Reputation: 31
The program I am trying to create involves writing a method called monster_fight(monster1,monster2) to have the two monsters "Fight". However I am having one issue retrieving the damage value stored in each of the monster object dictionaries named 'self.attacks'.
I am trying to retrieve a value from dictionary 'monster1.attacks' to reduce the hp of the monster2 object. However with the current code I have in place, the program does not recognize the value of the keys when I call the dictionary. Can anybody show what I am doing wrong?
Thanks!
class Monster():
def __init__(self, name, max_hp = 20, hp=20):
self.name = name
self.type = type
self.current_hp = max_hp
self.attacks = {'wait': 0}
self.possible_attacks = {'sneak_attack': 1,
'slash': 2,
'ice_storm': 3,
'fire_storm': 3,
'whirlwind': 3,
'earthquake': 2,
'double_hit': 4,
'wait': 0}
self.exp = 0
def add_attack(self, attack_name):
if attack_name in self.possible_attacks:
self.attacks[attack_name] = self.possible_attacks.get(attack_name)
return True
else:
return False
if attack_name in self.attacks:
return False
def remove_attack(self, attack_name):
if attack_name in self.attacks:
self.attacks.pop(attack_name)
if len(self.attacks) == 0:
self.attacks['wait'] = 0
return True
else:
return False
def win_fight(self):
self.exp += 5
self.current_hp = self.max_hp
def lose_fight(self):
self.exp += 1
self.current_hp = self.max_hp
def monster_fight(monster1,monster2):
round1 = 0
moves1 = []
moves2 = []
list1 = []
list2 = []
for i in monster1.attacks:
values = ''
values = monster1.attacks.get(i)
list1.append(values)
for i in range(0,len(monster2.attacks)):
values = monster2.attacks.get(i)
list2.append(values)
while monster1.current_hp > 0 or monster2.current_hp > 0:
round1 += 1
monster1_attack = int(monster1.attacks[list1[(round1-1)%len(list1)]])
monster2.current_hp -= monster1_attack
moves1.append(list1[(round1-1)%len(list1)])
monster2_attack= monster2.attacks[list2[(round1-1)%len(list2)]]
monster1.current_hp -= monster2_attack
moves2.append(list2[(round1-1)%len(list2)])
if monster1.current_hp <= 0:
monster1.lose_fight()
monster2.win_fight()
return round1, monster2.name, moves2
elif monster1.current_hp <= 0:
monster2.lose_fight()
monster1.win_fight()
return round1,monster1.name, moves1
else:
return -1,"None","None"
a = Monster("a", 9)
b = Monster("b", 9)
a.add_attack("ice_storm")
b.add_attack("ice_storm")
b.remove_attack("wait")
a.remove_attack("wait")
round1, winner, moves = monster_fight(a, b)
print(round1)
print(winner.name)
print(moves)
monster1_attack = int(monster1.attacks[list1[(round1-1)%len(list1)]])
KeyError: 3
Upvotes: 0
Views: 174
Reputation: 312630
Well, let's see. You're calling:
monster1_attack = int(monster1.attacks[list1[(round1-1)%len(list1)]])
monster1
is a Monster
object, created from:
a = Monster("a", 9)
a.add_attack("ice_storm")
a.remove_attack("wait")
So monster1.attacks
looks like:
{
'ice_storm': 3,
}
You're trying to access that dictionary using key dervied from list1[(round1-1)%len(list1)]
.
list1
is set here:
for i in monster1.attacks:
values = ''
values = monster1.attacks.get(i)
list1.append(values)
After the above code runs, list1
is a list that looks like:
[3]
(Because you ask for monster1.attacks.get(i)
, which will return the value associated with key i
.)
So when you ask for list1[(round1-1)%len(list1)]
, you get the value 3
, which means you're asking for monster1.attacks[3]
.
There is no key named 3
in monster1.attacks
. As we saw earlier, the only key is ice_storm
which has the value 3
. It looks like you're trying to figure out the damage that monster1
's attack will do. That is actually what you have in list1
, so in theory you could just write:
monster1_attack = list1[(round1-1)%len(list1)]
monster2.current_hp -= monster1_attack
I think your logic here may be a bit convoluted. You should probably think carefully about exactly what you're trying to accomplish and try to simplify your code with that goal in mind. Using the Python debugger to see the value of your variables prior to the error -- or using print
statements to accomplish the same thing -- can help diagnose this sort of problem.
Upvotes: 4