Stack Overflow Asked on November 22, 2021
I’m doing Georgia Tech’s CS1301xII course, and I have been stumped by this question.
I am supposed to make a function called count_letters. If find_consonants is true, then it counts consonants, or vowels if it is false. It should only return vowels and consonants, no uppercase letters, spaces, numbers, or punctuation. The output I get is 0 consonants, then 1 vowel. I expected 14, then 7.
def count_letters(string, find_consonants):
if find_consonants == True:
count = 0
for i in string:
if i == ("q" or "w" or"r" or "t" or "y" or "p" or "s" or "d" or "f" or "g" or "h" or "j" or "k" or "l" or "z" or "x" or "c" or "v" or "b" or "n" or "m"):
count += 1
return count
else:
count = 0
for i in string:
if i == ("a" or "e" or "i" or "o" or "u"):
count += 1
return count
(The next section is just for me to test myself, the autograder changes the string)
a_string = "up with the white and gold"
print(count_letters(a_string, True))
print(count_letters(a_string, False))
The or
operation in python is lazy evaluation, like:
>>> ('a' or 'c')
'a'
>>> ('c' or 'b' or 'a')
'c'
so i == ("a" or "e" or "i" or "o" or "u")
equivalent to i == 'a'
, is not the result you want.
You can change it like
This is crazy...though
count = 0
for i in string:
if (i == "a") or (i == "e") or (i == "i") or (i == "o") or (i == "u"):
count += 1
This one is more elegant.
count = 0
for i in string:
if i in 'aeiou':
count += 1
And this is Python
len([x for x in string if x in 'aeiou'])
Answered by K. Prot on November 22, 2021
@ObseleteAwareProduce's solution is correct and addresses the concept of checking if an item matches any item in a specific list. But as far as finding vowels and consonants, we can use re library as well
import re
regex = re.compile("a|e|i|o|u", flags=re.I)
vowels_count = len(regex.finditer(input_string))
Answered by vestronge on November 22, 2021
The test if find_consonants == True
is a bit overengineered. if find_consonants
suffices. You might also want to make use of list comprehensions, to avoid the explicit loops.
This should work for instance:
def count_letters(s, find_consonants):
if find_consonants:
return len([l for l in s if l in "bcdfghjklmnpqrstvwxyz"])
return len([l for l in s if l in "aeiou"])
Answered by karlson on November 22, 2021
Sadly, you can't use if var == "foo" or "bar"
; you'd have to write if var == "foo" or var == "bar"
.
For your function, try something like:
def count(string, count_consonants):
vowelcount = 0
consonantcount = 0
for letter in string:
if letter in "aeiou": #Detect if letter is in vowel list
vowelcount += 1
elif letter in "bcdfghjklmnpqrstvwxyz":
consonantcount += 1
if count_consonants:
return consonantcount
else:
return vowelcount
Answered by ObsoleteAwareProduce on November 22, 2021
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