python|闲谈2048小游戏和数组的旋转及翻转和转置

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:

python 复制代码
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:

python 复制代码
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

测试代码

python 复制代码
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]])


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