本文主要是介绍python｜闲谈2048小游戏和数组的旋转及翻转和转置，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
目录
2048
生成数组
n阶方阵
方阵旋转
顺时针旋转
逆时针旋转
mxn矩阵
矩阵旋转
测试代码
测试结果
翻转和转置
2048
《2048》是一款比较流行的数字游戏，最早于2014年3月20日发行。原版2048由Gabriele Cirulli首先在GitHub上发布，后被移植到各个平台，并且衍生出不计其数的版本。但在网上看到，居说它也不算是原创，是基于《1024》和《小3传奇》的玩法开发而成的；还有一说，它来源于另一款游戏《Threes!》，由Asher Vollmer和Greg Wohlwend合作开发，于2014年2月6日在App Store上架。
2048游戏规则很简单，游戏开始时在4x4的方格中随机出现数字2，每次可以选择上下左右其中一个方向去滑动，每滑动一次，所有的数字方块都会往滑动的方向靠拢外，相邻的相同数字在靠拢时会相加，系统也会在空白的格子里随机增加一个数字2或4。玩家要想办法在这16格范围中，不断上下左右滑动相加数字，从而凑出“2048”这个数字方块。
实际上，这个游戏就是在操作一个4x4的二维数组，数组的元素只要1-11就行了，因为2的11次方就是2048。同样，相邻相同数字的累加就变成了相邻相同指数的递增1。
在编写这个2048游戏前，先来谈谈4x4数组的操作，对python来说虽然也有数组，但通常会用列表来操作。以下就在IDLE shell上流水账操作：
生成数组
16个数字的列表推导式：
>>> [i for i in range(16)]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
用*解包更pythonic：
>>> [*range(16)]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
分割成4x4二维列表：
>>> [[*range(16)][i*4:i*4+4] for i in range(4)]
[[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14, 15]]
只是数列如此写法可能更好：
>>> [[*range(i*4,i*4+4)] for i in range(4)]
[[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14, 15]]
全0列表：
>>> [[0]*4 for _ in range(4)]
[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
n阶方阵
从4阶方阵扩展到n阶：
>>> matrix = lambda n:[[*range(i*n,i*n+n)] for i in range(n)]
>>> matrix(4)
[[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14, 15]]
>>> matrix(5)
[[0, 1, 2, 3, 4], [5, 6, 7, 8, 9], [10, 11, 12, 13, 14], [15, 16, 17, 18, 19], [20, 21, 22, 23, 24]]
>>> matrix(6)
[[0, 1, 2, 3, 4, 5], [6, 7, 8, 9, 10, 11], [12, 13, 14, 15, 16, 17], [18, 19, 20, 21, 22, 23], [24, 25, 26, 27, 28, 29], [30, 31, 32, 33, 34, 35]]
随机生成数字1或2，比例为3:1：
>>> from random import sample as rnd
>>> rnd([1,1,1,2],1)
[1]
>>> rnd([1,1,1,2],1)
[2]
>>> rnd([1,1,1,2],1)
[2]
>>> rnd([1,1,1,2],1)
随机产生1或者2个“1”，比例为2:1：
>>> from random import sample as rnd
>>> x = 4
>>> rnd([0]*(x*x-2)+rnd([0,1,1],2),x*x)
[0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
>>> rnd([0]*(x*x-2)+rnd([0,1,1],2),x*x)
[0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0]
>>> rnd([0]*(x*x-2)+rnd([0,1,1],2),x*x)
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0]
x = 5
rnd([0]*(x*x-2)+rnd([0,1,1],2),x*x)
[0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
rnd([0]*(x*x-2)+rnd([0,1,1],2),x*x)
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0]
rnd([0]*(x*x-2)+rnd([0,1,1],2),x*x)
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1]
rnd([0]*(x*x-2)+rnd([0,1,1],2),x*x)
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
方阵旋转
