CODEWITHSUNDEEP
pythonstringlist
Zopyrus
Zopyrus

Reputation: 147

Python string and lists

Been sitting on this minor problem a few days now, I don't know if I have it all wrong or just missed out on something.

The Objective: From each word in a sentence - Find the first vowel, remove the letters after that vowel from the word and multiply the remaining letters by 3.

The Example: If I have the sentence: "Hello World" the wanted output should be "HeHeHe WoWoWo".

My Code:

def bebis(inrad):
    utrad = ""
    inrad = inrad.split()
    for tkn in inrad:
        for tkn1 in tkn: #Eftersom tkn ar ordlista nu.
            if tkn1 in vokaler:
                count = len(tkn1)
                utrad += tkn1
            elif tkn1 in konsonanter:
                utrad += tkn1
    return utrad[:count+1]*3

print("Bebisspraket:",bebis(inrad))

My Thoughts: I split the sentence into a lists of words using split(). Then I use two for loops, one that should go through each word and the other one that should go through each letter in every word. If it finds a vowel, count where it is and then return the letters to the first vowel of the word.

My Problem: The output only gives me the first WORD in a sentence and breaks from there. So "Hello World" yields "HeHeHe" leaving me super frustrated. Why does it not go through the rest of the sentence?

Upvotes: 3

Views: 1741

Answers (3)

Adam Parkin
Adam Parkin

Reputation: 18670

Just as an alternative to the regular expression approach, I did:

def find_vowel_index(word):
    vows = set(["a", "e", "i", "o", "u"])
    for i, letter in enumerate(word):
            if letter in vows:
                    return i
    return -1 

def bebis(s, repeat=3):
    return " ".join([word[0:find_vowel_index(word)+1] * repeat for word in s.split() if find_vowel_index(word) >= 0])

Thoughts:

  • I did a set for the vowels as the "in" test on a set is a constant time operation (O(1))
  • added an optional repeat argument to bebis, so that if you want the word to repeat some number of times other than 3 it's easy to do so.
  • I don't like the multiple calls to find_vowel_index in bebis, it could be structured better.
  • a trade-off over the regex version is that this will get slower as words get longer (and have vowels "deeper" in the word). OTOH, I'd guess that on short words the overhead of the regex might be a bit costly.

Upvotes: 0

arshajii
arshajii

Reputation: 129497

How about something like this:

import re

def bebis_word(word):
    first_vowel = re.search("[aeiou]", word, re.IGNORECASE)

    if first_vowel:
        return word[0:first_vowel.start() + 1] * 3
    else:
        return ''    

def bebis(sentence):
    words = [bebis_word(word) for word in sentence.split()]

    return " ".join(words)

print bebis("Hello World")

Output:

HeHeHe WoWoWo

Upvotes: 4

Pierre GM
Pierre GM

Reputation: 20339

Your approach seems correct (splitting the sentence into words and iterating on the words to find the first vowel).

The problem is that your tkn1 variable is a letter, so len(tkn1) is always 1, so count=1

Here's a potential solution:

def bebis(inrad):
    utrad = ""
    inrad = inrad.split()
    # Loop on words 
    for tkn in inrad:
        # Loop on letters in the word
        for (i,tkn1) in enumerate(tkn): #Eftersom tkn ar ordlista nu.
            if tkn1 in vokaler:
                utrad += tkn[:i+1] * 3
                break
        utrad += " "
    return utrad

Here, we use the enumerate function that will give a list of tuples (index,letter). Once we tested that the current letter is a vowel, we take the first letters of the word (tkn[:i+1]), repeat them three times, store them in utrad and move to the next word (with the break statement that leaves the current loop). We just have to add an extra space between words.

Upvotes: 0

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