python｜闲谈2048小游戏和数组的旋转及翻转和转置

Hann Yang2024-02-29 9:18

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(\[\[ 0, 4, 8, 12, 16\], \[ 1, 5, 9, 13, 17\], \[ 2, 6, 10, 14, 18\], \[ 3, 7, 11, 15, 19\]\]) *** ** * ** *** 完

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