numpy有现成的函数rot90()，表示顺时针旋转数组90度。
>>> import numpy as np
>>> np.array(range(16))
array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15])
>>> np.array([[*range(i*4,i*4+4)] for i in range(4)])
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15]])
>>> array = np.array([[*range(i*4,i*4+4)] for i in range(4)])
逆时针旋转，参数k为正数：
>>> np.rot90(array)
array([[ 3, 7, 11, 15],
[ 2, 6, 10, 14],
[ 1, 5, 9, 13],
[ 0, 4, 8, 12]])
>>> np.rot90(array, k=2)
array([[15, 14, 13, 12],
[11, 10, 9, 8],
[ 7, 6, 5, 4],
[ 3, 2, 1, 0]])
>>> np.rot90(array, k=3)
array([[12, 8, 4, 0],
[13, 9, 5, 1],
[14, 10, 6, 2],
[15, 11, 7, 3]])
顺时针旋转，参数k为负数：
>>> np.rot90(array, k=-1)
array([[12, 8, 4, 0],
[13, 9, 5, 1],
[14, 10, 6, 2],
[15, 11, 7, 3]])
>>> np.rot90(array, k=-2)
array([[15, 14, 13, 12],
[11, 10, 9, 8],
[ 7, 6, 5, 4],
[ 3, 2, 1, 0]])
>>> np.rot90(array, k=-3)
array([[ 3, 7, 11, 15],
[ 2, 6, 10, 14],
[ 1, 5, 9, 13],
[ 0, 4, 8, 12]])
不使用numpy，只用列表推导式也能实现旋转：
顺时针旋转
>>> matrix = lambda n:[[*range(i*n,i*n+n)] for i in range(n)]
>>> mat4 = matrix(4)
>>> mat4
[[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14, 15]]
>>> [[mat[len(mat[0])-j-1][i] for j in range(len(mat[0]))] for i in range(len(mat))]
[[12, 8, 4, 0], [13, 9, 5, 1], [14, 10, 6, 2], [15, 11, 7, 3]]
写一个模拟np.array的__repr__方法来检测旋转效果：
class List():# 仅支持二维数组的展示def __init__(self, lst):self.x = lstdef __repr__(self):n = len(str(max(sum(self.x,[]))))res = []for mat in self.x:res.append(', '.join(f'{x:>{n}}' for x in mat))return '],\n\t['.join(res).join(['Array([ [','] ])'])检测结果如下：
>>> matrix = lambda n:[[*range(i*n,i*n+n)] for i in range(n)]
>>> rotate = lambda m: [[m[len(m)-j-1][i] for j in range(len(m))] for i in range(len(m[0]))]
>>> mat4 =matrix(4)
>>> List(mat4)
Array([ [ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15] ])
>>> List(rotate(mat4))
Array([ [12, 8, 4, 0],
[13, 9, 5, 1],
[14, 10, 6, 2],
[15, 11, 7, 3] ])
>>> List(rotate(rotate(mat4)))
Array([ [15, 14, 13, 12],
[11, 10, 9, 8],
[ 7, 6, 5, 4],
[ 3, 2, 1, 0] ])
>>> List(rotate(rotate(rotate(mat4))))
Array([ [ 3, 7, 11, 15],
[ 2, 6, 10, 14],
[ 1, 5, 9, 13],
[ 0, 4, 8, 12] ])
>>> List(rotate(rotate(rotate(rotate(mat4)))))
Array([ [ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15] ])
结果符合预期，旋转4次恢复原样；同样更高阶方阵也符合：
>>> List(matrix(5))
Array([ [ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19],
[20, 21, 22, 23, 24] ])
>>> List(rotate(matrix(5)))
Array([ [20, 15, 10, 5, 0],
[21, 16, 11, 6, 1],
[22, 17, 12, 7, 2],
[23, 18, 13, 8, 3],
[24, 19, 14, 9, 4] ])
逆时针旋转
>>> matrix = lambda n:[[*range(i*n,i*n+n)] for i in range(n)]
