Determine the common prefix of multiple strings

Question

I have a set of strings, e.g.

my_prefix_what_ever
my_prefix_what_so_ever
my_prefix_doesnt_matter

I simply want to find the longest common portion of these strings, here the prefix. In the above the result should be

my_prefix_

The strings

my_prefix_what_ever
my_prefix_what_so_ever
my_doesnt_matter

should result in the prefix

my_

Is there a relatively painless way in Python to determine the prefix (without having to iterate over each character manually)?

PS: I'm using Python 2.6.3.

Answer 1

Posting yet another solution as its more concise than most of the solutions here

def common_prefix(strings: list[str]) -> str:
    shortest = min(strings, key=len)
    for i, c in enumerate(shortest):
        if any(x[i] != c for x in strings):
            return shortest[:i]
    return shortest

assert common_prefix(["Abc01", "Abc02", "Abc03"]) == "Abc0"
assert common_prefix(["Abc01", "Abc01"]) == "Abc01"
assert common_prefix(["", "Abc01"]) == ""
assert common_prefix(["Xyz01", "Abc01"]) == ""

It will throw on an empty list but trivial to add a check for this if needed

Answer 2

import os
os.path.commonprefix(list_name) ## this will do the work

OS module link

Answer 3

I had a slight variation of the problem which I think will be useful to document:

I have a list like:

• my_prefix_what_ever
• my_prefix_what_so_ever
• my_prefix_doesnt_matter
• some_noise
• some_other_noise

So I would expect my_prefix to be returned. That can be done with:

from collections import Counter

def get_longest_common_prefix(values, min_length):
    substrings = [value[0: i-1] for value in values for i in range(min_length, len(value))]
    counter = Counter(substrings)
    # remove count of 1
    counter -= Counter(set(substrings))
    return max(counter, key=len)

Answer 4

numpy matrix approach:

import numpy as np


def common_prefix(strings: list[str]) -> str:
    # common prefix cannot be larger than the smallest str
    min_length = min(len(string) for string in strings)
    strings = [string[:min_length] for string in strings]

    array = np.array([list(x) for x in strings])  # covert to numpy matrix for column-wise operations
    for i in range(min_length):
        # for every column check if all char values are the same (same as first)
        if not all(array[:, i] == array[0][i]):
            # if not return the substring before the first char difference
            return strings[0][:i]

    # the common prefix is the full (shortest) str
    return strings[0]


assert common_prefix(["str1", "str2", "str3"]) == "str"
assert common_prefix(["s", "st", "str"]) == "s"
assert common_prefix(["1str", "str", "str"]) == ""

Answer 5

Find the common prefix in all words from the given input string, if there is no common prefix print -1

stringList = ['my_prefix_what_ever', 'my_prefix_what_so_ever', 'my_prefix_doesnt_matter']
len2 = len( stringList )
if len2 != 0:
    # let shortest word is prefix
    prefix = min( stringList )
    for i in range( len2 ):
        word = stringList[ i ]
        len1 = len( prefix )
        # slicing each word as lenght of prefix
        word = word[ 0:len1 ]
        for j in range( len1 ):
            # comparing each letter of word and prefix
            if word[ j ] != prefix[ j ]:
                # if letter does not match slice the prefix
                prefix = prefix[ :j ]
                break # after getting comman prefix move to next word
    if len( prefix ) != 0:
        print("common prefix: ",prefix)
    else:
        print("-1")
else:
     print("string List is empty") 

Answer 6

In one line without using itertools, for no particular reason, although it does iterate through each character:

''.join([z[0] for z in zip(*(list(s) for s in strings)) if all(x==z[0] for x in z)])

Answer 7

Here's a simple clean solution. The idea is to use zip() function to line up all the characters by putting them in a list of 1st characters, list of 2nd characters,...list of nth characters. Then iterate each list to check if they contain only 1 value.

a = ["my_prefix_what_ever", "my_prefix_what_so_ever", "my_prefix_doesnt_matter"]

list = [all(x[i] == x[i+1] for i in range(len(x)-1)) for x in zip(*a)]

print a[0][:list.index(0) if list.count(0) > 0 else len(list)]

output: my_prefix_

Answer 8

The second line of this employs the reduce function on each character in the input strings. It returns a list of N+1 elements where N is length of the shortest input string.

