在python数组中移动行和列

Question

我正在构建一个基本的映射机器人，其中所有数据都存储在2D python数组中。 但是，我找不到任何方法可以移动整个行或列，然后在其中插入空白行/列。例如：

['#','0','0']        
['0','1','0']                                            
['#','0','0']

如果移到右边，看起来像：

['0','#','0']        
['0','0','1']                                            
['0','#','0']

要么

['#','0','#']         
['0','1','0']                                     
['0','0','0']                                     

如果向下移动看起来像：

['0','0','0']         
['#','0','#']                                     
['0','1','0']

我已经想出了每当在预定义数组之外检测到某些东西时如何扩展数组，但是我无法像上面演示的那样移动行和列。

任何帮助将不胜感激。 谢谢 ：）

Answer 1

您是否为此查看过numpy和numpy.roll？

import numpy as np
a = np.array([['#','0','0'],
['0','1','0'],                                           
['#','0','0']])

那么您可以右移：

a = np.roll(a,1)
a[:,0] = 0

左移：

a = np.roll(a,-1)
a[:,-1] = 0

上移：

a = np.roll(a,-1,axis = 0)
a[-1,:] = 0

降档：

a = np.roll(a,1,axis = 0)
a[0,:] = 0

Answer 2

您可以创建一个类来处理这种运动，如下所示：

class Board(object):
    def __init__(self, rows):
        self.rows = rows
        self.print_status()

    def print_status(self):
        for row in self.rows:
            print(row)

    def right(self):
        new_rows = []
        for row in self.rows:
            row = row[-1:] + row[:len(row)-1]
            new_rows.append(row)
        self.rows = new_rows
        self.print_status()

    def left(self):
        new_rows = []
        for row in self.rows:
            row = row[1:] + row[:1]
            new_rows.append(row)
        self.rows = new_rows
        self.print_status()

    def up(self):
        new_rows = []
        for row in self.rows[1:]:
            new_rows.append(row)
        new_rows.append(self.rows[0])
        self.rows = new_rows
        self.print_status()

    def down(self):
        new_rows = []
        new_rows.append(self.rows[-1])
        for row in self.rows[:-1]:
            new_rows.append(row)
        self.rows = new_rows
        self.print_status()

例：

>>> a = Board([[1,2,3],[4,5,6],[7,8,9]])
[1, 2, 3]
[4, 5, 6]
[7, 8, 9]

>>> a.down()
[7, 8, 9]
[1, 2, 3]
[4, 5, 6]

>>> a.up()
[1, 2, 3]
[4, 5, 6]
[7, 8, 9]

>>> a.right()
[3, 1, 2]
[6, 4, 5]
[9, 7, 8]

Answer 3

这可以通过numpy的“ roll”功能轻松完成
您的示例：

import numpy as np
a = np.array([['#','0','0'], ['0','1','0'], ['#','0','0']])

输出将是：

array([['#', '0', '0'],
   ['0', '1', '0'],
   ['#', '0', '0']], dtype='<U1')

对于第一个用例：

np.roll(a, 1, 1)

输出将是

array([['0', '#', '0'],
   ['0', '0', '1'],
   ['0', '#', '0']], dtype='<U1')

对于第二种情况，您可以简单地转置：

a.T

和输出将是

array([['#', '0', '#'],
   ['0', '1', '0'],
   ['0', '0', '0']], dtype='<U1')

第三种情况是对转置矩阵应用numpy roll操作

np.roll(a.T, 1, 0)

产量

array([['0', '0', '0'],
   ['#', '0', '#'],
   ['0', '1', '0']], dtype='<U1')

Answer 4

numpy解决方案效果很好，但是这里是纯Python解决方案，没有导入。 请注意，大多数代码仅打印结果-每卷仅使用一行代码。 我还添加了shiftedup ，将矩阵向上旋转，然后将最后一行替换为全零（尽管比这更有效）。

myarray = [
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9],
]
print('\nOriginal:')
for row in myarray:
    print(row)

rolledup = myarray[1:] + myarray[:1]
print('\nRolled up:')
for row in rolledup:
    print(row)

rolleddown = myarray[-1:] + myarray[:-1]
print('\nRolled down:')
for row in rolleddown:
    print(row)

rolledleft = [row[1:] + row[:1] for row in myarray]
print('\nRolled left:')
for row in rolledleft:
    print(row)

rolledright = [row[-1:] + row[:-1] for row in myarray]
print('\nRolled right:')
for row in rolledright:
    print(row)

shiftedup= myarray[1:] + [[0] * len(myarray[0])]
print('\nShifted up:')
for row in shiftedup:
    print(row)

从中打印输出为：

Original:
[1, 2, 3]
[4, 5, 6]
[7, 8, 9]

Rolled up:
[4, 5, 6]
[7, 8, 9]
[1, 2, 3]

Rolled down:
[7, 8, 9]
[1, 2, 3]
[4, 5, 6]

Rolled left:
[2, 3, 1]
[5, 6, 4]
[8, 9, 7]

Rolled right:
[3, 1, 2]
[6, 4, 5]
[9, 7, 8]

Shifted up:
[4, 5, 6]
[7, 8, 9]
[0, 0, 0]