陷入人生编码游戏的愚蠢阶段

Question

我一直在尝试自己编写生活游戏代码，因为我想学习 python，我认为这可能是一个很好的做法，我刚刚完成，除了我需要将一个变量传递给另一种方法，但我真的不知道我怎么知道它看起来很愚蠢但是稍等一下，除非对我来说，否则代码并不像听起来那么简单

import random,time
f=3
c=3
contador = 0
tablero = []
tablero_dibujado = []

class juego(object):

    def tablero(self): #Create the Board
        for i in range(f):
            tablero.append([0]*c)
            tablero_dibujado.append([0]*c)

    def rellenar_tablero_random(self): #Fill the board with random 0 and 1
        for i in range(f):
            for j in range(c):
                tablero[i][j] = random.randint(0,1)
        print(tablero)

    def rellenar_tablero_manual(self): #Just to fill the board manually if I want for some reason
        tablero[0][1] = 1
        for i in range(2):
            tablero[1][i] = 1
        print(tablero)

    def distancia_vecino(self,cell_f,cell_c):   #Measure Distance for possible neighbours
        distancia_vecino = [(-1,-1),(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1)]
        for i in range(8):
            string = distancia_vecino[i]
            comparar_string = str(string)
            x,y = comparar_string.split(",")
            x = str(x).replace(",","")
            y = str(y).replace(",","")
            x = x.replace("(","")
            y = y.replace(")","")
            y = y.replace(" ","")
            x = int(x) + cell_f
            y = int(y) + cell_c

            if x>=f or x<0:
                continue
            else:
                if y>=c or y<0:
                    continue
                else:
                    game.detectar_vecino(cell_f,cell_c,x,y)
        game.vida(cell_f,cell_c)

    def detectar_vecino(self, cell_f, cell_c, x, y):    #Check how many neighboards do I have
        vecinos = 0

        if tablero[cell_f][cell_c] == 1:
            if tablero[cell_f][cell_c] == tablero[x][y]:
                vecinos+=1
        else:
            if tablero[cell_f][cell_c] != tablero[x][y]:
                vecinos+=1
        contador = contador + vecinos

    def iterar_tablero(self): #Just to iterate all the board to check the values
        for i in range(f):
            for j in range(c):
                game.distancia_vecino(i,j)
        a = input() #In order to the screen dont close after executing it

    def vida(self, cell_f, cell_c): #Check if they are alive and if they should be alive or dead

        print(contador)
        if tablero[cell_f][cell_c] == 1:
            if contador > 3 or contador < 2:
                tablero[cell_f][cell_c] = 0
        else:
            if contador == 3:
                tablero[cell_f][cell_c] = 1

        game.dibujar_tablero()

    def dibujar_tablero(self): #Draw the board in a more clearly way
        contador1 = 0
        for i in range(f):
            for j in range(c):
                if tablero[i][j] == 1:
                     tablero_dibujado[i][j] = "§"
                else:
                     tablero_dibujado[i][j] = "■"

        for i in tablero_dibujado:
            print(" ")
            for j in i:
                print(j, end=" ")
        print("")
        time.sleep(2)


game = juego()
game.tablero()
game.rellenar_tablero_manual()
game.iterar_tablero()

我需要的是，在 detectar_vecino 中，必须获取面板中一个单元格的所有邻居，问题是这样做我得到了：局部变量 'contador' 在赋值之前被引用。我知道为什么会这样 happen.However 我找不到任何替代方法所以如果有人知道我该如何解决它请拜托。

我只是想澄清一下，这不是来自任何地方的任何工作，我只是出于业余爱好而自己做这件事，最后我只想感谢你抽出时间，我很感激

Answer 1

像下面这样添加这一行global contador

def detectar_vecino(self, cell_f, cell_c, x, y):
    global contador   # Add this line.
    vecinos = 0

    if tablero[cell_f][cell_c] == 1:
        if tablero[cell_f][cell_c] == tablero[x][y]:
            vecinos+=1
    else:
        if tablero[cell_f][cell_c] != tablero[x][y]:
            vecinos+=1
    contador = contador + vecinos

陷入人生编码游戏的愚蠢阶段

Stuck in a stupid phase of coding game of life

python

conways-game-of-life