How to split text without spaces into list of words


Input: "tableapplechairtablecupboard..." many words

What would be an efficient algorithm to split such text to the list of words and get:

Output: ["table", "apple", "chair", "table", ["cupboard", ["cup", "board"]], ...]

First thing that cames to mind is to go through all possible words (starting with first letter) and find the longest word possible, continue from position=word_position+len(word)

We have a list of all possible words.
Word “cupboard” can be “cup” and “board”, select longest.
Language: python, but main thing is the algorithm itself.

Asked By: Sergey



It seems like fairly mundane backtracking will do. Start at the beggining of the string. Scan right until you have a word. Then, call the function on the rest of the string. Function returns “false” if it scans all the way to the right without recognizing a word. Otherwise, returns the word it found and the list of words returned by the recursive call.

Example: “tableapple”. Finds “tab”, then “leap”, but no word in “ple”. No other word in “leapple”. Finds “table”, then “app”. “le” not a word, so tries apple, recognizes, returns.

To get longest possible, keep going, only emitting (rather than returning) correct solutions; then, choose the optimal one by any criterion you choose (maxmax, minmax, average, etc.)

Answered By: Patrick87

You need to identify your vocabulary – perhaps any free word list will do.

Once done, use that vocabulary to build a suffix tree, and match your stream of input against that:

Answered By: Marcin

Using a trie data structure, which holds the list of possible words, it would not be too complicated to do the following:

  1. Advance pointer (in the concatenated string)
  2. Lookup and store the corresponding node in the trie
  3. If the trie node has children (e.g. there are longer words), go to 1.
  4. If the node reached has no children, a longest word match happened; add the word (stored in the node or just concatenated during trie traversal) to the result list, reset the pointer in the trie (or reset the reference), and start over
Answered By: miku

If you precompile the wordlist into a DFA (which will be very slow), then the time it takes to match an input will be proportional to the length of the string (in fact, only a little slower than just iterating over the string).

This is effectively a more general version of the trie algorithm which was mentioned earlier. I only mention it for completeless — as of yet, there’s no DFA implementation you can just use. RE2 would work, but I don’t know if the Python bindings let you tune how large you allow a DFA to be before it just throws away the compiled DFA data and does NFA search.

Answered By: Devin Jeanpierre

Here is solution using recursive search:

def find_words(instring, prefix = '', words = None):
    if not instring:
        return []
    if words is None:
        words = set()
        with open('/usr/share/dict/words') as f:
            for line in f:
    if (not prefix) and (instring in words):
        return [instring]
    prefix, suffix = prefix + instring[0], instring[1:]
    solutions = []
    # Case 1: prefix in solution
    if prefix in words:
            solutions.append([prefix] + find_words(suffix, '', words))
        except ValueError:
    # Case 2: prefix not in solution
        solutions.append(find_words(suffix, prefix, words))
    except ValueError:
    if solutions:
        return sorted(solutions,
                      key = lambda solution: [len(word) for word in solution],
                      reverse = True)[0]
        raise ValueError('no solution')

print(find_words('tableprechaun', words = set(['tab', 'table', 'leprechaun'])))


['table', 'apple', 'chair', 'table', 'cupboard']
['tab', 'leprechaun']
Answered By: unutbu

A naive algorithm won’t give good results when applied to real-world data. Here is a 20-line algorithm that exploits relative word frequency to give accurate results for real-word text.

(If you want an answer to your original question which does not use word frequency, you need to refine what exactly is meant by “longest word”: is it better to have a 20-letter word and ten 3-letter words, or is it better to have five 10-letter words? Once you settle on a precise definition, you just have to change the line defining wordcost to reflect the intended meaning.)

The idea

The best way to proceed is to model the distribution of the output. A good first approximation is to assume all words are independently distributed. Then you only need to know the relative frequency of all words. It is reasonable to assume that they follow Zipf’s law, that is the word with rank n in the list of words has probability roughly 1/(n log N) where N is the number of words in the dictionary.