Each element in lot is either (a) the input character, if all input strings match at that position, or (b) None. lot.index(None) is the position of the first None in lot: the length of the common prefix. out is that common prefix.

val = ["axc", "abc", "abc"]
lot = [reduce(lambda a, b: a if a == b else None, x) for x in zip(*val)] + [None]
out = val[0][:lot.index(None)]

Answer 9

Never rewrite what is provided to you: os.path.commonprefix does exactly this:

Return the longest path prefix (taken character-by-character) that is a prefix of all paths in list. If list is empty, return the empty string (''). Note that this may return invalid paths because it works a character at a time.

For comparison to the other answers, here's the code:

# Return the longest prefix of all list elements.
def commonprefix(m):
    "Given a list of pathnames, returns the longest common leading component"
    if not m: return ''
    s1 = min(m)
    s2 = max(m)
    for i, c in enumerate(s1):
        if c != s2[i]:
            return s1[:i]
    return s1

Answer 10

Here is another way of doing this using OrderedDict with minimal code.

import collections
import itertools

def commonprefix(instrings):
    """ Common prefix of a list of input strings using OrderedDict """

    d = collections.OrderedDict()

    for instring in instrings:
        for idx,char in enumerate(instring):
            # Make sure index is added into key
            d[(char, idx)] = d.get((char,idx), 0) + 1

    # Return prefix of keys while value == length(instrings)
    return ''.join([k[0] for k in itertools.takewhile(lambda x: d[x] == len(instrings), d)])

Answer 11

Just out of curiosity I figured out yet another way to do this:

def common_prefix(strings):

    if len(strings) == 1:#rule out trivial case
        return strings[0]

    prefix = strings[0]

    for string in strings[1:]:
        while string[:len(prefix)] != prefix and prefix:
            prefix = prefix[:len(prefix)-1]
        if not prefix:
            break

    return prefix

strings = ["my_prefix_what_ever","my_prefix_what_so_ever","my_prefix_doesnt_matter"]

print common_prefix(strings)
#Prints "my_prefix_"

As Ned pointed out it's probably better to use os.path.commonprefix, which is a pretty elegant function.

Answer 12

Ned Batchelder is probably right. But for the fun of it, here's a more efficient version of phimuemue's answer using itertools.

import itertools

strings = ['my_prefix_what_ever', 
           'my_prefix_what_so_ever', 
           'my_prefix_doesnt_matter']

def all_same(x):
    return all(x[0] == y for y in x)

char_tuples = itertools.izip(*strings)
prefix_tuples = itertools.takewhile(all_same, char_tuples)
''.join(x[0] for x in prefix_tuples)

As an affront to readability, here's a one-line version :)

>>> from itertools import takewhile, izip
>>> ''.join(c[0] for c in takewhile(lambda x: all(x[0] == y for y in x), izip(*strings)))
'my_prefix_'

Answer 13

Here's my solution:

a = ["my_prefix_what_ever", "my_prefix_what_so_ever", "my_prefix_doesnt_matter"]

prefix_len = len(a[0])
for x in a[1 : ]:
    prefix_len = min(prefix_len, len(x))
    while not x.startswith(a[0][ : prefix_len]):
        prefix_len -= 1

prefix = a[0][ : prefix_len]

Answer 14

The following is an working, but probably quite inefficient solution.

a = ["my_prefix_what_ever", "my_prefix_what_so_ever", "my_prefix_doesnt_matter"]
b = zip(*a)
c = [x[0] for x in b if x==(x[0],)*len(x)]
result = "".join(c)

For small sets of strings, the above is no problem at all. But for larger sets, I personally would code another, manual solution that checks each character one after another and stops when there are differences.

Algorithmically, this yields the same procedure, however, one might be able to avoid constructing the list c.