Detecting syllables in a word

I need to find a fairly efficient way to detect syllables in a word. E.g.,

Invisible -> in-vi-sib-le

There are some syllabification rules that could be used:

V
CV
VC
CVC
CCV
CCCV
CVCC

*where V is a vowel and C is a consonant.
E.g.,

Pronunciation (5 Pro-nun-ci-a-tion; CV-CVC-CV-V-CVC)

I've tried few methods, among which were using regex (which helps only if you want to count syllables) or hard coded rule definition (a brute force approach which proves to be very inefficient) and finally using a finite state automata (which did not result with anything useful).

The purpose of my application is to create a dictionary of all syllables in a given language. This dictionary will later be used for spell checking applications (using Bayesian classifiers) and text to speech synthesis.

I would appreciate if one could give me tips on an alternate way to solve this problem besides my previous approaches.

I work in Java, but any tip in C/C++, C#, Python, Perl... would work for me.

14 Answers
14

Read about the TeX approach to this problem for the purposes of hyphenation. Especially see Frank Liang's thesis dissertation Word Hy-phen-a-tion by Com-put-er. His algorithm is very accurate, and then includes a small exceptions dictionary for cases where the algorithm does not work.

I stumbled across this page looking for the same thing, and found a few implementations of the Liang paper here:
https://github.com/mnater/hyphenator

That is unless you're the type that enjoys reading a 60 page thesis instead of adapting freely available code for non-unique problem. :)

Here is a solution using NLTK:

from nltk.corpus import cmudict
d = cmudict.dict()
def nsyl(word):
return [len(list(y for y in x if y[-1].isdigit())) for x in d[word.lower()]]


I'm trying to tackle this problem for a program that will calculate the flesch-kincaid and flesch reading score of a block of text. My algorithm uses what I found on this website: http://www.howmanysyllables.com/howtocountsyllables.html and it gets reasonably close. It still has trouble on complicated words like invisible and hyphenation, but I've found it gets in the ballpark for my purposes.

It has the upside of being easy to implement. I found the "es" can be either syllabic or not. It's a gamble, but I decided to remove the es in my algorithm.

private int CountSyllables(string word)
{
char vowels = { 'a', 'e', 'i', 'o', 'u', 'y' };
string currentWord = word;
int numVowels = 0;
bool lastWasVowel = false;
foreach (char wc in currentWord)
{
bool foundVowel = false;
foreach (char v in vowels)
{
//don't count diphthongs
if (v == wc && lastWasVowel)
{
foundVowel = true;
lastWasVowel = true;
break;
}
else if (v == wc && !lastWasVowel)
{
numVowels++;
foundVowel = true;
lastWasVowel = true;
break;
}
}

//if full cycle and no vowel found, set lastWasVowel to false;
if (!foundVowel)
lastWasVowel = false;
}
//remove es, it's _usually? silent
if (currentWord.Length > 2 &&
currentWord.Substring(currentWord.Length - 2) == "es")
numVowels--;
// remove silent e
else if (currentWord.Length > 1 &&
currentWord.Substring(currentWord.Length - 1) == "e")
numVowels--;

return numVowels;
}


This is a particularly difficult problem which is not completely solved by the LaTeX hyphenation algorithm. A good summary of some available methods and the challenges involved can be found in the paper Evaluating Automatic Syllabification Algorithms for English (Marchand, Adsett, and Damper 2007).

Perl has Lingua::Phonology::Syllable module. You might try that, or try looking into its algorithm. I saw a few other older modules there, too.

I don't understand why a regular expression gives you only a count of syllables. You should be able to get the syllables themselves using capture parentheses. Assuming you can construct a regular expression that works, that is.

Thanks Joe Basirico, for sharing your quick and dirty implementation in C#. I've used the big libraries, and they work, but they're usually a bit slow, and for quick projects, your method works fine.

