首页 > Python资料 博客日记
python|闲谈2048小游戏和数组的旋转及翻转和转置
2024-02-29 09:00:03Python资料围观164次
目录
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 = lst
def __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 %= 4
if 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 matrix
return 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 %= 4
if 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 matrix
return res
def 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
标签:
相关文章
最新发布
- 【Python】selenium安装+Microsoft Edge驱动器下载配置流程
- Python 中自动打开网页并点击[自动化脚本],Selenium
- Anaconda基础使用
- 【Python】成功解决 TypeError: ‘<‘ not supported between instances of ‘str’ and ‘int’
- manim边学边做--三维的点和线
- CPython是最常用的Python解释器之一,也是Python官方实现。它是用C语言编写的,旨在提供一个高效且易于使用的Python解释器。
- Anaconda安装配置Jupyter(2024最新版)
- Python中读取Excel最快的几种方法!
- Python某城市美食商家爬虫数据可视化分析和推荐查询系统毕业设计论文开题报告
- 如何使用 Python 批量检测和转换 JSONL 文件编码为 UTF-8
点击排行
- 版本匹配指南:Numpy版本和Python版本的对应关系
- 版本匹配指南:PyTorch版本、torchvision 版本和Python版本的对应关系
- Python 可视化 web 神器:streamlit、Gradio、dash、nicegui;低代码 Python Web 框架:PyWebIO
- 相关性分析——Pearson相关系数+热力图(附data和Python完整代码)
- Python与PyTorch的版本对应
- Anaconda版本和Python版本对应关系(持续更新...)
- Python pyinstaller打包exe最完整教程
- Windows上安装 Python 环境并配置环境变量 (超详细教程)