Once you have fixed the model, you can use dynamic programming to infer the position of the spaces. The most likely sentence is the one that maximizes the product of the probability of each individual word, and it’s easy to compute it with dynamic programming. Instead of directly using the probability we use a cost defined as the logarithm of the inverse of the probability to avoid overflows.

The code

from math import log

# Build a cost dictionary, assuming Zipf's law and cost = -math.log(probability).
words = open("words-by-frequency.txt").read().split()
wordcost = dict((k, log((i+1)*log(len(words)))) for i,k in enumerate(words))
maxword = max(len(x) for x in words)

def infer_spaces(s):
    """Uses dynamic programming to infer the location of spaces in a string
    without spaces."""

    # Find the best match for the i first characters, assuming cost has
    # been built for the i-1 first characters.
    # Returns a pair (match_cost, match_length).
    def best_match(i):
        candidates = enumerate(reversed(cost[max(0, i-maxword):i]))
        return min((c + wordcost.get(s[i-k-1:i], 9e999), k+1) for k,c in candidates)

    # Build the cost array.
    cost = [0]
    for i in range(1,len(s)+1):
        c,k = best_match(i)

    # Backtrack to recover the minimal-cost string.
    out = []
    i = len(s)
    while i>0:
        c,k = best_match(i)
        assert c == cost[i]
        i -= k

    return " ".join(reversed(out))

which you can use with

s = 'thumbgreenappleactiveassignmentweeklymetaphor'

The results

I am using this quick-and-dirty 125k-word dictionary I put together from a small subset of Wikipedia.

Before: thumbgreenappleactiveassignmentweeklymetaphor.
After: thumb green apple active assignment weekly metaphor.

Before: thereismassesoftextinformationofpeoplescommentswhichisparsedfromhtmlbuttherearen

After: there is masses of text information of peoples comments which is parsed from html but there are no delimited characters in them for example thumb green apple active assignment weekly metaphor apparently there are thumb green apple etc in the string i also have a large dictionary to query whether the word is reasonable so what s the fastest way of extraction thx a lot.

Before: itwasadarkandstormynighttherainfellintorrentsexceptatoccasionalintervalswhenitwascheckedbyaviolentgustofwindwhichsweptupthestreetsforitisinlondonthatoursceneliesrattlingalongthehousetopsandfiercelyagitatingthescantyflameofthelampsthatstruggledagainstthedarkness.

After: it was a dark and stormy night the rain fell in torrents except at occasional intervals when it was checked by a violent gust of wind which swept up the streets for it is in london that our scene lies rattling along the housetops and fiercely agitating the scanty flame of the lamps that struggled against the darkness.

As you can see it is essentially flawless. The most important part is to make sure your word list was trained to a corpus similar to what you will actually encounter, otherwise the results will be very bad.


The implementation consumes a linear amount of time and memory, so it is reasonably efficient. If you need further speedups, you can build a suffix tree from the word list to reduce the size of the set of candidates.

If you need to process a very large consecutive string it would be reasonable to split the string to avoid excessive memory usage. For example you could process the text in blocks of 10000 characters plus a margin of 1000 characters on either side to avoid boundary effects. This will keep memory usage to a minimum and will have almost certainly no effect on the quality.

Answered By: Generic Human

Unutbu’s solution was quite close but I find the code difficult to read, and it didn’t yield the expected result. Generic Human’s solution has the drawback that it needs word frequencies. Not appropriate for all use case.

Here’s a simple solution using a Divide and Conquer algorithm.

  1. It tries to minimize the number of words E.g. find_words('cupboard') will return ['cupboard'] rather than ['cup', 'board'] (assuming that cupboard, cup and board are in the dictionnary)
  2. The optimal solution is not unique, the implementation below returns a solution. find_words('charactersin') could return ['characters', 'in'] or maybe it will return ['character', 'sin'] (as seen below). You could quite easily modify the algorithm to return all optimal solutions.
  3. In this implementation solutions are memoized so that it runs in a reasonable time.

