字典 Python 变量

Question

I want use a variable as dictionnary name for find and use fastly some values. (For example i want the agressivity of my fighter one). I use a variable because the fighter can change and i want a dynamic script.

fighter1 = {"agressivity": 6, "agility": 2}
agressivity = 0
arenafighter = "fighter1"
agressivity = arenafighter["agressivity"]

But i have the same problem "TypeError: string indices must be integers"

Answer 1

You can use a dict in a dict

fighters = {
    "fighter1": {"agressivity": 6, "agility": 2},
    "fighter2": {"agressivity": 8, "agility": 4},
}

arenafighter = "fighter1"
agressivity = fighters[arenafighter]["agressivity"]

Answer 2

I believe the problem with your code is that when defining arenafighter you put fighter1 in quotes, which makes the value of arenafighter into the word "fighter1" because it is a string.

Try this:

fighter1 = {"agressivity": 6, "agility": 2}
agressivity = 0
arenafighter = fighter1
agressivity = arenafighter["agressivity"]

Answer 3

Why don't you make a dictionary with all fighters? It could look like this:

fighters = {
   "fighter_1": {"agressivity": 6, "agility": 2},
   "fighter_2": {"agressivity": 5, "agility": 4}
}

by doing that now you can access fighters like this:

fighter = "fighter_1"
print(fighters[fighter]["agressivity"])

Answer 4

What you need is some simple class and object creation. B'coz you say the fighter can change and you want to dynamically assign values for each fighter..see if this code helps

class Fighter:
    
    def __init__(self,name,agressivity,agility):
        self.name = name
        self.agressivity = agressivity
        self.agility = agility
        

        
fighter1 = Fighter(fighter1, 6,  2)
print(fighter1.agressivity)
arenafighter = Fighter(arenafighter, 0,  2)
print(arenafighter.agressivity)