How can I tell if a string repeats itself in Python?

The Question :

356 people think this question is useful

I’m looking for a way to test whether or not a given string repeats itself for the entire string or not.


    '0045662100456621004566210045662100456621',             # '00456621'
    '0072992700729927007299270072992700729927',             # '00729927'
    '001443001443001443001443001443001443001443',           # '001443'
    '037037037037037037037037037037037037037037037',        # '037'
    '047619047619047619047619047619047619047619',           # '047619'
    '002457002457002457002457002457002457002457',           # '002457'
    '001221001221001221001221001221001221001221',           # '001221'
    '001230012300123001230012300123001230012300123',        # '00123'
    '0013947001394700139470013947001394700139470013947',    # '0013947'
    '001001001001001001001001001001001001001001001001001',  # '001'
    '001406469760900140646976090014064697609',              # '0014064697609'

are strings which repeat themselves, and


are examples of ones that do not.

The repeating sections of the strings I’m given can be quite long, and the strings themselves can be 500 or more characters, so looping through each character trying to build a pattern then checking the pattern vs the rest of the string seems awful slow. Multiply that by potentially hundreds of strings and I can’t see any intuitive solution.

I’ve looked into regexes a bit and they seem good for when you know what you’re looking for, or at least the length of the pattern you’re looking for. Unfortunately, I know neither.

How can I tell if a string is repeating itself and if it is, what the shortest repeating subsequence is?

The Question Comments :
  • looping through each character trying to build a pattern then checking the pattern vs the rest of the string seems awful slow – but is it?
  • possible duplicate of Writing a regex to detect repeat-characters
  • @AvinashRaj That’s only matching part of a string, not the full thing.
  • @AvinashRaj The OP is asking about all possible solutions. The question you link to only accepts regex solution. Note that regex may be able to solve the problem but in much more time than necessary. For example an optimal solution (i.e. linear time) would use the suffix tree of the text. You just have to find the longest repeated substring and do some checks on the lengths.
  • @TigerhawkT3 The real dataset is far too large and unwieldy, but the examples in the question are part of it, and if you wish, here’s some more.

The Answer 1

580 people think this answer is useful

Here’s a concise solution which avoids regular expressions and slow in-Python loops:

def principal_period(s):
    i = (s+s).find(s, 1, -1)
    return None if i == -1 else s[:i]

See the Community Wiki answer started by @davidism for benchmark results. In summary,

David Zhang’s solution is the clear winner, outperforming all others by at least 5x for the large example set.

(That answer’s words, not mine.)

This is based on the observation that a string is periodic if and only if it is equal to a nontrivial rotation of itself. Kudos to @AleksiTorhamo for realizing that we can then recover the principal period from the index of the first occurrence of s in (s+s)[1:-1], and for informing me of the optional start and end arguments of Python’s string.find.

The Answer 2

183 people think this answer is useful

Here’s a solution using regular expressions.

import re

REPEATER = re.compile(r"(.+?)\1+$")

def repeated(s):
    match = REPEATER.match(s)
    return if match else None

Iterating over the examples in the question:

examples = [

for e in examples:
    sub = repeated(e)
    if sub:
        print("%r: %r" % (e, sub))
        print("%r does not repeat." % e)

… produces this output:

'0045662100456621004566210045662100456621': '00456621'
'0072992700729927007299270072992700729927': '00729927'
'001443001443001443001443001443001443001443': '001443'
'037037037037037037037037037037037037037037037': '037'
'047619047619047619047619047619047619047619': '047619'
'002457002457002457002457002457002457002457': '002457'
'001221001221001221001221001221001221001221': '001221'
'001230012300123001230012300123001230012300123': '00123'
'0013947001394700139470013947001394700139470013947': '0013947'
'001001001001001001001001001001001001001001001001001': '001'
'001406469760900140646976090014064697609': '0014064697609'
'004608294930875576036866359447' does not repeat.
'00469483568075117370892018779342723' does not repeat.
'004739336492890995260663507109' does not repeat.
'001508295625942684766214177978883861236802413273' does not repeat.
'007518796992481203' does not repeat.
'0071942446043165467625899280575539568345323741' does not repeat.
'0434782608695652173913' does not repeat.
'0344827586206896551724137931' does not repeat.
'002481389578163771712158808933' does not repeat.
'002932551319648093841642228739' does not repeat.
'0035587188612099644128113879' does not repeat.
'003484320557491289198606271777' does not repeat.
'00115074798619102416570771' does not repeat.