The code:

words = set()
with open('/usr/share/dict/words') as f:
    for line in f:

solutions = {}
def find_words(instring):
    # First check if instring is in the dictionnary
    if instring in words:
        return [instring]
    # No... But maybe it's a result we already computed
    if instring in solutions:
        return solutions[instring]
    # Nope. Try to split the string at all position to recursively search for results
    best_solution = None
    for i in range(1, len(instring) - 1):
        part1 = find_words(instring[:i])
        part2 = find_words(instring[i:])
        # Both parts MUST have a solution
        if part1 is None or part2 is None:
        solution = part1 + part2
        # Is the solution found "better" than the previous one?
        if best_solution is None or len(solution) < len(best_solution):
            best_solution = solution
    # Remember (memoize) this solution to avoid having to recompute it
    solutions[instring] = best_solution
    return best_solution

This will take about about 5sec on my 3GHz machine:

result = find_words("thereismassesoftextinformationofpeoplescommentswhichisparsedfromhtmlbuttherearenodelimitedcharactersinthemforexamplethumbgreenappleactiveassignmentweeklymetaphorapparentlytherearethumbgreenappleetcinthestringialsohavealargedictionarytoquerywhetherthewordisreasonablesowhatsthefastestwayofextractionthxalot")
assert(result is not None)
print ' '.join(result)

the reis masses of text information of peoples comments which is parsed from h t m l but there are no delimited character sin them for example thumb green apple active assignment weekly metaphor apparently there are thumb green apple e t c in the string i also have a large dictionary to query whether the word is reasonable so whats the fastest way of extraction t h x a lot

Answered By: Rems

Based on unutbu’s solution I’ve implemented a Java version:

private static List<String> splitWordWithoutSpaces(String instring, String suffix) {
    if(isAWord(instring)) {
        if(suffix.length() > 0) {
            List<String> rest = splitWordWithoutSpaces(suffix, "");
            if(rest.size() > 0) {
                List<String> solutions = new LinkedList<>();
                return solutions;
        } else {
            List<String> solutions = new LinkedList<>();
            return solutions;

    if(instring.length() > 1) {
        String newString = instring.substring(0, instring.length()-1);
        suffix = instring.charAt(instring.length()-1) + suffix;
        List<String> rest = splitWordWithoutSpaces(newString, suffix);
        return rest;
    return Collections.EMPTY_LIST;

Input: "tableapplechairtablecupboard"

Output: [table, apple, chair, table, cupboard]

Input: "tableprechaun"

Output: [tab, leprechaun]

The answer by Generic Human is great. But the best implementation of this I’ve ever seen was written Peter Norvig himself in his book ‘Beautiful Data’.

Before I paste his code, let me expand on why Norvig’s method is more accurate (although a little slower and longer in terms of code).

  1. The data is a bit better – both in terms of size and in terms of precision (he uses a word count rather than a simple ranking)
  2. More importantly, it’s the logic behind n-grams that really makes the approach so accurate.

The example he provides in his book is the problem of splitting a string ‘sitdown’. Now a non-bigram method of string split would consider p(‘sit’) * p (‘down’), and if this less than the p(‘sitdown’) – which will be the case quite often – it will NOT split it, but we’d want it to (most of the time).

However when you have the bigram model you could value p(‘sit down’) as a bigram vs p(‘sitdown’) and the former wins. Basically, if you don’t use bigrams, it treats the probability of the words you’re splitting as independent, which is not the case, some words are more likely to appear one after the other. Unfortunately those are also the words that are often stuck together in a lot of instances and confuses the splitter.

Here’s the link to the data (it’s data for 3 separate problems and segmentation is only one. Please read the chapter for details):

and here’s the link to the code:

These links have been up a while, but I’ll copy paste the segmentation part of the code here anyway

import re, string, random, glob, operator, heapq
from collections import defaultdict
from math import log10

def memo(f):
    "Memoize function f."
    table = {}
    def fmemo(*args):
        if args not in table:
            table[args] = f(*args)
        return table[args]
    fmemo.memo = table
    return fmemo

def test(verbose=None):
    """Run some tests, taken from the chapter.
    Since the hillclimbing algorithm is randomized, some tests may fail."""
    import doctest
    print 'Running tests...'
    doctest.testfile('ngrams-test.txt', verbose=verbose)

