python｜闲谈2048小游戏和数组的旋转及翻转和转置

2048

生成数组

n阶方阵

方阵旋转

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$ ,

$1, 5, 9, 13, 17$ ,

$2, 6, 10, 14, 18$ ,

$3, 7, 11, 15, 19$ ])

完