The regular expression (.+?)\1+$ is divided into three parts:

  1. (.+?) is a matching group containing at least one (but as few as possible) of any character (because +? is non-greedy).

  2. \1+ checks for at least one repetition of the matching group in the first part.

  3. $ checks for the end of the string, to ensure that there’s no extra, non-repeating content after the repeated substrings (and using re.match() ensures that there’s no non-repeating text before the repeated substrings).

In Python 3.4 and later, you could drop the $ and use re.fullmatch() instead, or (in any Python at least as far back as 2.3) go the other way and use with the regex ^(.+?)\1+$, all of which are more down to personal taste than anything else.

The Answer 3

90 people think this answer is useful

You can make the observation that for a string to be considered repeating, its length must be divisible by the length of its repeated sequence. Given that, here is a solution that generates divisors of the length from 1 to n / 2 inclusive, divides the original string into substrings with the length of the divisors, and tests the equality of the result set:

from math import sqrt, floor

def divquot(n):
    if n > 1:
        yield 1, n
    swapped = []
    for d in range(2, int(floor(sqrt(n))) + 1):
        q, r = divmod(n, d)
        if r == 0:
            yield d, q
            swapped.append((q, d))
    while swapped:
        yield swapped.pop()

def repeats(s):
    n = len(s)
    for d, q in divquot(n):
        sl = s[0:d]
        if sl * q == s:
            return sl
    return None

EDIT: In Python 3, the / operator has changed to do float division by default. To get the int division from Python 2, you can use the // operator instead. Thank you to @TigerhawkT3 for bringing this to my attention.

The // operator performs integer division in both Python 2 and Python 3, so I’ve updated the answer to support both versions. The part where we test to see if all the substrings are equal is now a short-circuiting operation using all and a generator expression.

UPDATE: In response to a change in the original question, the code has now been updated to return the smallest repeating substring if it exists and None if it does not. @godlygeek has suggested using divmod to reduce the number of iterations on the divisors generator, and the code has been updated to match that as well. It now returns all positive divisors of n in ascending order, exclusive of n itself.

Further update for high performance: After multiple tests, I’ve come to the conclusion that simply testing for string equality has the best performance out of any slicing or iterator solution in Python. Thus, I’ve taken a leaf out of @TigerhawkT3 ‘s book and updated my solution. It’s now over 6x as fast as before, noticably faster than Tigerhawk’s solution but slower than David’s.

The Answer 4

88 people think this answer is useful

Here are some benchmarks for the various answers to this question. There were some surprising results, including wildly different performance depending on the string being tested.

Some functions were modified to work with Python 3 (mainly by replacing / with // to ensure integer division). If you see something wrong, want to add your function, or want to add another test string, ping @ZeroPiraeus in the Python chatroom.

In summary: there’s about a 50x difference between the best- and worst-performing solutions for the large set of example data supplied by OP here (via this comment). David Zhang’s solution is the clear winner, outperforming all others by around 5x for the large example set.

A couple of the answers are very slow in extremely large “no match” cases. Otherwise, the functions seem to be equally matched or clear winners depending on the test.

Here are the results, including plots made using matplotlib and seaborn to show the different distributions:

Corpus 1 (supplied examples – small set)

mean performance:
 0.0003  david_zhang
 0.0009  zero
 0.0013  antti
 0.0013  tigerhawk_2
 0.0015  carpetpython
 0.0029  tigerhawk_1
 0.0031  davidism
 0.0035  saksham
 0.0046  shashank
 0.0052  riad
 0.0056  piotr

median performance:
 0.0003  david_zhang
 0.0008  zero
 0.0013  antti
 0.0013  tigerhawk_2
 0.0014  carpetpython
 0.0027  tigerhawk_1
 0.0031  davidism
 0.0038  saksham
 0.0044  shashank
 0.0054  riad
 0.0058  piotr

Corpus 1 graph

Corpus 2 (supplied examples – large set)

mean performance:
 0.0006  david_zhang
 0.0036  tigerhawk_2
 0.0036  antti
 0.0037  zero
 0.0039  carpetpython
 0.0052  shashank
 0.0056  piotr
 0.0066  davidism
 0.0120  tigerhawk_1
 0.0177  riad
 0.0283  saksham

median performance:
 0.0004  david_zhang
 0.0018  zero
 0.0022  tigerhawk_2
 0.0022  antti
 0.0024  carpetpython
 0.0043  davidism
 0.0049  shashank
 0.0055  piotr
 0.0061  tigerhawk_1
 0.0077  riad
 0.0109  saksham