################ Word Segmentation (p. 223)

def segment(text):
    "Return a list of words that is the best segmentation of text."
    if not text: return []
    candidates = ([first]+segment(rem) for first,rem in splits(text))
    return max(candidates, key=Pwords)

def splits(text, L=20):
    "Return a list of all possible (first, rem) pairs, len(first)<=L."
    return [(text[:i+1], text[i+1:]) 
            for i in range(min(len(text), L))]

def Pwords(words): 
    "The Naive Bayes probability of a sequence of words."
    return product(Pw(w) for w in words)

#### Support functions (p. 224)

def product(nums):
    "Return the product of a sequence of numbers."
    return reduce(operator.mul, nums, 1)

class Pdist(dict):
    "A probability distribution estimated from counts in datafile."
    def __init__(self, data=[], N=None, missingfn=None):
        for key,count in data:
            self[key] = self.get(key, 0) + int(count)
        self.N = float(N or sum(self.itervalues()))
        self.missingfn = missingfn or (lambda k, N: 1./N)
    def __call__(self, key): 
        if key in self: return self[key]/self.N  
        else: return self.missingfn(key, self.N)

def datafile(name, sep='t'):
    "Read key,value pairs from file."
    for line in file(name):
        yield line.split(sep)

def avoid_long_words(key, N):
    "Estimate the probability of an unknown word."
    return 10./(N * 10**len(key))

N = 1024908267229 ## Number of tokens

Pw  = Pdist(datafile('count_1w.txt'), N, avoid_long_words)

#### segment2: second version, with bigram counts, (p. 226-227)

def cPw(word, prev):
    "Conditional probability of word, given previous word."
        return P2w[prev + ' ' + word]/float(Pw[prev])
    except KeyError:
        return Pw(word)

P2w = Pdist(datafile('count_2w.txt'), N)

def segment2(text, prev='<S>'): 
    "Return (log P(words), words), where words is the best segmentation." 
    if not text: return 0.0, [] 
    candidates = [combine(log10(cPw(first, prev)), first, segment2(rem, first)) 
                  for first,rem in splits(text)] 
    return max(candidates) 

def combine(Pfirst, first, (Prem, rem)): 
    "Combine first and rem results into one (probability, words) pair." 
    return Pfirst+Prem, [first]+rem 

Based on the excellent work in the top answer, I’ve created a pip package for easy use.

>>> import wordninja
>>> wordninja.split('derekanderson')
['derek', 'anderson']

To install, run pip install wordninja.

The only differences are minor. This returns a list rather than a str, it works in python3, it includes the word list and properly splits even if there are non-alpha chars (like underscores, dashes, etc).

Thanks again to Generic Human!

Answered By: keredson

For German language there is CharSplit which uses machine learning and works pretty good for strings of a few words.

Answered By: Karl Adler

Expanding on @miku’s suggestion to use a Trie, an append-only Trie is relatively straight-forward to implement in python:

class Node:
    def __init__(self, is_word=False):
        self.children = {}
        self.is_word = is_word

class TrieDictionary:
    def __init__(self, words=tuple()):
        self.root = Node()
        for word in words:

    def add(self, word):
        node = self.root
        for c in word:
            node = node.children.setdefault(c, Node())
        node.is_word = True

    def lookup(self, word, from_node=None):
        node = self.root if from_node is None else from_node
        for c in word:
                node = node.children[c]
            except KeyError:
                return None

        return node

We can then build a Trie-based dictionary from a set of words:

dictionary = {"a", "pea", "nut", "peanut", "but", "butt", "butte", "butter"}
trie_dictionary = TrieDictionary(words=dictionary)

Which will produce a tree that looks like this (* indicates beginning or end of a word):

* -> a*
  \-> p -> e -> a*
   \              -> n -> u -> t*
     \-> b -> u -> t*
      \             -> t*
       \                 -> e*
        \                     -> r*
          -> n -> u -> t*

We can incorporate this into a solution by combining it with a heuristic about how to choose words. For example we can prefer longer words over shorter words:

def using_trie_longest_word_heuristic(s):
    node = None
    possible_indexes = []

