我需要找到一种相当有效的方法来检测单词中的音节。例如，
隐形 -> in-vi-sib-le
有一些可以使用的音节规则：
V CV VC CVC CCV CCCV CVCC
*其中 V 是元音，C 是辅音。例如，
发音（5 Pro-nun-ci-a-tion；CV-CVC-CV-V-CVC）
我尝试了几种方法，其中使用正则表达式（仅在您想计算音节时才有帮助）或硬编码规则定义（证明非常低效的蛮力方法），最后使用有限状态自动机（确实没有任何有用的结果）。
我的应用程序的目的是创建给定语言的所有音节的字典。该词典稍后将用于拼写检查应用程序（使用贝叶斯分类器）和文本到语音合成。
除了我以前的方法之外，如果有人能给我关于解决此问题的替代方法的提示，我将不胜感激。
我在 Java 中工作，但 C/C++、C#、Python、Perl 中的任何技巧都对我有用。
阅读有关此问题的 TeX 方法以进行断字。尤其是Frank Liang 的论文 Word Hy-phen-a-tion by Computer。他的算法非常准确，然后包含一个小的异常字典，用于算法不起作用的情况。
我偶然发现这个页面寻找相同的东西，并在这里找到了 Liang 论文的一些实现： https ://github.com/mnater/hyphenator或继任者：https ://github.com/mnater/Hyphenopoly
那是除非你是喜欢阅读 60 页论文而不是为非唯一问题改编免费可用代码的类型。:)
这是使用NLTK的解决方案：
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()]]
我正在尝试通过一个程序来解决这个问题，该程序将计算一段文本的 flesch-kincaid 和 flesch 阅读分数。我的算法使用我在这个网站上找到的内容：http: //www.howmanysyllables.com/howtocountsyllables.html并且它相当接近。它仍然无法处理诸如隐形和连字符之类的复杂单词，但我发现它对我的目的来说是可行的。
它具有易于实施的优点。我发现“es”可以是音节也可以不是。这是一场赌博，但我决定删除算法中的 es。
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;
}
这是一个特别困难的问题，LaTeX 断字算法并没有完全解决。在Evaluating Automatic Syllabification Algorithms for English（Marchand、Adsett 和 Damper 2007）一文中可以找到对一些可用方法和所涉及挑战的一个很好的总结。
为什么要计算呢？每个在线词典都有此信息。http://dictionary.reference.com/browse/invisible in·vis·i·ble
碰撞@Tihamer 和@joe-basirico。非常有用的功能，并不完美，但对大多数中小型项目都很好。乔，我用 Python 重写了你的代码实现：
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
希望有人觉得这很有用！
感谢 Joe Basirico，分享了您在 C# 中快速而肮脏的实现。我使用过大型库，它们可以工作，但它们通常有点慢，对于快速项目，你的方法效果很好。
这是您的 Java 代码以及测试用例：
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));
}
结果符合预期（它对 Flesch-Kincaid 来说足够好）：
txt=what countSyllables=1
txt=super countSyllables=2
txt=Maryland countSyllables=3
txt=American countSyllables=3
txt=disenfranchized countSyllables=5
txt=Sophia countSyllables=2
不久前我遇到了同样的问题。
我最终使用CMU 发音词典快速准确地查找大多数单词。对于字典中没有的单词，我使用了一种机器学习模型，该模型在预测音节计数方面的准确率约为 98%。
我在这里将整个事情包装在一个易于使用的 python 模块中：https ://github.com/repp/big-phoney
安装：
pip install big-phoney
计数音节：
from big_phoney import BigPhoney
phoney = BigPhoney()
phoney.count_syllables('triceratops') # --> 4
如果您不使用 Python，并且想尝试基于 ML 模型的方法，我对音节计数模型如何在 Kaggle 上工作进行了非常详细的文章。
Perl 有Lingua::Phonology::Syllable模块。您可以尝试一下，或者尝试研究它的算法。我在那里也看到了其他一些较旧的模块。
我不明白为什么正则表达式只给你一个音节计数。您应该能够使用捕获括号自己获取音节。假设您可以构造一个有效的正则表达式，那就是。
今天我发现了这个带有英语或德语模式的 Frank Liang 的连字算法的 Java 实现，它工作得很好，可以在 Maven Central 上找到。
Cave：删除模式文件的最后几行很重要.tex，否则这些文件无法在 Maven Central 上以当前版本加载。
要加载和使用hyphenator，您可以使用以下 Java 代码片段。texTable是.tex包含所需模式的文件的名称。这些文件可在项目 github 站点上找到。
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;
}
之后Hyphenator就可以使用了。要检测音节，基本思想是在提供的连字符处拆分术语。
String hyphenedTerm = hyphenator.hyphenate(term);
String hyphens[] = hyphenedTerm.split("\u00AD");
int syllables = hyphens.length;
您需要拆分"\u00AD“，因为 API 不会返回正常的"-".
