Amethyst Megan G. #20
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| # helper function to create one big list of strings | ||
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| def get_one_list_of_letters(): |
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Yessss! I love how you are making use of helper functions here! It allows how us make more concise and readable code! Also great job on leaving a comment about what this function does.
| def get_one_list_of_letters(): | ||
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| total_letters_list = [] | ||
| for elem in LETTER_POOL: |
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I really like how you created a nested data structure and then you used a nested loop to build a big list for you to pull a random letter from. You definitely are showing proficiency when it comes to this topic! ⭐️
| i = 0 | ||
| while i < 10: | ||
| # randomize letter choice | ||
| random_letter = total_letters_list[random.randint(0, 97)] |
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Small comment, one thing that we want to try to avoid as developers is hard coding values inside for code base. So, instead using 97, we could maybe do len(total_letters_list) that way if anyone wants to change the weights of the letter then they won't have change the 97 as well! Again, just a small change to make your code more scalable.
| if hand.count(random_letter) < total_letters_list.count(random_letter): | ||
| hand.append(random_letter) | ||
| else: | ||
| # use continue to end this iteration and immediately start next iteration | ||
| continue | ||
| i += 1 |
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count is a great function to use in a situation like this! That way you don't have worry about manipulating any data and removing things! An alternative refactor we could do to make the code a bit more concise is this:
while len(hand) < 10:
random_letter = total_letters_list[random.randint(0, 97)]
if hand.count(random_letter) < total_letters_list.count(random_letter):
hand.append(random_letter)This code will run the same as yours, this code just doesn't keep track of a counter since the times we append to hand is equal to len(hand)
| def uses_available_letters(word, letter_bank): | ||
| pass | ||
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| # made copy of list as to not modify the original copy | ||
| letter_bank_copy = copy.deepcopy(letter_bank) | ||
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| # iterate through each letter in the word and make letters all uppercase | ||
| for letter in word.upper(): | ||
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| if letter in letter_bank_copy: | ||
| # remove letter from letter bank so it can't be reused | ||
| letter_bank_copy.remove(letter) | ||
| else: | ||
| return False | ||
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| return True |
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Nice use of the deepcopy function here, it lets you get a copy of the letter_bank without changing the object itself. This shows me that you have a good understanding of variables and memory. An alternative approach that doesn't mutate the original letter_bank is this here:
def uses_available_letters(word, letter_bank):
cap_word = word.upper()
for letter in cap_word:
if letter_bank.count(letter) == 0:
return False
if cap_word.count(letter) > letter_bank.count(letter):
return False
return TrueIt is basically the same thing as what you had in wave 1, using the count function to count the instances of a letter in both the word and letter_bank.
| for key in letter_values: | ||
| if letter in key: |
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Awesome, Megan! You used tuples as the keys for your dictionary and then used the if letter in key to add up the score! ⭐️
| if len(word) > 6: | ||
| score += 8 | ||
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| return score | ||
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| def get_highest_word_score(word_list): |
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Great function! I know we talked about this logic in co-working and you implemented very well! 🤩
| winning_word = word | ||
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| return winning_word, highest_score |
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Megan, you completed your first official Ada project! Congrats! Your code was a pleasure to look through! Very easy to follow and quite concise! Please feel free to reach out if you need any clarifications on the comments I left! You did great! Can't wait to review more of your code! 😝✨
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