Python中的音节计数

我需要写一个函数来读取单词中的音节（例如，HAIRY是2个音节）。我的代码显示在底部，我相信它在大多数情况下都能工作，因为它适用于我所做的所有其他测试，但不适用于“毛茸茸”的测试，因为它只能读一个音节

def syllable_count(word):
    count = 0
    vowels = "aeiouy"
    if word[0] in vowels:
        count += 1
    for index in range(1, len(word)):
        if word[index] in vowels and word[index - 1] not in vowels:
            count += 1
            if word.endswith("e"):
                count -= 1
    if count == 0:
        count += 1
    return count

---

试验

print(syllable_count("HAIRY"))

预期：2

---

Received:1

问题是您给它一个大写字符串，但您只比较小写值。这可以通过在函数开头添加

word=word.lower（）

来解决

def syllable_count(word):
    word = word.lower()
    count = 0
    vowels = "aeiouy"
    if word[0] in vowels:
        count += 1
    for index in range(1, len(word)):
        if word[index] in vowels and word[index - 1] not in vowels:
            count += 1
    if word.endswith("e"):
        count -= 1
    if count == 0:
        count += 1
    return count

print(syllable_count('HAIRY'))  # prints "2"

---

当以小写形式给出任何内容时，您的代码似乎运行良好。但是，如果您用大写字母传递一个单词，它将始终返回1。这是因为你测试的是“aeiou”，而不是“aeiouAEIOU”。 你可以用几种方法解决这个问题

例1：

vowels = "aeiouyAEIOUY"

例2：

print(syllable_count("HAIRY".lower()))

示例3：在“音节计数”函数的开头添加这行代码

word = word.lower()

---

音节检测有某些规则，您可以从网站上查看这些规则：

下面是python代码：

import re
def sylco(word) :
    word = word.lower()

    # exception_add are words that need extra syllables
    # exception_del are words that need less syllables

    exception_add = ['serious','crucial']
    exception_del = ['fortunately','unfortunately']

    co_one = ['cool','coach','coat','coal','count','coin','coarse','coup','coif','cook','coign','coiffe','coof','court']
    co_two = ['coapt','coed','coinci']

    pre_one = ['preach']

    syls = 0 #added syllable number
    disc = 0 #discarded syllable number

    #1) if letters < 3 : return 1
    if len(word) <= 3 :
        syls = 1
        return syls

    #2) if doesn't end with "ted" or "tes" or "ses" or "ied" or "ies", discard "es" and "ed" at the end.
    # if it has only 1 vowel or 1 set of consecutive vowels, discard. (like "speed", "fled" etc.)

    if word[-2:] == "es" or word[-2:] == "ed" :
        doubleAndtripple_1 = len(re.findall(r'[eaoui][eaoui]',word))
        if doubleAndtripple_1 > 1 or len(re.findall(r'[eaoui][^eaoui]',word)) > 1 :
            if word[-3:] == "ted" or word[-3:] == "tes" or word[-3:] == "ses" or word[-3:] == "ied" or word[-3:] == "ies" :
                pass
            else :
                disc+=1

    #3) discard trailing "e", except where ending is "le"  

    le_except = ['whole','mobile','pole','male','female','hale','pale','tale','sale','aisle','whale','while']

    if word[-1:] == "e" :
        if word[-2:] == "le" and word not in le_except :
            pass

        else :
            disc+=1

    #4) check if consecutive vowels exists, triplets or pairs, count them as one.

    doubleAndtripple = len(re.findall(r'[eaoui][eaoui]',word))
    tripple = len(re.findall(r'[eaoui][eaoui][eaoui]',word))
    disc+=doubleAndtripple + tripple

    #5) count remaining vowels in word.
    numVowels = len(re.findall(r'[eaoui]',word))

    #6) add one if starts with "mc"
    if word[:2] == "mc" :
        syls+=1

    #7) add one if ends with "y" but is not surrouned by vowel
    if word[-1:] == "y" and word[-2] not in "aeoui" :
        syls +=1

    #8) add one if "y" is surrounded by non-vowels and is not in the last word.

    for i,j in enumerate(word) :
        if j == "y" :
            if (i != 0) and (i != len(word)-1) :
                if word[i-1] not in "aeoui" and word[i+1] not in "aeoui" :
                    syls+=1

    #9) if starts with "tri-" or "bi-" and is followed by a vowel, add one.

    if word[:3] == "tri" and word[3] in "aeoui" :
        syls+=1

    if word[:2] == "bi" and word[2] in "aeoui" :
        syls+=1

    #10) if ends with "-ian", should be counted as two syllables, except for "-tian" and "-cian"

    if word[-3:] == "ian" : 
    #and (word[-4:] != "cian" or word[-4:] != "tian") :
        if word[-4:] == "cian" or word[-4:] == "tian" :
            pass
        else :
            syls+=1

    #11) if starts with "co-" and is followed by a vowel, check if exists in the double syllable dictionary, if not, check if in single dictionary and act accordingly.

    if word[:2] == "co" and word[2] in 'eaoui' :

        if word[:4] in co_two or word[:5] in co_two or word[:6] in co_two :
            syls+=1
        elif word[:4] in co_one or word[:5] in co_one or word[:6] in co_one :
            pass
        else :
            syls+=1

    #12) if starts with "pre-" and is followed by a vowel, check if exists in the double syllable dictionary, if not, check if in single dictionary and act accordingly.

    if word[:3] == "pre" and word[3] in 'eaoui' :
        if word[:6] in pre_one :
            pass
        else :
            syls+=1

    #13) check for "-n't" and cross match with dictionary to add syllable.

    negative = ["doesn't", "isn't", "shouldn't", "couldn't","wouldn't"]

    if word[-3:] == "n't" :
        if word in negative :
            syls+=1
        else :
            pass   

    #14) Handling the exceptional words.

    if word in exception_del :
        disc+=1

    if word in exception_add :
        syls+=1     

    # calculate the output
    return numVowels - disc + syls

重新导入
def sylco（word）：
word=word.lower（）
#例外情况是需要额外音节的单词
#例外情况是需要较少音节的单词
异常添加=[“严重”、“关键”]
例外情况=幸运、不幸
COU one=[“酷”，“coach”，“coat”，“coal”，“count”，“coin”，“cool”，“cout”，“coif”，“coif”，“cook”，“coign”，“coiffe”，“coof”，“coof”，“court”]
cou_two=['coapt'，'coed'，'coinci']
前一节=[“布道”]
syls=0#增加音节数
disc=0#丢弃的音节数
#1） 如果字母<3：返回1
如果len（word）您也可以使用lambda映射

fun_check = lambda x: 1 if x in ["a","i","e","o","u","y","A","E","I","O","U","y"] else 0
sum(list(map(fun_check,"your_string")))

单行

    sum(list(map(lambda x: 1 if x in ["a","i","e","o","u","y","A","E","I","O","U","y"] else 0,"your string")))

多个示例失败，“announcement”返回4（应该是3），“columbia”返回3（应该是4），…更多失败示例：“trued”，“course”，对于“tickle”，函数返回1，但应该是2。我认为添加这个应该会有所帮助-word.endswith（“e”）而不是word.endswith（“le”）