这种方法优于 Joe Basirico 的答案，因为它支持许多不同的语言并且更准确地检测德语连字符。
谢谢@joe-basirico 和@tihamer。我已将@tihamer 的代码移植到 Lua 5.1、5.2 和 luajit 2（很可能也会在其他版本的 lua 上运行）：
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
还有一些有趣的测试来确认它是否有效（尽可能多）：
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.")
我找不到合适的方法来计算音节，所以我自己设计了一个方法。
你可以在这里查看我的方法：https ://stackoverflow.com/a/32784041/2734752
我使用字典和算法的组合方法来计算音节。
你可以在这里查看我的图书馆：https ://github.com/troywatson/Lawrence-Style-Checker
我刚刚测试了我的算法，获得了 99.4% 的命中率！
Lawrence lawrence = new Lawrence();
System.out.println(lawrence.getSyllable("hyphenation"));
System.out.println(lawrence.getSyllable("computer"));
输出：
4
3
在进行了大量测试并尝试了连字符包之后，我根据一些示例编写了自己的包。我还尝试了与连字符字典接口的pyhyphenandpyphen包，但在许多情况下它们会产生错误的音节数量。对于这个用例来说，这个nltk包实在是太慢了。
我在 Python 中的实现是我编写的一个类的一部分，音节计数例程粘贴在下面。它有点高估了音节的数量，因为我还没有找到一种解释无声词尾的好方法。
该函数返回每个单词的音节比率，因为它用于 Flesch-Kincaid 可读性分数。这个数字不一定要准确，只要足够接近就可以估计。
在我的第 7 代 i7 CPU 上，这个函数需要 1.1-1.2 毫秒来处理一个 759 字的示例文本。
def _countSyllablesEN(self, theText):
cleanText = ""
for ch in theText:
if ch in "abcdefghijklmnopqrstuvwxyz'’":
cleanText += ch
else:
cleanText += " "
asVow = "aeiouy'’"
dExep = ("ei","ie","ua","ia","eo")
theWords = cleanText.lower().split()
allSylls = 0
for inWord in theWords:
nChar = len(inWord)
nSyll = 0
wasVow = False
wasY = False
if nChar == 0:
continue
if inWord[0] in asVow:
nSyll += 1
wasVow = True
wasY = inWord[0] == "y"
for c in range(1,nChar):
isVow = False
if inWord[c] in asVow:
nSyll += 1
isVow = True
if isVow and wasVow:
nSyll -= 1
if isVow and wasY:
nSyll -= 1
if inWord[c:c+2] in dExep:
nSyll += 1
wasVow = isVow
wasY = inWord[c] == "y"
if inWord.endswith(("e")):
nSyll -= 1
if inWord.endswith(("le","ea","io")):
nSyll += 1
if nSyll < 1:
nSyll = 1
# print("%-15s: %d" % (inWord,nSyll))
allSylls += nSyll
return allSylls/len(theWords)
你可以试试Spacy Sylables。这适用于 Python 3.9：
设置：
pip install spacy
pip install spacy_syllables
python -m spacy download en_core_web_md
代码：
import spacy
from spacy_syllables import SpacySyllables
nlp = spacy.load('en_core_web_md')
syllables = SpacySyllables(nlp)
nlp.add_pipe('syllables', after='tagger')
def spacy_syllablize(word):
token = nlp(word)[0]
return token._.syllables
for test_word in ["trampoline", "margaret", "invisible", "thought", "Pronunciation", "couldn't"]:
print(f"{test_word} -> {spacy_syllablize(test_word)}")
输出：
trampoline -> ['tram', 'po', 'line']
margaret -> ['mar', 'garet']
invisible -> ['in', 'vis', 'i', 'ble']
thought -> ['thought']
Pronunciation -> ['pro', 'nun', 'ci', 'a', 'tion']
couldn't -> ['could']
我包括一个在 R 中“可以”工作的解决方案。远非完美。
countSyllablesInWord = function(words)
{
#word = "super";
n.words = length(words);
result = list();
for(j in 1:n.words)
{
word = words[j];
vowels = c("a","e","i","o","u","y");
word.vec = strsplit(word,"")[[1]];
word.vec;
n.char = length(word.vec);
is.vowel = is.element(tolower(word.vec), vowels);
n.vowels = sum(is.vowel);
# nontrivial problem
if(n.vowels <= 1)
{
syllables = 1;
str = word;
} else {
# syllables = 0;
previous = "C";