>>> rotate2 = lambda m:[[m[j][len(m[0])-i-1] for j in range(len(m))] for i in range(len(m[0]))]
>>> List(rotate2(matrix(4)))
Array([ [ 3, 7, 11, 15],
[ 2, 6, 10, 14],
[ 1, 5, 9, 13],
[ 0, 4, 8, 12] ])
>>> List(rotate2(rotate2(matrix(4))))
Array([ [15, 14, 13, 12],
[11, 10, 9, 8],
[ 7, 6, 5, 4],
[ 3, 2, 1, 0] ])
>>> List(rotate2(rotate2(rotate2(matrix(4)))))
Array([ [12, 8, 4, 0],
[13, 9, 5, 1],
[14, 10, 6, 2],
[15, 11, 7, 3] ])
>>> List(rotate2(rotate2(rotate2(rotate2(matrix(4))))))
Array([ [ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15] ])
>>> List(rotate2(matrix(5)))
Array([ [ 4, 9, 14, 19, 24],
[ 3, 8, 13, 18, 23],
[ 2, 7, 12, 17, 22],
[ 1, 6, 11, 16, 21],
[ 0, 5, 10, 15, 20] ])
>>> List(rotate2(rotate2(matrix(5))))
Array([ [24, 23, 22, 21, 20],
[19, 18, 17, 16, 15],
[14, 13, 12, 11, 10],
[ 9, 8, 7, 6, 5],
[ 4, 3, 2, 1, 0] ])
mxn矩阵
把方阵拓展到矩阵：
>>> matrix = lambda m, n: [[i * n + j for j in range(n)] for i in range(m)]
>>> List(matrix(3,4))
Array([ [ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11] ])
>>> List(matrix(5,4))
Array([ [ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15],
[16, 17, 18, 19] ])
>>> List(matrix(5,5))
Array([ [ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19],
[20, 21, 22, 23, 24] ])
矩阵旋转
rotate顺时针旋转，rotate2逆时针旋转
>>> matrix = lambda m, n: [[i * n + j for j in range(n)] for i in range(m)]
>>> rotate = lambda m: [[m[len(m)-j-1][i] for j in range(len(m))] for i in range(len(m[0]))]
>>> rotate2 = lambda m:[[m[j][len(m[0])-i-1] for j in range(len(m))] for i in range(len(m[0]))]
>>> List(matrix(3,4))
Array([ [ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11] ])
>>> List(rotate(matrix(3,4)))
Array([ [ 8, 4, 0],
[ 9, 5, 1],
[10, 6, 2],
[11, 7, 3] ])
>>> List(rotate2(rotate2(rotate2(matrix(3,4)))))
Array([ [ 8, 4, 0],
[ 9, 5, 1],
[10, 6, 2],
[11, 7, 3] ])
>>> List(rotate(rotate(matrix(3,4))))
Array([ [11, 10, 9, 8],
[ 7, 6, 5, 4],
[ 3, 2, 1, 0] ])
>>> List(rotate2(rotate2(matrix(3,4))))
Array([ [11, 10, 9, 8],
[ 7, 6, 5, 4],
[ 3, 2, 1, 0] ])
>>> List(rotate(rotate(rotate(matrix(3,4)))))
Array([ [ 3, 7, 11],
[ 2, 6, 10],
[ 1, 5, 9],
[ 0, 4, 8] ])
>>> List(rotate2(matrix(3,4)))
Array([ [ 3, 7, 11],
[ 2, 6, 10],
[ 1, 5, 9],
[ 0, 4, 8] ])
>>> List(rotate(rotate(rotate(rotate(matrix(3,4))))))
Array([ [ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11] ])
List(rotate2(rotate2(rotate2(rotate2(matrix(3,4))))))
Array([ [ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11] ])
旋转函数还能写成如下形式，只是坐标与range参数的互调形式：
