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I am trying to write a function that checks if I can generate the required word/phrase using the characters provided. The phrase created can contain any characters including special characters, capital letters, numbers, and spaces.

I can generate the phrase if the frequency of unique characters in the characters string is equal or greater than the frequency in the phrase.

What I tried to do was use a counter and then compare the dictionaries that it produced but not sure how to find if the characters in the character string are more in frequency than in the phrase.

This is my code:

from collections import Counter

def generate_phrase(characters, phrase):
  new_characters = characters.lower()
  new_phrase = phrase.lower()
  a = Counter(new_characters)
  b = Counter(new_phrase)
  if a == b:
    return True 
  else:
    return False

This example should produce false, as there are less characters in the character string than the phrase

characters = "cbacba"
phrase = "aabbccc"

generate_phrase(characters, phrase)

Ouput:
False

This example should produce True as although it has other letters in it, that are not in the phrase. It does contain all characters of the phrase and the right frequency of them.

characters = "Magiciansktb!"
phrase = "m!aagsnici"

generate_phrase(characters, phrase)

Ideal Ouput:
True

My current code output:
False
Karl Knechtel
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Jennifer
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    Just to make it clearer can you give an example of `characters` and `phrase`. – Maurice Meyer Aug 01 '22 at 18:48
  • You will need to implement the comparison with a loop; for each character in the phrase, are there enough instances in that slot in the other counter? – tripleee Aug 01 '22 at 18:52
  • I have updated my question to give 2 examples of what it should be :) – Jennifer Aug 01 '22 at 19:00
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    `if a == b:` checks whether the two counters have **the same** count for each letter. This would only be the case if the two inputs are anagrams of each other. `collections.Counter` does not implement a "subset" operation; you have to do it yourself - see the linked duplicate. – Karl Knechtel Aug 01 '22 at 19:38
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    @DanielHao thank you for posting the associated question, it is has been extremely helpful :) – Jennifer Aug 01 '22 at 21:31
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    Please note that this is **not a discussion forum** - do not edit questions to indicate that they are "solved" or to put a solution into the question. Instead, use upvotes and/or accept markings as appropriate. Please see https://meta.stackoverflow.com/help/someone-answers for more details. – Karl Knechtel Aug 01 '22 at 22:11
  • where can i add my answer to the question ? – Jennifer Aug 01 '22 at 22:22
  • I have found an alternative way to solve the question and would also like to show it so that others can view it as well – Jennifer Aug 01 '22 at 22:23

2 Answers2

0

Alternatively, you can try this code to see if it helps you:

It tries to handle some edge cases here, if errors caught earlier, and it bails out right away, will not go further process. Just make a note of it.


def generate_phrase(letters: str, phrase: str) -> bool:
    phrase_cnt = Counter(phrase.lower())
    chars_cnt  = Counter(letters.lower())

    if not phrase_cnt and not chars_cnt:      # both are empty?
        return True
        
    #if not letters:  return False
        
    if len(phrase) >  len(letters):       # not enought letters?
        return False


    for ch, cnt in phrase_cnt.items():
     
        if ch in chars_cnt and cnt <= chars_cnt.get(ch):
            continue
        else:
            return False 
        
    return True


if __name__ == '__main__':
    characters = "cbacba"
    phrase = "aabbccc"

    print(generate_phrase(characters, phrase))          # False

    print(generate_phrase('footeatips', 'tattoo'))      # False

    print(generate_phrase('gooseteatipstim', 'tattoo'))  # True

    characters2 = "Magiciansktb!"
    phrase2 = "m!aagsnici"
    print(generate_phrase(characters2, phrase2))   # True
Daniel Hao
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    Thank you so much, this method has helped a lot :) – Jennifer Aug 01 '22 at 21:33
  • As @KarlKnechtel suggested earlier ^^^, you could read it in the posted comments. BTW, I did *not* send out any link or other info. beyond this answer. – Daniel Hao Aug 01 '22 at 22:48
  • ahh im sorry, im getting confused as there was a lot of comments, @KarlKnechtel commented with a link to a suggested article - after reading i was able to try out that method as well on the posted question. – Jennifer Aug 01 '22 at 22:54
-1

Use all() to test if every count from characters is at least the count of the corresponding character in phrase.

def generate_phrase(characters, phrase):
    new_characters = characters.lower()
    new_phrase = phrase.lower()
    char_counts = Counter(new_characters)
    phrase_counts = Counter(new_phrase)
    return all(char_counts.get(char, 0) >= count for char, count in phrase_counts.items())
Barmar
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