# on average ?
str = "";
n.hyphen = 0;
for(i in 1:n.char)
{
my.char = word.vec[i];
my.vowel = is.vowel[i];
if(my.vowel)
{
if(previous == "C")
{
if(i == 1)
{
str = paste0(my.char, "-");
n.hyphen = 1 + n.hyphen;
} else {
if(i < n.char)
{
if(n.vowels > (n.hyphen + 1))
{
str = paste0(str, my.char, "-");
n.hyphen = 1 + n.hyphen;
} else {
str = paste0(str, my.char);
}
} else {
str = paste0(str, my.char);
}
}
# syllables = 1 + syllables;
previous = "V";
} else { # "VV"
# assume what ? vowel team?
str = paste0(str, my.char);
}
} else {
str = paste0(str, my.char);
previous = "C";
}
#
}
syllables = 1 + n.hyphen;
}
result[[j]] = list("syllables" = syllables, "vowels" = n.vowels, "word" = str);
}
if(n.words == 1) { result[[1]]; } else { result; }
}
以下是一些结果：
my.count = countSyllablesInWord(c("America", "beautiful", "spacious", "skies", "amber", "waves", "grain", "purple", "mountains", "majesty"));
my.count.df = data.frame(matrix(unlist(my.count), ncol=3, byrow=TRUE));
colnames(my.count.df) = names(my.count[[1]]);
my.count.df;
