当前位置 博文首页 > python实现2048小游戏

    python实现2048小游戏

    作者:admin 时间:2021-07-28 18:49

    2048的python实现。修改自某网友的代码,解决了原网友版本的两个小bug:

    1. 原版游戏每次只消除一次,而不是递归消除。如 [2 ,2 ,2 ,2] 左移动的话应该是 [4, 4, 0, 0] , 而不是[8 , 0 , 0 ,0]
    2. 对游戏结束的侦测有bug,已经改正。

    2048game.py

    # -*- coding: utf-8 -*-
    """
    Created on Tue Jul 1 14:15:39 2014
     
    @author: kelvin
    """
     
    import random
     
    class game2048:
     totalScore = 0
     v = [[2, 8, 8, 2],
       [4, 2, 4, 8],
       [2, 4, 2, 0],
       [4, 2, 4, 0]]
     '''
     v = [[0, 0, 0, 0],
       [0, 0, 0, 0],
       [0, 0, 0, 0],
       [0, 0, 0, 0]]
     '''
     def __init__(self):
      for i in range(4):
       self.v[i] = [random.choice([0,0,0,2,2,4]) for x in range(4)]
     
     
     def display(self):
      print('{0:4} {1:4} {2:4} {3:4}'.format(self.v[0][0], self.v[0][1], self.v[0][2], self.v[0][3]))
      print('{0:4} {1:4} {2:4} {3:4}'.format(self.v[1][0], self.v[1][1], self.v[1][2], self.v[1][3]))
      print('{0:4} {1:4} {2:4} {3:4}'.format(self.v[2][0], self.v[2][1], self.v[2][2], self.v[2][3]))
      print('{0:4} {1:4} {2:4} {3:4}'.format(self.v[3][0], self.v[3][1], self.v[3][2], self.v[3][3]))
      print('得分为:{0:4}'.format(self.totalScore))
      print('游戏是否结束:{0:4}'.format(self.isOver()))
     #重新排列
     def align(self,vList, direction):
      for i in range(vList.count(0)):
       vList.remove(0)
      zeros = [0 for x in range(4-len(vList))]
      if direction == 'left':
       vList.extend(zeros)
      else:
       vList[:0] = zeros
     #将相同的元素相加,返回新增积分
     def addSame(self,vList, direction):
      increment=0
      if direction == 'left':
       for i in [0,1,2]:
        if vList[i]==vList[i+1] and vList[i+1]!=0:
         vList[i] *= 2
         vList[i+1] = 0
         increment += vList[i]
      else:
       for i in [3,2,1]:
        if vList[i]==vList[i-1] and vList[i-1]!=0:
         vList[i] *= 2
         vList[i-1] = 0
         increment += vList[i]
      return increment
     #处理行和方向,返回新增积分
     def handle(self, vList, direction):
      self.align(vList, direction)
      increment = self.addSame(vList, direction)
      self.align(vList, direction)
      self.totalScore += increment #直接加到总值
      return increment
     #判断游戏是否结束
     def judge(self):
       
      if self.isOver():
       print('你输了,游戏结束!')
       return False
      else:
       if self.totalScore >= 2048:
        print('你赢了,游戏结束!但是你还可以继续玩。')
       return True
     #判断游戏是否真正结束
     def isOver(self):
      N = self.calcCharNumber(0)
      if N!=0:
       return False
      else:
       for row in range(4):
        flag = self.isListOver(self.v[row])
        if flag==False:
         return False 
       for col in range(4):
        # 将矩阵中一列复制到一个列表中然后处理
        vList = [self.v[row][col] for row in range(4)]
        flag = self.isListOver(vList)
        if flag==False:
         return False
      return True
      
     #判断一个列表是否还可以合并
     def isListOver(self, vList):
      for i in [0,1,2]:
       if vList[i]==vList[i+1] and vList[i+1]!=0:
        return False
      return True
     def calcCharNumber(self, char):
      n = 0
      for q in self.v:
       n += q.count(char)
      return n
     def addElement(self):
      # 统计空白区域数目 N
      N = self.calcCharNumber(0)
      if N!=0:
       # 按2和4出现的几率为3/1来产生随机数2和4
       num = random.choice([2, 2, 2, 4]) 
       # 产生随机数k,上一步产生的2或4将被填到第k个空白区域
       k = random.randrange(1, N+1) #k的范围为[1,N]
       n = 0
       for i in range(4):
        for j in range(4):
         if self.v[i][j] == 0:
          n += 1
          if n == k:
           self.v[i][j] = num
           return
     
         
     def moveLeft(self):
      self.moveHorizontal('left')
     def moveRight(self):
      self.moveHorizontal('right')
     def moveHorizontal(self, direction):
      for row in range(4):
       self.handle(self.v[row], direction)
     
     def moveUp(self):
      self.moveVertical('left')
     def moveDown(self):
      self.moveVertical('right')
     def moveVertical(self, direction):
      for col in range(4):
       # 将矩阵中一列复制到一个列表中然后处理
       vList = [self.v[row][col] for row in range(4)]
       self.handle(vList, direction)
       # 从处理后的列表中的数字覆盖原来矩阵中的值
       for row in range(4):
        self.v[row][col] = vList[row]
         
     #主要的处理函数
     def operation(self):
      op = input('operator:')
      if op in ['a', 'A']: # 向左移动
       self.moveLeft()
       self.addElement()
      elif op in ['d', 'D']: # 向右移动
       self.moveRight()
       self.addElement()
      elif op in ['w', 'W']: # 向上移动
       self.moveUp()
       self.addElement()
      elif op in ['s', 'S']: # 向下移动
       self.moveDown()
       self.addElement()
      else:
       print('错误的输入。请输入 [W, S, A, D] 或者是其小写') 
     
    #开始
    print('输入:W(上移) S(下移) A(左移) D(右移), press <CR>.')
    g =game2048()
    flag = True
    while True:
     g.display()
     flag = g.judge()
     g.operation()
     flag = g.judge()

    演示图

    以上所述就是本文的全部内容了,希望大家能够喜欢。

    jsjbwy
    下一篇:没有了