Here is your code in Java, along with test cases:

public static int countSyllables(String word)
{
char vowels = { 'a', 'e', 'i', 'o', 'u', 'y' };
char currentWord = word.toCharArray();
int numVowels = 0;
boolean lastWasVowel = false;
for (char wc : currentWord) {
boolean foundVowel = false;
for (char v : vowels)
{
//don't count diphthongs
if ((v == wc) && lastWasVowel)
{
foundVowel = true;
lastWasVowel = true;
break;
}
else if (v == wc && !lastWasVowel)
{
numVowels++;
foundVowel = true;
lastWasVowel = true;
break;
}
}
// If full cycle and no vowel found, set lastWasVowel to false;
if (!foundVowel)
lastWasVowel = false;
}
// Remove es, it's _usually? silent
if (word.length() > 2 &&
word.substring(word.length() - 2) == "es")
numVowels--;
// remove silent e
else if (word.length() > 1 &&
word.substring(word.length() - 1) == "e")
numVowels--;
return numVowels;
}

public static void main(String args) {
String txt = "what";
System.out.println("txt="+txt+" countSyllables="+countSyllables(txt));
txt = "super";
System.out.println("txt="+txt+" countSyllables="+countSyllables(txt));
txt = "Maryland";
System.out.println("txt="+txt+" countSyllables="+countSyllables(txt));
txt = "American";
System.out.println("txt="+txt+" countSyllables="+countSyllables(txt));
txt = "disenfranchized";
System.out.println("txt="+txt+" countSyllables="+countSyllables(txt));
txt = "Sophia";
System.out.println("txt="+txt+" countSyllables="+countSyllables(txt));
}


The result was as expected (it works good enough for Flesch-Kincaid):

txt=what countSyllables=1
txt=super countSyllables=2
txt=Maryland countSyllables=3
txt=American countSyllables=3
txt=disenfranchized countSyllables=5
txt=Sophia countSyllables=2


Bumping @Tihamer and @joe-basirico. Very useful function, not perfect, but good for most small-to-medium projects. Joe, I have re-written an implementation of your code in Python:

def countSyllables(word):
vowels = "aeiouy"
numVowels = 0
lastWasVowel = False
for wc in word:
foundVowel = False
for v in vowels:
if v == wc:
if not lastWasVowel: numVowels+=1 #don't count diphthongs
foundVowel = lastWasVowel = True
break
if not foundVowel: #If full cycle and no vowel found, set lastWasVowel to false
lastWasVowel = False
if len(word) > 2 and word[-2:] == "es": #Remove es - it's "usually" silent (?)
numVowels-=1
elif len(word) > 1 and word[-1:] == "e": #remove silent e
numVowels-=1
return numVowels


Hope someone finds this useful!

Today I found this Java implementation of Frank Liang's hyphenation algorithmn with pattern for English or German, which works quite well and is available on Maven Central.

Cave: It is important to remove the last lines of the .tex pattern files, because otherwise those files can not be loaded with the current version on Maven Central.

.tex

To load and use the hyphenator, you can use the following Java code snippet. texTable is the name of the .tex files containing the needed patterns. Those files are available on the project github site.

hyphenator

texTable

.tex

private Hyphenator createHyphenator(String texTable) {
Hyphenator hyphenator = new Hyphenator();
hyphenator.setErrorHandler(new ErrorHandler() {
public void debug(String guard, String s) {
logger.debug("{},{}", guard, s);
}

public void info(String s) {
logger.info(s);
}

public void warning(String s) {
logger.warn("WARNING: " + s);
}

public void error(String s) {
logger.error("ERROR: " + s);
}

public void exception(String s, Exception e) {
logger.error("EXCEPTION: " + s, e);
}

public boolean isDebugged(String guard) {
return false;
}
});

BufferedReader table = null;

try {
table = new BufferedReader(new InputStreamReader(Thread.currentThread().getContextClassLoader()
.getResourceAsStream((texTable)), Charset.forName("UTF-8")));
hyphenator.loadTable(table);
} catch (Utf8TexParser.TexParserException e) {
logger.error("error loading hyphenation table: {}", e.getLocalizedMessage(), e);
throw new RuntimeException("Failed to load hyphenation table", e);
} finally {
if (table != null) {
try {
table.close();
} catch (IOException e) {
logger.error("Closing hyphenation table failed", e);
}
}
}

return hyphenator;
}


Afterwards the Hyphenator is ready to use. To detect syllables, the basic idea is to split the term at the provided hyphens.