# syllables vowels word
# 1 4 4 A-me-ri-ca
# 2 4 5 be-auti-fu-l
# 3 3 4 spa-ci-ous
# 4 2 2 ski-es
# 5 2 2 a-mber
# 6 2 2 wa-ves
# 7 2 2 gra-in
# 8 2 2 pu-rple
# 9 3 4 mo-unta-ins
# 10 3 3 ma-je-sty
我没有意识到这是一个多大的“兔子洞”，看起来很容易。
################ hackathon #######
# https://en.wikipedia.org/wiki/Gunning_fog_index
# THIS is a CLASSIFIER PROBLEM ...
# https://stackoverflow.com/questions/405161/detecting-syllables-in-a-word
# http://www.speech.cs.cmu.edu/cgi-bin/cmudict
# http://www.syllablecount.com/syllables/
# https://enchantedlearning.com/consonantblends/index.shtml
# start.digraphs = c("bl", "br", "ch", "cl", "cr", "dr",
# "fl", "fr", "gl", "gr", "pl", "pr",
# "sc", "sh", "sk", "sl", "sm", "sn",
# "sp", "st", "sw", "th", "tr", "tw",
# "wh", "wr");
# start.trigraphs = c("sch", "scr", "shr", "sph", "spl",
# "spr", "squ", "str", "thr");
#
#
#
# end.digraphs = c("ch","sh","th","ng","dge","tch");
#
# ile
#
# farmer
# ar er
#
# vowel teams ... beaver1
#
#
# # "able"
# # http://www.abcfastphonics.com/letter-blends/blend-cial.html
# blends = c("augh", "ough", "tien", "ture", "tion", "cial", "cian",
# "ck", "ct", "dge", "dis", "ed", "ex", "ful",
# "gh", "ng", "ous", "kn", "ment", "mis", );
#
# glue = c("ld", "st", "nd", "ld", "ng", "nk",
# "lk", "lm", "lp", "lt", "ly", "mp", "nce", "nch",
# "nse", "nt", "ph", "psy", "pt", "re", )
#
#
# start.graphs = c("bl, br, ch, ck, cl, cr, dr, fl, fr, gh, gl, gr, ng, ph, pl, pr, qu, sc, sh, sk, sl, sm, sn, sp, st, sw, th, tr, tw, wh, wr");
#
# # https://mantra4changeblog.wordpress.com/2017/05/01/consonant-digraphs/
# digraphs.start = c("ch","sh","th","wh","ph","qu");
# digraphs.end = c("ch","sh","th","ng","dge","tch");
# # https://www.education.com/worksheet/article/beginning-consonant-blends/
# blends.start = c("pl", "gr", "gl", "pr",
#
# blends.end = c("lk","nk","nt",
#
#
# # https://sarahsnippets.com/wp-content/uploads/2019/07/ScreenShot2019-07-08at8.24.51PM-817x1024.png
# # Monte Mon-te
# # Sophia So-phi-a
# # American A-mer-i-can
#
# n.vowels = 0;
# for(i in 1:n.char)
# {
# my.char = word.vec[i];
#
#
#
#
#
# n.syll = 0;
# str = "";
#
# previous = "C"; # consonant vs "V" vowel
#
# for(i in 1:n.char)
# {
# my.char = word.vec[i];
#
# my.vowel = is.element(tolower(my.char), vowels);
# if(my.vowel)
# {
# n.vowels = 1 + n.vowels;
# if(previous == "C")
# {
# if(i == 1)
# {
# str = paste0(my.char, "-");
# } else {
# if(n.syll > 1)
# {
# str = paste0(str, "-", my.char);
# } else {
# str = paste0(str, my.char);
# }
# }
# n.syll = 1 + n.syll;
# previous = "V";
# }
#
# } else {
# str = paste0(str, my.char);
# previous = "C";
# }
# #
# }
#
#
#
#
## https://jzimba.blogspot.com/2017/07/an-algorithm-for-counting-syllables.html
# AIDE 1
# IDEA 3
# IDEAS 2
# IDEE 2
# IDE 1
# AIDA 2
# PROUSTIAN 3
# CHRISTIAN 3
# CLICHE 1
# HALIDE 2
# TELEPHONE 3
# TELEPHONY 4
# DUE 1
# IDEAL 2
# DEE 1
# UREA 3
# VACUO 3
# SEANCE 1
# SAILED 1
# RIBBED 1
# MOPED 1
# BLESSED 1
# AGED 1
# TOTED 2
# WARRED 1
# UNDERFED 2
# JADED 2
# INBRED 2
# BRED 1
# RED 1
# STATES 1
# TASTES 1
# TESTES 1
# UTILIZES 4
为了更好地衡量，一个简单的 kincaid 可读性函数...... syllables 是从第一个函数返回的计数列表......
由于我的函数有点偏向于更多的音节，这将给出一个夸大的可读性分数......现在这很好......如果目标是使文本更具可读性，这不是最糟糕的事情。
computeReadability = function(n.sentences, n.words, syllables=NULL)
{
n = length(syllables);
n.syllables = 0;
for(i in 1:n)
{
my.syllable = syllables[[i]];
n.syllables = my.syllable$syllables + n.syllables;
}
# Flesch Reading Ease (FRE):
FRE = 206.835 - 1.015 * (n.words/n.sentences) - 84.6 * (n.syllables/n.words);
# Flesh-Kincaid Grade Level (FKGL):
FKGL = 0.39 * (n.words/n.sentences) + 11.8 * (n.syllables/n.words) - 15.59;
# FKGL = -0.384236 * FRE - 20.7164 * (n.syllables/n.words) + 63.88355;
# FKGL = -0.13948 * FRE + 0.24843 * (n.words/n.sentences) + 13.25934;
list("FRE" = FRE, "FKGL" = FKGL);
}
我曾经使用 jsoup 来执行此操作。这是一个示例音节解析器：
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;
}