I wrote this function to count the syllables in a word using Python to complete an exercise.
The exercise's text is:
Return the number of syllables in the given word, according to this rule: Each contiguous sequence of one or more vowels is a syllable, with the following exception: a lone "e" at the end of a word is not considered a syllable unless the word has no other syllables. You should consider y a vowel.
I also disagree with the text because using the given instructions the syllable count for the word "example" are 2 and should be 3.
But, as discussed in the comments, the exercise text is the one that I implemented and would like to get suggestions and improvements for.
def count_syllables(word):
word = word.lower()
counter = 0
is_previous_vowel = False
for index, value in enumerate(word):
if value in ["a", "e", "i", "o", "u", "y"]:
if index == len(word) - 1:
if value == "e":
if counter == 0:
counter += 1
else:
counter += 1
else:
if is_previous_vowel == True:
counter += 1
is_previous_vowel = False
break
is_previous_vowel = True
else:
if is_previous_vowel == True:
counter += 1
is_previous_vowel = False
print ("In \"{}\" I count {} syllables".format(word, counter))
count_syllables("coding")
1 Answer 1
As a good practice to promote code reuse, the function should return a numeric result. The print() should be done by the caller.
Constantly referring to is_previous_vowel is tedious. There are a couple of solutions you could use involving itertools. In particular, groupby() and zip_longest() could be useful.
Ultimately, the simplest solution is to use a regular expression to declare what you are looking for: one or more consecutive vowels ('[aeiouy]+'), but not an e at the end ('(?!e$)).
import re
def count_syllables(word):
return len(
re.findall('(?!e$)[aeiouy]+', word, re.I) +
re.findall('^[^aeiouy]*e$', word, re.I)
)
-
\$\begingroup\$ This is a masterpiece \$\endgroup\$Pitto– Pitto2019年07月15日 07:50:09 +00:00Commented Jul 15, 2019 at 7:50
ordfunction you can convert any character to its 8 bit integer representation. Then you could write instead offoranwhile looptill end character is found (0x00), where your counter would then becounter += 1 if (ord(word[i]) & 97 == 97 else 0. You can see what integer representations of your vowel letters are by going into console and justlist(map(ord, 'aeiouy')). With while loop you can save in speed by not assigning length of string. GKeyWords: Bitwise operators. \$\endgroup\$eforms a syllable; treatment ofyis too simplistic (say,beyond). However I am curious how did you come up with 4 syllables inexample. Nevertheless, the problem statement is what has to be implemented. \$\endgroup\$