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I have the following string:

S = "to be or not to be, that is the question?"

I want to be able to create a dictionary that has the output of

{'question': 4, 'is': 1, 'be,': 1, 'or': 1, 'the': 1, 'that': 1, 'be': 1, 'to': 1, 'not': 1}

where I get the number of vowels in each word next to the word, not the count of each word itself. So far I have:

{x:y for x in S.split() for y in [sum(1 for char in word if char.lower() in set('aeiou')) for word in S.split()]}

with an output of: 

{'or': 4, 'the': 4, 'question?': 4, 'be,': 4, 'that': 4, 'to': 4, 'be': 4, 'is': 4, 'not': 4}

How do I get a dictionary from a string where the values are the vowel counts from each word?

jpmc26
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python_newbie
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  • `{'tell':1, 'me':1, 'what':1, 'I':1, 'tell':1, 'you,':2, 'to':1, 'you':2}` is not a valid dictionary, because there's keys in there multiple times. – Marcus Müller Sep 14 '15 at 16:41
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    Nikki, welcome to StackOverflow I didn't think this is a -6 question, so I upvoted it. In the future, try to clearly set apart your question and state it in the form of a question so that you don't get this reception again. If you accept an answer, it will give you plus two to your rep. Cheers. I'll try to help you restate the question here. – Russia Must Remove Putin Sep 14 '15 at 17:01

2 Answers2

1

number of vowels in each word next to the word, not the count of each word itself?

>>> s = "to be or not to be, that is the question"

first remove punctuation:

>>> new_s = s.translate(None, ',?!.')
>>> new_s
'to be or not to be that is the question'

then split on the whitespace:

>>> split = new_s.split()
>>> split
['to', 'be', 'or', 'not', 'to', 'be', 'that', 'is', 'the', 'question']

Now count the vowels in a dictionary. Note there are no redundant counts:

>>> vowel_count = {i: sum(c.lower() in 'aeiou' for c in i) for i in split}
>>> vowel_count
{'be': 1, 'that': 1, 'is': 1, 'question': 4, 'to': 1, 'not': 1, 'the': 1, 'or': 1}
Russia Must Remove Putin
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-1

You can use re (regex module) to find all the valid words (\w+ - does not includes spaces and commas), and use Counter to check the frequencies: 

import re

from collections import Counter
s = "tell me what I tell you, to you"
print Counter(re.findall(r'\w+', s))

OUTPUT

Counter({'you': 2, 'tell': 2, 'me': 1, 'what': 1, 'I': 1, 'to': 1})
Nir Alfasi
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