Hyphenator

String hyphenedTerm = hyphenator.hyphenate(term);

String hyphens = hyphenedTerm.split("u00AD");

int syllables = hyphens.length;


You need to split on "u00AD", since the API does not return a normal "-".

"u00AD

"-"

This approach outperforms the answer of Joe Basirico, since it supports many different languages and detects German hyphenation more accurate.

Why calculate it? Every online dictionary has this info. http://dictionary.reference.com/browse/invisible
in·vis·i·ble

Thank you @joe-basirico and @tihamer. I have ported @tihamer's code to Lua 5.1, 5.2 and luajit 2 (most likely will run on other versions of lua as well):

countsyllables.lua

countsyllables.lua

function CountSyllables(word)
local vowels = { 'a','e','i','o','u','y' }
local numVowels = 0
local lastWasVowel = false

for i = 1, #word do
local wc = string.sub(word,i,i)
local foundVowel = false;
for _,v in pairs(vowels) do
if (v == string.lower(wc) and lastWasVowel) then
foundVowel = true
lastWasVowel = true
elseif (v == string.lower(wc) and not lastWasVowel) then
numVowels = numVowels + 1
foundVowel = true
lastWasVowel = true
end
end

if not foundVowel then
lastWasVowel = false
end
end

if string.len(word) > 2 and
string.sub(word,string.len(word) - 1) == "es" then
numVowels = numVowels - 1
elseif string.len(word) > 1 and
string.sub(word,string.len(word)) == "e" then
numVowels = numVowels - 1
end

return numVowels
end


And some fun tests to confirm it works (as much as it's supposed to):

countsyllables.tests.lua

countsyllables.tests.lua

require "countsyllables"

tests = {
{ word = "what", syll = 1 },
{ word = "super", syll = 2 },
{ word = "Maryland", syll = 3},
{ word = "American", syll = 4},
{ word = "disenfranchized", syll = 5},
{ word = "Sophia", syll = 2},
{ word = "End", syll = 1},
{ word = "I", syll = 1},
{ word = "release", syll = 2},
{ word = "same", syll = 1},
}

for _,test in pairs(tests) do
local resultSyll = CountSyllables(test.word)
assert(resultSyll == test.syll,
"Word: "..test.word.."n"..
"Expected: "..test.syll.."n"..
"Result: "..resultSyll)
end

print("Tests passed.")


I could not find an adequate way to count syllables, so I designed a method myself.

You can view my method here: https://stackoverflow.com/a/32784041/2734752

I use a combination of a dictionary and algorithm method to count syllables.

You can view my library here: https://github.com/troywatson/Lawrence-Style-Checker

I just tested my algorithm and had a 99.4% strike rate!

Lawrence lawrence = new Lawrence();

System.out.println(lawrence.getSyllable("hyphenation"));
System.out.println(lawrence.getSyllable("computer"));


Output:

4
3


I ran into this exact same issue a little while ago.

I ended up using the CMU Pronunciation Dictionary for quick and accurate lookups of most words. For words not in the dictionary, I fell back to a machine learning model that's ~98% accurate at predicting syllable counts.

I wrapped the whole thing up in an easy-to-use python module here: https://github.com/repp/big-phoney

Install:
pip install big-phoney

pip install big-phoney

Count Syllables:

from big_phoney import BigPhoney
phoney = BigPhoney()
phoney.count_syllables('triceratops') # --> 4


If you're not using Python and you want to try the ML-model-based approach, I did a pretty detailed write up on how the syllable counting model works on Kaggle.

I used jsoup to do this once. Here's a sample syllable parser:

public String syllables(String text){
String url = "https://www.merriam-webster.com/dictionary/" + text;
String relHref;
try{
Document doc = Jsoup.connect(url).get();
Element link = doc.getElementsByClass("word-syllables").first();
if(link == null){return new String{text};}
relHref = link.html();
}catch(IOException e){
relHref = text;
}
String syl = relHref.split("·");
return syl;
}


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.