CODEWITHSUNDEEP
pythonpython-3.xlistfor-loop
Jim421616
Jim421616

Reputation: 1536

Counting how many times each vowel appears

I've written a little program to count how many times each vowel appears in a list, but it's not returning the correct count, and I can't see why:

vowels = ['a', 'e', 'i', 'o', 'u']
vowelCounts = [aCount, eCount, iCount, oCount, uCount] = (0,0,0,0,0)
wordlist = ['big', 'cats', 'like', 'really']

for word in wordlist:
    for letter in word:
        if letter == 'a':
            aCount += 1
        if letter == 'e':
            eCount += 1
        if letter == 'i':
            iCount += 1
        if letter == 'o':
            oCount += 1
        if letter == 'u':
            uCount += 1
for vowel, count in zip(vowels, vowelCounts):
    print('"{0}" occurs {1} times.'.format(vowel, count))

The output is

"a" occurs 0 times.
"e" occurs 0 times.
"i" occurs 0 times.
"o" occurs 0 times.
"u" occurs 0 times.

However, if I type aCount in the Python shell, it gives me 2, which is correct, so my code has indeed updated the aCount variable and stored it correctly. Why isn't it printing the correct output?

Upvotes: 2

Views: 801

Answers (4)

Anton vBR
Anton vBR

Reputation: 18916

You can also use collections counter (which is the natural go-to function when counting things, it returns a dictionary):

from collections import Counter

vowels = list('aeiou')
wordlist = ['big', 'cats', 'like', 'really']

lettersum = Counter(''.join(wordlist))

print('\n'.join(['"{}" occurs {} time(s).'.format(i,lettersum.get(i,0)) for i in vowels]))

Returns:

"a" occurs 2 time(s).
"e" occurs 2 time(s).
"i" occurs 2 time(s).
"o" occurs 0 time(s).
"u" occurs 0 time(s).

lettersum:

Counter({'l': 3, 'a': 2, 'e': 2, 'i': 2, 'c': 1, 'b': 1, 
         'g': 1, 'k': 1, 's': 1, 'r': 1, 't': 1, 'y': 1})

Upvotes: 4

Moustafa Essam
Moustafa Essam

Reputation: 1

In addition to what @jpp said simple datatypes like integers are returned by value not by reference so when you assign it to something and change that something it doesn't get affected

a = 10
b = a #b=a=10
b = 11 #b=11, a=10
print a, b
--> 10 11

I'd have made this a comment to his but I need reputation to do that :D

Upvotes: 0

Ajax1234
Ajax1234

Reputation: 71461

You can use a dictionary comprehension:

vowels = ['a', 'e', 'i', 'o', 'u']
wordlist = ['big', 'cats', 'like', 'really']
new_words = ''.join(wordlist)
new_counts = {i:sum(i == a for a in new_words) for i in vowels}

Output:

{'a': 2, 'e': 2, 'i': 2, 'o': 0, 'u': 0}

Upvotes: 1

jpp
jpp

Reputation: 164773

The problem is with this line:

vowelCounts = [aCount, eCount, iCount, oCount, uCount] = (0,0,0,0,0)

vowelCounts does not get updated if you start incrementing aCount later.

Setting a = [b, c] = (0, 0) is equivalent to a = (0, 0) and [b, c] = (0, 0). The latter is equivalent to setting b = 0 and c = 0.

Reorder your logic as below and it will work:

aCount, eCount, iCount, oCount, uCount = (0,0,0,0,0)

for word in wordlist:
    for letter in word:
        # logic 

vowelCounts = [aCount, eCount, iCount, oCount, uCount]

for vowel, count in zip(vowels, vowelCounts):
    print('"{0}" occurs {1} times.'.format(vowel, count))

Upvotes: 6

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