Find the common beginning in a list of strings (Python recipe) by Stephan Diehl

ActiveState Code (http://code.activestate.com/recipes/252177/)

problem with the proof. I have a small problem with the proof you let to the user.

I'am even asking if it exist considering the example

hipop

hapap

which leads to hpp and so become:

hipop

hapap

hpp

which leads to hp which seems wrong.

You are right. I've changed the code to get a right result in these kind of cases (hope, I got it right this time)

How about this ?

import operator

def allsame(seq):
    "Determines whether all the elements in a sequence are the same."

    # Compare all the elements to the first in the sequence,
    # then do a logical and (min) to see if they all matched.
    return min([elem==seq[0] for elem in seq]+[True])

def getcommonstart(seqlist):
    """Returns the common prefix of a list of sequences.

    Raises an exception if the list is empty.
    """
    # if not seqlist: return None
    m = [allsame(seq) for seq in zip(*seqlist)] # Map the matching elements
    m.append(False)                             # Guard in case all the sequences match
    return seqlist[0][0:m.index(False)]         # Truncate before first mismatch

import operator not needed. And you don't even need the operator module.

I originally had

def allsame(seq):
   return reduce(operator.and_,[elem==seq[0] for elem in seq],True)

but I changed it to use min() instead and forgot to remove the import.

That's nice. What about a combined approach?:

def getcommonletters(strlist):
    def booltrigger(expr):
        if not booltrigger.truth: return False
        if not expr:booltrigger.truth = False
        return expr
    booltrigger.truth = True
    return ''.join([x[0] for x in zip(*strlist) if \
                booltrigger(min([x[0]==elem for elem in x]))])

The 'booltrigger' is doing what should be possible with list comprehension anyway: Stop the iteration.

Or you could use itertools. I believe that itertools.takewhile does something close enough to what you were trying to achieve with booltrigger.

import itertools

def allsame(seq):
    return min([elem==seq[0] for elem in seq]+[True])

def getcommonstart(seqlist):
    letters = itertools.izip(*seqlist)                 # let's be lazy
    common = itertools.takewhile(allsame, letters)     # still lazy
    return ''.join([letters[0] for letters in common]) # merge back

What about ... <p>... this:</p>

>>> import os
>>> os.path.commonprefix(["hippie", "hippo"])
'hipp'
>>>

Lazy 'allsame'

def allsame(seq):
    return not seq or False not in itertools.imap(seq[0].__eq__, seq)

Know thy libraries! Hmm, that's why I thought this should have been solved somewhere already. I'm happy though that this was not the first comment :-) Just for the interested, the posixpath.commonprefix looks like this:

def commonprefix(m):
    "Given a list of pathnames, returns the longest common leading component"
    if not m: return ''
    prefix = m[0]
    for item in m:
        for i in range(len(prefix)):
            if prefix[:i+1] != item[:i+1]:
                prefix = prefix[:i]
                if i == 0:
                    return ''
                break
    return prefix

One-line version of this code. Not that it is very interesting, but I amused myself by creating a one-line version of findcommonstart as follows:

def findcommonstart(strlist):
    return strlist[0][:([min([x[0]==elem for elem in x]) \
                         for x in zip(*strlist)]+[0]).index(0)]
<pre>

</pre>

corrected (and optimized) version.

def getcommonstart(seq):
    if not seq:return ""
    seq.sort()
    s1, s2 = seq[0], seq[-1]
    l = min(len(s1), len(s2))
    if l == 0 :
        return ""
    for i in xrange(l) :
        if s1[i] != s2[i] :
            return s1[0:i]
    return s1[0:l]

Avoid sort(). The sort() could be expensive, and isn't really needed, since you're only interested in the first and last element of the sorted sequence, i.e. the minimum and maximum elements of the sequence. So, change

seq.sort()
s1, s2 = seq[0], seq[-1]

to

s1 = min(seq)
s2 = max(seq)

Avoid sort(). [Somehow, this comment went elsewhere in the comment-tree yesterday.]

The sort() could be expensive, and isn't really needed, since you're only interested in the first and last element of the sorted sequence, i.e. the minimum and maximum elements of the sequence. So, change

seq.sort()
s1, s2 = seq[0], seq[-1]

to

s1 = min(seq)
s2 = max(seq)

Binary search version. There are even fewer trips around the Python eval loop if the longest match is found using binary search instead of sequential comparison.

def commonprefix(m):
    "Given a list of pathnames, returns the longest common leading component"
    if not m: return ''
    a, b = min(m), max(m)
    lo, hi = 0, min(len(a), len(b))
    while lo < hi:
        mid = (lo+hi)//2 + 1
        if a[lo:mid] == b[lo:mid]:
            lo = mid
        else:
            hi = mid - 1
    return a[:hi]

sort not bad. Why should sort be slower than computing a min and a max? As far as my computational knowledge goes, in order to compute some minimum or maximum, the first thing to do is to sort anyway. Computing separate min and max actually slows things down.

I've done some tests and it looks like Raymonds version with sort instead of min,max is indeed faster. (And by the way, is the quickest of all the shown algorithms)

correction. max,min seems indeed to be faster. I did the first test with presorted lists.

one = [0, 1, 4]
two = [0, 1, 3]

both = [x if x[0] == x[1] else None for x in zip(one, two)] + [None]

common = one[:both.index(None)]