Marlyn Lopez - Luz Andrea Garcia Zapata /C17 - Otters Class#53
Marlyn Lopez - Luz Andrea Garcia Zapata /C17 - Otters Class#53andrygzt wants to merge 4 commits intoada-c17:masterfrom
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kendallatada
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kendallatada
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Hi Marlyn and Andrea! This has been scored as yellow due to the logical errors in two of your functions. Please check your code to see the comments I left. Let me know if you have any questions. Nice work overall! :)
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| import random | ||
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| #from constants import * |
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Careful not to leave unused code hanging around. It looks like adagrams/constants.py and tests/constants.py are not being used. I know neither of these modules are being used, but try to avoid duplicate code. These modules are exactly the same, they just live in different directories. So, make sure to only have one copy of a module that gets imported by all other modules / tests that may need it.
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| def draw_letters(): | ||
| pass | ||
| letter_pool = LETTER_POOL.copy() |
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Nice use of the copy method to avoid modifying the LETTER_POOL constant ✨
| upper_word = word.upper() | ||
| for i in upper_word: | ||
| counter += 1 | ||
| if upper_word.count(i) == letter_bank.count(i) and counter == len(upper_word): |
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Love your usage of the count method! 💕 I would reconsider how you have structured the conditionals in this loop though.
- This function passes all of the provided tests but there are some logical errors that will lead to incorrect behavior. Consider if you call the function with these arguments:
uses_available_letters("cog", ["D", "O", "G"])the function will returnTrue. Think about why that is and how you can update your code to solve this logical error. - If we only want to return True/False after checking all letters in the word, then we don't need to check if we're at the end of the word every loop iteration. Think about how you can accomplish this without checking this conditional each loop iteration. With this new approach, is the counter necessary?
- It's possible to implement this solution with only 1 conditional statement. Think about how you could accomplish that.
| # allow user enter upper and lower case letters | ||
| upper_word = word.upper() | ||
| score = 0 | ||
| for letter in upper_word: |
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| def get_highest_word_score(word_list): |
| best_word = word | ||
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| if len(temp_word) == len(word): | ||
| best_word = word_list[0] |
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Make sure to read the requirements carefully and that your interpretation aligns with the overall goal. The wording in the README for this specific case could be interpreted as return the first item in the list if there are two items of the same length and the same score. However, if we go with that interpretation and we run the function with this input:
get_highest_word_score(["A", "AAAAAAAAAA", "EEEEEEEEEE"])
The output will be: ('A', 18)
'A' is not the word with the highest score and 18 is not the correct score for the word 'A'. I would categorize this as a logical error. Think about how you could update your code to handle this case correctly.
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