Syllable Count In Python
Question:
I need to write a function that will read syllables in a word (for example, HAIRY is 2 syllables). I have my code shown on the bottom and I’m confident it works in most cases, because it works with every other test I’ve done, but not with “HAIRY” where it only reads as 1 syllable.
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
TEST
print(syllable_count("HAIRY"))
Expected: 2
Received: 1
Answers:
The problem is that you’re giving it an uppercase string, but you only compare to lowercase values. This can be fixed by adding word = word.lower() to the start of your function.
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"
Your code seems to be working fine when given anything in lower case. However if you pass it a word in all upper case it will always return 1. This is because you are testing against “aeiou” and not “aeiouAEIOU”.
You can fix this in a few ways.
Example 1:
vowels = "aeiouyAEIOUY"
Example 2:
print(syllable_count("HAIRY".lower()))
Example 3: add this line of code at the start of the ‘syllable_count’ function
word = word.lower()
There are certain rules for syllable detection, you can view the rules from the website: Counting Syllables in the English Language Using Python
Here’s the python code:
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
you could use this as well using lambda map
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")))
in single line
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")))
For the best results, you might want to use a dictionary-based implementation. Packages like PyHyphen provide such functionality.
Hyphenator('en_US').syllables('beautiful') # = ['beau', 'ti', 'ful']
(Though, when I test this library using "hairy" it outputs only one syllable because this word is currently not in the default dictionary)
I need to write a function that will read syllables in a word (for example, HAIRY is 2 syllables). I have my code shown on the bottom and I’m confident it works in most cases, because it works with every other test I’ve done, but not with “HAIRY” where it only reads as 1 syllable.
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
TEST
print(syllable_count("HAIRY"))
Expected: 2
Received: 1
The problem is that you’re giving it an uppercase string, but you only compare to lowercase values. This can be fixed by adding word = word.lower() to the start of your function.
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"
Your code seems to be working fine when given anything in lower case. However if you pass it a word in all upper case it will always return 1. This is because you are testing against “aeiou” and not “aeiouAEIOU”.
You can fix this in a few ways.
Example 1:
vowels = "aeiouyAEIOUY"
Example 2:
print(syllable_count("HAIRY".lower()))
Example 3: add this line of code at the start of the ‘syllable_count’ function
word = word.lower()
There are certain rules for syllable detection, you can view the rules from the website: Counting Syllables in the English Language Using Python
Here’s the python code:
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
you could use this as well using lambda map
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")))
in single line
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")))
For the best results, you might want to use a dictionary-based implementation. Packages like PyHyphen provide such functionality.
Hyphenator('en_US').syllables('beautiful') # = ['beau', 'ti', 'ful']
(Though, when I test this library using "hairy" it outputs only one syllable because this word is currently not in the default dictionary)