>>> rotate = lambda m: [[m[j][i] for j in range(len(m)-1,-1,-1)] for i in range(len(m[0]))]
>>> rotate2 = lambda m: [[m[j][i] for j in range(len(m))] for i in range(len(m[0])-1,-1,-1)]
lambda匿名函数虽然很简洁，但没有普通函数易懂，我们把lambda函数改成模拟np.rot90()的普通函数rotate(matrix, k=1)，其中参数k为90度的倍数，正数顺时针旋转，负数则逆时针旋转：
def rotate(matrix, k=1):rows = len(matrix)cols = len(matrix[0])res = [[0]*rows for _ in range(cols)]k %= 4if k==1:for i in range(rows):for j in range(cols):res[j][rows-i-1] = matrix[i][j]elif k==2:res = [[0]*cols for _ in range(rows)]for i in range(rows):for j in range(cols):res[rows-i-1][cols-j-1] = matrix[i][j]elif k==3:for i in range(rows):for j in range(cols):res[cols-j-1][i] = matrix[i][j]else:return matrixreturn res测试代码
def rotate(matrix, k=1):rows = len(matrix)cols = len(matrix[0])res = [[0]*rows for _ in range(cols)]k %= 4if k==1:for i in range(rows):for j in range(cols):res[j][rows-i-1] = matrix[i][j]elif k==2:res = [[0]*cols for _ in range(rows)]for i in range(rows):for j in range(cols):res[rows-i-1][cols-j-1] = matrix[i][j]elif k==3:for i in range(rows):for j in range(cols):res[cols-j-1][i] = matrix[i][j]else:return matrixreturn resdef show(matrix):n = len(str(max(sum(matrix,[]))))res = []for mat in matrix:res.append(', '.join(f'{x:>{n}}' for x in mat))print('],\n\t['.join(res).join(['Array([ [','] ])']))matrix = lambda m, n: [[i * n + j for j in range(n)] for i in range(m)]for i in range(-4,5):show(rotate(matrix(4,4), i))for i in range(-4,5):show(rotate(matrix(5,3), i))测试结果
Array([ [ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15] ])
Array([ [12, 8, 4, 0],
[13, 9, 5, 1],
[14, 10, 6, 2],
[15, 11, 7, 3] ])
Array([ [15, 14, 13, 12],
[11, 10, 9, 8],
[ 7, 6, 5, 4],
[ 3, 2, 1, 0] ])
Array([ [ 3, 7, 11, 15],
[ 2, 6, 10, 14],
[ 1, 5, 9, 13],
[ 0, 4, 8, 12] ])
Array([ [ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15] ])
Array([ [12, 8, 4, 0],
[13, 9, 5, 1],
[14, 10, 6, 2],
[15, 11, 7, 3] ])
Array([ [15, 14, 13, 12],
[11, 10, 9, 8],
[ 7, 6, 5, 4],
[ 3, 2, 1, 0] ])
Array([ [ 3, 7, 11, 15],
[ 2, 6, 10, 14],
[ 1, 5, 9, 13],
[ 0, 4, 8, 12] ])
Array([ [ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15] ])
Array([ [ 0, 1, 2],
[ 3, 4, 5],
[ 6, 7, 8],
[ 9, 10, 11],
[12, 13, 14] ])
Array([ [12, 9, 6, 3, 0],
[13, 10, 7, 4, 1],
[14, 11, 8, 5, 2] ])
Array([ [14, 13, 12],
[11, 10, 9],
[ 8, 7, 6],
[ 5, 4, 3],
[ 2, 1, 0] ])
Array([ [ 2, 5, 8, 11, 14],
[ 1, 4, 7, 10, 13],
[ 0, 3, 6, 9, 12] ])
Array([ [ 0, 1, 2],
[ 3, 4, 5],
[ 6, 7, 8],
[ 9, 10, 11],
[12, 13, 14] ])
Array([ [12, 9, 6, 3, 0],
[13, 10, 7, 4, 1],
[14, 11, 8, 5, 2] ])
