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Kunzite: G Quinn #109

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Kunzite: G Quinn #109

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0522bc9
Completed Adagrams
genesisquinn Mar 24, 2023
6a86681
Adagrams Revisions
genesisquinn Mar 24, 2023
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124 changes: 120 additions & 4 deletions adagrams/game.py
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Original file line number Diff line number Diff line change
@@ -1,11 +1,127 @@
import random
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Good idea to add a line break between the imports and functions in a file like this.

def draw_letters():
pass
letters =[]
LETTER_POOL = {
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This dictionary could be moved outside this function into the main scope of the module so that draw_letters doesn't need to recreate this dictionary every time the function is run.

'A': 9,
'B': 2,
'C': 2,
'D': 4,
'E': 12,
'F': 2,
'G': 3,
'H': 2,
'I': 9,
'J': 1,
'K': 1,
'L': 4,
'M': 2,
'N': 6,
'O': 8,
'P': 2,
'Q': 1,
'R': 6,
'S': 4,
'T': 6,
'U': 4,
'V': 2,
'W': 2,
'X': 1,
'Y': 2,
'Z': 1
}
while len(letters) <10:
letter = random.choice(list(LETTER_POOL))
if LETTER_POOL[letter] > 0:
LETTER_POOL[letter] -= 1
else:
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One suggestion - rather than have an else block that skips the append operation on line 39, why not just append within the if block?

continue
letters.append(letter)
return letters





def uses_available_letters(word, letter_bank):
pass
new_word=word.upper()
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Careful with the spacing, should have space on either side of the equals sign in Python variable assignments:

new_word = word.upper()

letter_bank_copy =list(letter_bank)
for letter in new_word:
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This works perfectly well, but could be simplified a bit:

    for letter in word.upper():
        if letter in letter_bank_copy:
            letter_bank_copy.remove(letter)
        else:
            return False


    return True

if letter not in letter_bank_copy:
return False
elif letter in letter_bank_copy:
letter_bank_copy.remove(letter)
return True





def score_word(word):
pass
points =[]
score_chart = {
'A': 1,
'B': 3,
'C': 3,
'D': 2,
'E': 1,
'F': 4,
'G': 2,
'H': 4,
'I': 1,
'J': 8,
'K': 5,
'L': 1,
'M': 3,
'N': 1,
'O': 1,
'P': 3,
'Q': 10,
'R': 1,
'S': 1,
'T': 1,
'U': 1,
'V': 4,
'W': 4,
'X': 8,
'Y': 4,
'Z': 10
}
new_word=word.upper()
for letter in new_word:
if letter in score_chart:
points.append(score_chart[letter])
total= sum(points)
if len(new_word) ==7 or len(new_word)==8 or len(new_word)==9 or len(new_word)==10:
total+=8
return total

def get_highest_word_score(word_list):
pass
scores_dict={}
maximum = 0
max_key = None
word_list.sort(key=len)
for word in word_list:
score = score_word(word)
scores_dict[word] = score
for scores in scores_dict :
if scores_dict[scores] > maximum:
maximum = scores_dict[scores]
max_key = scores
#Line 111 & 112 doesn't work if there are 3 or more words that are the same length
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Still, this is close and passes the tests. Great job identifying where code can be improved.

elif len(word_list[-1]) == len(word_list[-2]) and score ==maximum:
max_key = word_list[-2]
elif len(word_list[-1]) == 10 and score == maximum:
max_key = word_list[-1]
return (max_key, maximum)

# how can I do this using max()?
# I originally had this but didn't how how to break ties according to len() using max().
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You'd probably need some additional conditional logic to make use of max in that way. There are a lot of different ways to accomplish the same solution, you'll see that time and again in programming!

#
# def get_highest_word_score(word_list):
# word_list.sort(key=len)
# score_dict={}
# for word in word_list:
# score = score_word(word)
# score_dict[word]=score
# winner = max(score_dict, key=score_dict.get)
# return(winner, score_dict[winner])