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Dan Garrison
22,457 PointsIs there a better way to handle this challenge (Chance Scoring - OOP)?
I'm specifically looking at Challenge Task 2 of 2 in the chance scoring challenge on OOP. My code works and it passes the challenges, but I feel like it's more complicated then it should be.
My code is attached, but to explain it a bit. I basically created a variable that used the value of index 0 in the hand. I then initiated an empty list. I then compared every dice to the check value and if it matched then I added it to the list. I then checked if the length of the list equaled 5 and if it did then I returned 50 and if it did not then I returned 0.
Is there a better way to handle the challenge?
class YatzyScoresheet:
def score_ones(self, hand):
return sum(hand.ones)
def _score_set(self, hand, set_size):
scores = [0]
for worth, count in hand._sets.items():
if count == set_size:
scores.append(worth*set_size)
return max(scores)
def score_one_pair(self, hand):
return self._score_set(hand, 2)
def score_chance(self, hand):
return sum(hand)
def score_yatzy(self, hand):
check_value = hand[0]
check_hand = []
for die in hand:
if die == check_value:
check_hand.append(die)
if len(check_hand) == 5:
return 50
else:
return 0
2 Answers
Paul Jenkins
Python Web Development Techdegree Graduate 21,149 PointsYes, it is. You can iterate through your hand using the hand._sets.items() then check if the count of each item in the hand is 5. If yes, return 50. Otherwise, after you've iterated through the hand and nothing = 5, you can return 0.
Adam Kielar
Python Web Development Techdegree Graduate 13,818 Points'''python def score_yatzy(self, hand): if hand[1:] == hand[:-1]: return 50 else: return 0 '''