Python新手实现2048小游戏
程序员文章站
2022-05-02 22:04:29
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接触 Python 不久,看到很多人写2048,自己也捣鼓了一个,主要是熟悉Python语法。
程序使用Python3 写的,代码150行左右,基于控制台,方向键使用输入字符模拟。
演示图片
2048.py
# -*- coding:UTF-8 -*- #! /usr/bin/python3 import random v = [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]] def display(v, score): '''显示界面 ''' print('{0:4} {1:4} {2:4} {3:4}'.format(v[0][0], v[0][1], v[0][2], v[0][3])) print('{0:4} {1:4} {2:4} {3:4}'.format(v[1][0], v[1][1], v[1][2], v[1][3])) print('{0:4} {1:4} {2:4} {3:4}'.format(v[2][0], v[2][1], v[2][2], v[2][3])) print('{0:4} {1:4} {2:4} {3:4}'.format(v[3][0], v[3][1], v[3][2], v[3][3]), ' Total score: ', score) def init(v): '''随机分布网格值 ''' for i in range(4): v[i] = [random.choice([0, 0, 0, 2, 2, 4]) for x in range(4)] def align(vList, direction): '''对齐非零的数字 direction == 'left':向左对齐,例如[8,0,0,2]左对齐后[8,2,0,0] direction == 'right':向右对齐,例如[8,0,0,2]右对齐后[0,0,8,2] ''' # 移除列表中的0 for i in range(vList.count(0)): vList.remove(0) # 被移除的0 zeros = [0 for x in range(4 - len(vList))] # 在非0数字的一侧补充0 if direction == 'left': vList.extend(zeros) else: vList[:0] = zeros def addSame(vList, direction): '''在列表查找相同且相邻的数字相加, 找到符合条件的返回True,否则返回False,同时还返回增加的分数 direction == 'left':从右向左查找,找到相同且相邻的两个数字,左侧数字翻倍,右侧数字置0 direction == 'right':从左向右查找,找到相同且相邻的两个数字,右侧数字翻倍,左侧数字置0 ''' score = 0 if direction == 'left': for i in [0, 1, 2]: if vList[i] == vList[i+1] != 0: vList[i] *= 2 vList[i+1] = 0 score += vList[i] return {'bool':True, 'score':score} else: for i in [3, 2, 1]: if vList[i] == vList[i-1] != 0: vList[i-1] *= 2 vList[i] = 0 score += vList[i-1] return {'bool':True, 'score':score} return {'bool':False, 'score':score} def handle(vList, direction): '''处理一行(列)中的数据,得到最终的该行(列)的数字状态值, 返回得分 vList: 列表结构,存储了一行(列)中的数据 direction: 移动方向,向上和向左都使用方向'left',向右和向下都使用'right' ''' totalScore = 0 align(vList, direction) result = addSame(vList, direction) while result['bool'] == True: totalScore += result['score'] align(vList, direction) result = addSame(vList, direction) return totalScore def operation(v): '''根据移动方向重新计算矩阵状态值,并记录得分 ''' totalScore = 0 gameOver = False direction = 'left' op = input('operator:') if op in ['a', 'A']: # 向左移动 direction = 'left' for row in range(4): totalScore += handle(v[row], direction) elif op in ['d', 'D']: # 向右移动 direction = 'right' for row in range(4): totalScore += handle(v[row], direction) elif op in ['w', 'W']: # 向上移动 direction = 'left' for col in range(4): # 将矩阵中一列复制到一个列表中然后处理 vList = [v[row][col] for row in range(4)] totalScore += 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): v[row][col] = vList[row] elif op in ['s', 'S']: # 向下移动 direction = 'right' for col in range(4): # 同上 vList = [v[row][col] for row in range(4)] totalScore += handle(vList, direction) for row in range(4): v[row][col] = vList[row] else: print('Invalid input, please enter a charactor in [W, S, A, D] or the lower') return {'gameOver':gameOver, 'score':totalScore} # 统计空白区域数目 N N = 0 for q in v: N += q.count(0) # 不存在剩余的空白区域时,游戏结束 if N == 0: gameOver = True return {'gameOver':gameOver, 'score':totalScore} # 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) n = 0 for i in range(4): for j in range(4): if v[i][j] == 0: n += 1 if n == k: v[i][j] = num break return {'gameOver':gameOver, 'score':totalScore} init(v) score = 0 print('Input:W(Up) S(Down) A(Left) D(Right), press.') while True: display(v, score) result = operation(v) if result['gameOver'] == True: print('Game Over, You failed!') print('Your total score:', score) else: score += result['score'] if score >= 2048: print('Game Over, You Win!!!') print('Your total score:', score)
以上所述就是本文给大家分享的全部代码了,希望能够对大家学习Python有所帮助。