    # O(1) short-circuit if whole string is a word, doesn't go against longest-word wins
    if s in dictionary:
        return [ s ]

    for i in range(len(s)):
        # traverse the trie, char-wise to determine intermediate words
        node = trie_dictionary.lookup(s[i], from_node=node)

        # no more words start this way
        if node is None:
            # iterate words we have encountered from biggest to smallest
            for possible in possible_indexes[::-1]:
                # recurse to attempt to solve the remaining sub-string
                end_of_phrase = using_trie_longest_word_heuristic(s[possible+1:])

                # if we have a solution, return this word + our solution
                if end_of_phrase:
                    return [ s[:possible+1] ] + end_of_phrase

            # unsolvable

        # if this is a leaf, append the index to the possible words list
        elif node.is_word:

    # empty string OR unsolvable case 
    return []

We can use this function like this:

>>> using_trie_longest_word_heuristic("peanutbutter")
[ "peanut", "butter" ]

Because we maintain our position in the Trie as we search for longer and longer words, we traverse the trie at most once per possible solution (rather than 2 times for peanut: pea, peanut). The final short-circuit saves us from walking char-wise through the string in the worst-case.

The final result is only a handful of inspections:

'peanutbutter' - not a word, go charwise
'p' - in trie, use this node
'e' - in trie, use this node
'a' - in trie and edge, store potential word and use this node
'n' - in trie, use this node
'u' - in trie, use this node
't' - in trie and edge, store potential word and use this node
'b' - not in trie from `peanut` vector
'butter' - remainder of longest is a word

A benefit of this solution is in the fact that you know very quickly if longer words exist with a given prefix, which spares the need to exhaustively test sequence combinations against a dictionary. It also makes getting to an unsolvable answer comparatively cheap to other implementations.

The downsides of this solution are a large memory footprint for the trie and the cost of building the trie up-front.

Answered By: Matthew Story

Here is the accepted answer translated to JavaScript (requires node.js, and the file “wordninja_words.txt” from

var fs = require("fs");

var splitRegex = new RegExp("[^a-zA-Z0-9']+", "g");
var maxWordLen = 0;
var wordCost = {};

fs.readFile("./wordninja_words.txt", 'utf8', function(err, data) {
    if (err) {
        throw err;
    var words = data.split('n');
    words.forEach(function(word, index) {
        wordCost[word] = Math.log((index + 1) * Math.log(words.length));
    words.forEach(function(word) {
        if (word.length > maxWordLen)
            maxWordLen = word.length;
    splitRegex = new RegExp("[^a-zA-Z0-9']+", "g");

function split(s) {
    var list = [];
    s.split(splitRegex).forEach(function(sub) {
        _split(sub).forEach(function(word) {
    return list;
module.exports = split;

function _split(s) {
    var cost = [0];

    function best_match(i) {
        var candidates = cost.slice(Math.max(0, i - maxWordLen), i).reverse();
        var minPair = [Number.MAX_SAFE_INTEGER, 0];
        candidates.forEach(function(c, k) {
            if (wordCost[s.substring(i - k - 1, i).toLowerCase()]) {
                var ccost = c + wordCost[s.substring(i - k - 1, i).toLowerCase()];
            } else {
                var ccost = Number.MAX_SAFE_INTEGER;
            if (ccost < minPair[0]) {
                minPair = [ccost, k + 1];
        return minPair;

    for (var i = 1; i < s.length + 1; i++) {

    var out = [];
    i = s.length;
    while (i > 0) {
        var c = best_match(i)[0];
        var k = best_match(i)[1];
        if (c == cost[i])
            console.log("Alert: " + c);

        var newToken = true;
        if (s.slice(i - k, i) != "'") {
            if (out.length > 0) {
                if (out[-1] == "'s" || (Number.isInteger(s[i - 1]) && Number.isInteger(out[-1][0]))) {
                    out[-1] = s.slice(i - k, i) + out[-1];
                    newToken = false;

        if (newToken) {
            out.push(s.slice(i - k, i))

        i -= k

    return out.reverse();
Answered By: Christian Ayscue

If you have an exhaustive list of the words contained within the string:

word_list = ["table", "apple", "chair", "cupboard"]

Using list comprehension to iterate over the list to locate the word and how many times it appears.

string = "tableapplechairtablecupboard"

def split_string(string, word_list):

    return ("".join([(item + " ")*string.count(item.lower()) for item in word_list if item.lower() in string])).strip()

The function returns a string output of words in order of the list table table apple chair cupboard

Answered By: zainsharif1

Many Thanks for the help in

A small contribution of the same in Java from my side.

