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Python

Aldo Rivadeneira
Aldo Rivadeneira
3,241 Points

Hands Challenge 2/3 i don't understand the following error message

I'm getting this message when i check the code

Bummer: Didn't get True for .doubles on a pair

I try in different ways but i don't get where's my error :/

hands.py
from dice import D6


class Hand(list):
    def __init__(self, size=2, die_class=D6, *args, **kwargs):
        if not die_class:
            raise ValueError("You must provide a die class")
        super().__init__()

        for _ in range(size):
            self.append(die_class())
        self.sort()

    def _by_value(self, value):
        dice = []
        for die in self:
            if die == value:
                dice.append(die)
        return dice


class CapitalismHand(Hand):
    @property
    def doubles(self):
        for i in self:
            if i == self:
                return True
            else:
                return False

    @property
    def ones(self):
        return self._by_value(1)

    @property
    def twos(self):
        return self._by_value(2)

    @property
    def threes(self):
        return self._by_value(3)

    @property
    def fours(self):
        return self._by_value(4)

    @property
    def fives(self):
        return self._by_value(5)

    @property
    def sixes(self):
        return self._by_value(6)

    @property
    def _sets(self):
        return {
            1: len(self.ones),
            2: len(self.twos),
            3: len(self.threes),
            4: len(self.fours),
            5: len(self.fives),
            6: len(self.sixes)
        }

1 Answer

Alex Koumparos
seal-mask
.a{fill-rule:evenodd;}techdegree
Alex Koumparos
Python Development Techdegree Student 36,862 Points

Hi Aldo,

Your problem is your comparison. You write for i in self, thus each i must be an element in self, yet on the next line you test if i is equal to self. An element of a list can never be equal to the list itself.

Remember that this class will always comprise exactly two elements, and so you can safely check if the element at index 0 is equal to the element at index 1.

Hope that helps

Cheers

Alex