Sphinx - Olga Karaivanska#38
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| @@ -1,11 +1,108 @@ | |||
| import random | |||
| LETTER_POOL = { | |||
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Great constant variable naming convention here! 🎉
| letter_count = {letter: 0 for letter in LETTER_POOL} | ||
| letter_list = list(LETTER_POOL.keys()) |
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Awesome use of comprehension here and use of the keys method and converting that into a list!
| hand_of_letters.append(letter) | ||
| letter_count[letter] += 1 | ||
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| return hand_of_letters |
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Great function, Olga! ⭐️ Only thing that's missing is the distribution. Read below:
Right now your code isn't actually utilizing the weight distribution of the letters. Currently, all letters have the same chance of being selected (they all can only appear their respective limits). randint is only randomly selecting from a pool of integers from 0 to 26. This means that Z is just as likely as likely to be selected as E. You would need it to select from a pool of integers from 0 to 98 to have the correct distribution. How could we change your letter pool to consider this weighted distribution?
| def uses_available_letters(word, letter_bank): | ||
| pass | ||
| letter_bank_copy = letter_bank [:] | ||
| word = word.upper() |
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Nice work on ensuring that all the letters where of the same casing. You could also use casefold, which is used for caseless matching. You can find more information on the method here.
Also be mindful of re-assigning the parameter of a function. It can be a side effect of your function if the caller doesn't expect for the data that they passed in to be altered.
| word = word.upper() | ||
| for letter in word: | ||
| for number, letters in scores.items(): | ||
| if letter in letters: |
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Depending on how we structure our dictionary we could avoid this loop here. What do you think the change would be? When you learn about Big O this kind of thinking will come in handy.
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| if len(word) < len(winner[0]) or len(word) == 10: | ||
| winner = (word, score) | ||
| return winner |
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Very concise and readable function, praises! 🎉
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