The public method splitContiguousWords could be embedded with the other 2 methods in the class having ninja_words.txt in same directory(or modified as per the choice of coder). And the method splitContiguousWords could be used for the purpose.

public List<String> splitContiguousWords(String sentence) {

    String splitRegex = "[^a-zA-Z0-9']+";
    Map<String, Number> wordCost = new HashMap<>();
    List<String> dictionaryWords = IOUtils.linesFromFile("ninja_words.txt",;
    double naturalLogDictionaryWordsCount = Math.log(dictionaryWords.size());
    long wordIdx = 0;
    for (String word : dictionaryWords) {
        wordCost.put(word, Math.log(++wordIdx * naturalLogDictionaryWordsCount));
    int maxWordLength = Collections.max(dictionaryWords, Comparator.comparing(String::length)).length();
    List<String> splitWords = new ArrayList<>();
    for (String partSentence : sentence.split(splitRegex)) {
        splitWords.add(split(partSentence, wordCost, maxWordLength));
    }"Split word for the sentence: {}", splitWords);
    return splitWords;

private String split(String partSentence, Map<String, Number> wordCost, int maxWordLength) {
    List<Pair<Number, Number>> cost = new ArrayList<>();
    cost.add(new Pair<>(Integer.valueOf(0), Integer.valueOf(0)));
    for (int index = 1; index < partSentence.length() + 1; index++) {
        cost.add(bestMatch(partSentence, cost, index, wordCost, maxWordLength));
    int idx = partSentence.length();
    List<String> output = new ArrayList<>();
    while (idx > 0) {
        Pair<Number, Number> candidate = bestMatch(partSentence, cost, idx, wordCost, maxWordLength);
        Number candidateCost = candidate.getKey();
        Number candidateIndexValue = candidate.getValue();
        if (candidateCost.doubleValue() != cost.get(idx).getKey().doubleValue()) {
            throw new RuntimeException("Candidate cost unmatched; This should not be the case!");
        boolean newToken = true;
        String token = partSentence.substring(idx - candidateIndexValue.intValue(), idx);
        if (token != "'" && output.size() > 0) {
            String lastWord = output.get(output.size() - 1);
            if (lastWord.equalsIgnoreCase("'s") ||
                    (Character.isDigit(partSentence.charAt(idx - 1)) && Character.isDigit(lastWord.charAt(0)))) {
                output.set(output.size() - 1, token + lastWord);
                newToken = false;
        if (newToken) {
        idx -= candidateIndexValue.intValue();
    return String.join(" ", Lists.reverse(output));

private Pair<Number, Number> bestMatch(String partSentence, List<Pair<Number, Number>> cost, int index,
                      Map<String, Number> wordCost, int maxWordLength) {
    List<Pair<Number, Number>> candidates = Lists.reverse(cost.subList(Math.max(0, index - maxWordLength), index));
    int enumerateIdx = 0;
    Pair<Number, Number> minPair = new Pair<>(Integer.MAX_VALUE, Integer.valueOf(enumerateIdx));
    for (Pair<Number, Number> pair : candidates) {
        String subsequence = partSentence.substring(index - enumerateIdx, index).toLowerCase();
        Number minCost = Integer.MAX_VALUE;
        if (wordCost.containsKey(subsequence)) {
            minCost = pair.getKey().doubleValue() + wordCost.get(subsequence).doubleValue();
        if (minCost.doubleValue() < minPair.getKey().doubleValue()) {
            minPair = new Pair<>(minCost.doubleValue(), enumerateIdx);
    return minPair;
Answered By: Arnab Das

This will help

from wordsegment import load, segment

Answered By: Emeka Boris Ama
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