.
You are encouraged to solve this task according to the task description, using any language you may know.
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>
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>S1$="/home/user1/tmp/coverage/test" S2$="/home/user1/tmp/covert/operator" S3$="/home/user1/tmp/coven/members" SOut$="" i=1
While (StringField(S1$,i,"/")=StringField(S2$,i,"/")) And (StringField(S1$,i,"/")=StringField(S3$,i,"/"))
SOut$+StringField(S1$,i,"/")+"/" i+1
Wend </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>
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