Corpus 1 graph

Corpus 3 (edge cases)

mean performance:
 0.0123  shashank
 0.0375  david_zhang
 0.0376  piotr
 0.0394  carpetpython
 0.0479  antti
 0.0488  tigerhawk_2
 0.2269  tigerhawk_1
 0.2336  davidism
 0.7239  saksham
 3.6265  zero
 6.0111  riad

median performance:
 0.0107  tigerhawk_2
 0.0108  antti
 0.0109  carpetpython
 0.0135  david_zhang
 0.0137  tigerhawk_1
 0.0150  shashank
 0.0229  saksham
 0.0255  piotr
 0.0721  davidism
 0.1080  zero
 1.8539  riad

Corpus 3 graph

The tests and raw results are available here.

The Answer 5

37 people think this answer is useful

Non-regex solution:

def repeat(string):
    for i in range(1, len(string)//2+1):
        if not len(string)%len(string[0:i]) and string[0:i]*(len(string)//len(string[0:i])) == string:
            return string[0:i]

Faster non-regex solution, thanks to @ThatWeirdo (see comments):

def repeat(string):
    l = len(string)
    for i in range(1, len(string)//2+1):
        if l%i: continue
        s = string[0:i]
        if s*(l//i) == string:
            return s

The above solution is very rarely slower than the original by a few percent, but it’s usually a good bit faster – sometimes a whole lot faster. It’s still not faster than davidism’s for longer strings, and zero’s regex solution is superior for short strings. It comes out to the fastest (according to davidism’s test on github – see his answer) with strings of about 1000-1500 characters. Regardless, it’s reliably second-fastest (or better) in all cases I tested. Thanks, ThatWeirdo.





The Answer 6

24 people think this answer is useful

First, halve the string as long as it’s a “2 part” duplicate. This reduces the search space if there are an even number of repeats. Then, working forwards to find the smallest repeating string, check if splitting the full string by increasingly larger sub-string results in only empty values. Only sub-strings up to length // 2 need to be tested since anything over that would have no repeats.

def shortest_repeat(orig_value):
    if not orig_value:
        return None

    value = orig_value

    while True:
        len_half = len(value) // 2
        first_half = value[:len_half]

        if first_half != value[len_half:]:

        value = first_half

    len_value = len(value)
    split = value.split

    for i in (i for i in range(1, len_value // 2) if len_value % i == 0):
        if not any(split(value[:i])):
            return value[:i]

    return value if value != orig_value else None

This returns the shortest match or None if there is no match.

The Answer 7

16 people think this answer is useful

The problem may also be solved in O(n) in worst case with prefix function.

Note, it may be slower in general case(UPD: and is much slower) than other solutions which depend on number of divisors of n, but usually find fails sooner, I think one of bad cases for them will be aaa....aab, where there are n - 1 = 2 * 3 * 5 * 7 ... *p_n - 1 a‘s

First of all you need to calculate prefix function

def prefix_function(s):
    n = len(s)
    pi = [0] * n
    for i in xrange(1, n):
        j = pi[i - 1]
        while(j > 0 and s[i] != s[j]):
            j = pi[j - 1]
        if (s[i] == s[j]):
            j += 1
        pi[i] = j;
    return pi

then either there’s no answer or the shortest period is

k = len(s) - prefix_function(s[-1])

and you just have to check if k != n and n % k == 0 (if k != n and n % k == 0 then answer is s[:k], else there’s no answer

You may check the proof here (in Russian, but online translator will probably do the trick)

def riad(s):
    n = len(s)
    pi = [0] * n
    for i in xrange(1, n):
        j = pi[i - 1]
        while(j > 0 and s[i] != s[j]):
            j = pi[j - 1]
        if (s[i] == s[j]):
            j += 1
        pi[i] = j;
    k = n - pi[-1]
    return s[:k] if (n != k and n % k == 0) else None

The Answer 8

16 people think this answer is useful

This version tries only those candidate sequence lengths that are factors of the string length; and uses the * operator to build a full-length string from the candidate sequence:

def get_shortest_repeat(string):
    length = len(string)
    for i in range(1, length // 2 + 1):
        if length % i:  # skip non-factors early

        candidate = string[:i]
        if string == candidate * (length // i):
            return candidate

    return None

Thanks to TigerhawkT3 for noticing that length // 2 without + 1 would fail to match the abab case.