Array([ [14, 13, 12],
[11, 10, 9],
[ 8, 7, 6],
[ 5, 4, 3],
[ 2, 1, 0] ])
Array([ [ 2, 5, 8, 11, 14],
[ 1, 4, 7, 10, 13],
[ 0, 3, 6, 9, 12] ])
Array([ [ 0, 1, 2],
[ 3, 4, 5],
[ 6, 7, 8],
[ 9, 10, 11],
[12, 13, 14] ])
翻转和转置
翻转可以是水平方向和重置方向的：
>>> matrix = lambda m, n: [[i * n + j for j in range(n)] for i in range(m)]
>>> flipH = lambda m: [[m[i][len(m[0])-j-1] for j in range(len(m[0]))] for i in range(len(m))]
>>> flipV = lambda m: [[m[len(m)-j-1][i] for i in range(len(m[0]))] for j in range(len(m))]
>>> List(flipH(matrix(4,4)))
Array([ [ 3, 2, 1, 0],
[ 7, 6, 5, 4],
[11, 10, 9, 8],
[15, 14, 13, 12] ])
>>> List(flipV(matrix(4,4)))
Array([ [12, 13, 14, 15],
[ 8, 9, 10, 11],
[ 4, 5, 6, 7],
[ 0, 1, 2, 3] ])
>>> List(flipH(matrix(3,5)))
Array([ [ 4, 3, 2, 1, 0],
[ 9, 8, 7, 6, 5],
[14, 13, 12, 11, 10] ])
>>> List(flipV(matrix(3,5)))
Array([ [10, 11, 12, 13, 14],
[ 5, 6, 7, 8, 9],
[ 0, 1, 2, 3, 4] ])
>>> List(flipH(matrix(5,4)))
Array([ [ 3, 2, 1, 0],
[ 7, 6, 5, 4],
[11, 10, 9, 8],
[15, 14, 13, 12],
[19, 18, 17, 16] ])
>>> List(flipV(matrix(5,4)))
Array([ [16, 17, 18, 19],
[12, 13, 14, 15],
[ 8, 9, 10, 11],
[ 4, 5, 6, 7],
[ 0, 1, 2, 3] ])
转置可以看作是翻转和旋转的组合，对方阵来说就是以对角线为轴的翻转：
>>> transpose = lambda m: [[m[j][i] for j in range(len(m))] for i in range(len(m[0]))]
>>> List(transpose(matrix(4,4)))
Array([ [ 0, 4, 8, 12],
[ 1, 5, 9, 13],
[ 2, 6, 10, 14],
[ 3, 7, 11, 15] ])
>>> List(transpose(transpose(matrix(4,4))))
Array([ [ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15] ])
>>> List(rotate(matrix(4,4)))
Array([ [12, 8, 4, 0],
[13, 9, 5, 1],
[14, 10, 6, 2],
[15, 11, 7, 3] ])
>>> List(flipH(rotate(matrix(4,4))))
Array([ [ 0, 4, 8, 12],
[ 1, 5, 9, 13],
[ 2, 6, 10, 14],
[ 3, 7, 11, 15] ])
>>> List(rotate2(matrix(4,4)))
Array([ [ 3, 7, 11, 15],
[ 2, 6, 10, 14],
[ 1, 5, 9, 13],
[ 0, 4, 8, 12] ])
>>> List(flipV(rotate2(matrix(4,4))))
Array([ [ 0, 4, 8, 12],
[ 1, 5, 9, 13],
[ 2, 6, 10, 14],
[ 3, 7, 11, 15] ])
在numpy中，转置由.T属性完成
>>> import numpy as np
>>> arr = np.array(matrix(3,4))
>>> arr
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
>>> arr.T
array([[ 0, 4, 8],
[ 1, 5, 9],
[ 2, 6, 10],
[ 3, 7, 11]])
>>> arr = np.array(matrix(4,4))
>>> arr.T
array([[ 0, 4, 8, 12],
[ 1, 5, 9, 13],
[ 2, 6, 10, 14],
[ 3, 7, 11, 15]])
>>> arr.T.T
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15]])
>>> arr = np.array(matrix(5,4))
>>> arr.T
array([[ 0, 4, 8, 12, 16],
[ 1, 5, 9, 13, 17],
[ 2, 6, 10, 14, 18],
[ 3, 7, 11, 15, 19]])
完
这篇关于python｜闲谈2048小游戏和数组的旋转及翻转和转置的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