The Answer 9

15 people think this answer is useful

Here’s a straight forward solution, without regexes.

For substrings of s starting from zeroth index, of lengths 1 through len(s), check if that substring, substr is the repeated pattern. This check can be performed by concatenating substr with itself ratio times, such that the length of the string thus formed is equal to the length of s. Hence ratio=len(s)/len(substr).

Return when first such substring is found. This would provide the smallest possible substring, if one exists.

def check_repeat(s):
    for i in range(1, len(s)):
        substr = s[:i]
        ratio = len(s)/len(substr)
        if substr * ratio == s:
            print 'Repeating on "%s"' % substr
    print 'Non repeating'

>>> check_repeat('254725472547')
Repeating on "2547"
>>> check_repeat('abcdeabcdeabcdeabcde')
Repeating on "abcde"

The Answer 10

9 people think this answer is useful

I started with more than eight solutions to this problem. Some were bases on regex (match, findall, split), some of string slicing and testing, and some with string methods (find, count, split). Each had benefits in code clarity, code size, speed and memory consumption. I was going to post my answer here when I noticed that execution speed was ranked as important, so I did more testing and improvement to arrive at this:

def repeating(s):
    size = len(s)
    incr = size % 2 + 1
    for n in xrange(1, size//2+1, incr):
        if size % n == 0:
            if s[:n] * (size//n) == s:
                return s[:n]

This answer seems similar to a few other answers here, but it has a few speed optimisations others have not used:

  • xrange is a little faster in this application,
  • if an input string is an odd length, do not check any even length substrings,
  • by using s[:n] directly, we avoid creating a variable in each loop.

I would be interested to see how this performs in the standard tests with common hardware. I believe it will be well short of David Zhang’s excellent algorithm in most tests, but should be quite fast otherwise.

I found this problem to be very counter-intuitive. The solutions I thought would be fast were slow. The solutions that looked slow were fast! It seems that Python’s string creation with the multiply operator and string comparisons are highly optimised.

The Answer 11

2 people think this answer is useful

This function runs very quickly (tested and it’s over 3 times faster than fastest solution here on strings with over 100k characters and the difference gets bigger the longer the repeating pattern is). It tries to minimise the number of comparisons needed to get the answer:

def repeats(string):
    n = len(string)
    tried = set([])
    best = None
    nums = [i for i in  xrange(2, int(n**0.5) + 1) if n % i == 0]
    nums = [n/i for i in nums if n/i!=i] + list(reversed(nums)) + [1]
    for s in nums:
        if all(t%s for t in tried):
            print 'Trying repeating string of length:', s
            if string[:s]*(n/s)==string:
                best = s
    if best:
        return string[:best]

Note that for example for string of length 8 it checks only fragment of size 4 and it does not have to test further because pattern of length 1 or 2 would result in repeating pattern of length 4:

>>> repeats('12345678')
Trying repeating string of length: 4

# for this one we need only 2 checks 
>>> repeats('1234567812345678')
Trying repeating string of length: 8
Trying repeating string of length: 4

The Answer 12

0 people think this answer is useful

In David Zhang’s answer if we have some sort of circular buffer this will not work: principal_period('6210045662100456621004566210045662100456621') due to the starting 621, where I would have liked it to spit out: 00456621.

Extending his solution we can use the following:

def principal_period(s):
    for j in range(int(len(s)/2)):
        idx = (s[j:]+s[j:]).find(s[j:], 1, -1)
        if idx != -1:
            # Make sure that the first substring is part of pattern
            if s[:j] == s[j:][:idx][-j:]:

    return None if idx == -1 else s[j:][:idx]

>>> '00456621'

The Answer 13

-1 people think this answer is useful

Here is the code in python that checks for repetition of sub string in the main string given by the user.

print "Enter a string...."
#mainstring = String given by user
    print "Invalid string"
#charlist = Character list of mainstring
print "Length of your string :",len(mainstring)
for i in range(0,len(mainstring)):
    count = 0
    for j in splitlist:
        if j =='':
    if count == len(splitlist):
if count == len(splitlist):
    if count == 2:
        print "No repeating Sub-String found in string %r"%(mainstring)

        print "Sub-String %r repeats in string %r"%(strarr,mainstring)
else :
    print "No repeating Sub-String found in string %r"%(mainstring)



Output :

Length of your string : 40

Sub-String ‘00456621’ repeats in string ‘0045662100456621004566210045662100456621’

Input :



Length of your string : 30

No repeating Sub-String found in string ‘004608294930875576036866359447’

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