Question

给定二维矩阵，例如

l = [[1,1,1],
     [2,5,2],
     [3,3,3]])

对列和行实施移位操作的最有效方法是什么？

E.g。

shift('up', l) 

[[2, 5, 2],
 [3, 3, 3],
 [1, 1, 1]]

但

shift('left', l) 

[[1, 1, 1],
 [5, 2, 2],
 [3, 3, 3]]

由于this answer，我在两个深度使用collections.deque但是'up'或'down'只需要1个班次，'left'或'right'需要N个班次（my实现是为每一行使用for循环。）

在C中我认为可以使用指针算法来改进（参见例如this answer）。

有更好的pythonic方式吗？

修改

通过有效我的意思是，如果有办法避免N班次。
我们可以假设矩阵是平方的。
转变可以到位。

感谢martineau指出了问题的这些重点。对不起，我之前没有指出过它们。

Answer 1

Numpy提供了一个名为roll（）的方法来移动条目。

>>> import numpy as np
>>> x = np.arange(9)
>>> x = x.reshape(3, 3)
>>> print(x)

[[0 1 2]
 [3 4 5]
 [6 7 8]]

>>> x = np.roll(x, -1, axis=0) # up
>>> print(x)

[[3 4 5]
 [6 7 8]
 [0 1 2]]

>>> x = np.roll(x, 1, axis=0) # down
>>> print(x)

[[0 1 2]
 [3 4 5]
 [6 7 8]]

>>> x = np.roll(x, 2, axis=1) # right
>>> print(x)

[[1 2 0]
 [4 5 3]
 [7 8 6]]

>>> x = np.roll(x, -2, axis=1) # left
>>> print(x)    

[[0 1 2]
 [3 4 5]
 [6 7 8]]

我认为与大多数解决方案相比，Numpy将非常有效在矩阵运算方面，您不会受到二维矩阵的约束。

Answer 2

这是一种非常有效的方法，可以使用非方形矩阵：

DIRS = NONE, UP, DOWN, LEFT, RIGHT = 'unshifted', 'up', 'down', 'left', 'right'

def shift(matrix, direction, dist):
    """ Shift a 2D matrix in-place the given distance of rows or columns in the
        specified (NONE, UP, DOWN, LEFT, RIGHT) direction and return it.
    """
    if dist and direction in (UP, DOWN, LEFT, RIGHT):
        n = 0
        if direction in (UP, DOWN):
            n = (dist % len(matrix) if direction == UP else -(dist % len(matrix)))
        elif direction in (LEFT, RIGHT):
            n = (dist % len(matrix[0]) if direction == LEFT else -(dist % len(matrix[0])))
            matrix[:] = list(zip(*matrix))  # Transpose rows and columns for shifting.

        h = matrix[:n]
        del matrix[:n]
        matrix.extend(h)

        if direction in (LEFT, RIGHT):
            matrix[:] = map(list, zip(*matrix))  # Undo previous transposition.

    return matrix


if __name__ == '__main__':

    # Some non-square test matrices.
    matrix1 = [[1, 2, 3],
               [4, 5, 6],
               [7, 8, 9],
               [10, 11, 12]]

    matrix2 = [[1, 2, 3, 4],
               [5, 6, 7, 8],
               [9, 10, 11, 12]]

    def shift_and_print(matrix, direction, dist):
        GAP =  2  # Plus one for a ":" character.
        indent = max(map(len, DIRS)) + GAP
        print(direction
                + ': ' + (indent-2-len(direction))*' '
                + ('\n'+indent*' ').join(map(str, shift(matrix, direction, dist)))
                + '\n')

    for matrix in matrix1, matrix2:
        for direction in DIRS:
            shift_and_print(matrix, direction, 1)  # Printed results are cumulative.

输出（注意，结果是累积的，因为操作是就地执行的，并且移位应用于上一次调用的结果）：

no shift: [1, 2, 3]
          [4, 5, 6]
          [7, 8, 9]
          [10, 11, 12]

up:       [4, 5, 6]
          [7, 8, 9]
          [10, 11, 12]
          [1, 2, 3]

down:     [1, 2, 3]
          [4, 5, 6]
          [7, 8, 9]
          [10, 11, 12]

left:     [2, 3, 1]
          [5, 6, 4]
          [8, 9, 7]
          [11, 12, 10]

right:    [1, 2, 3]
          [4, 5, 6]
          [7, 8, 9]
          [10, 11, 12]

no shift: [1, 2, 3, 4]
          [5, 6, 7, 8]
          [9, 10, 11, 12]

up:       [5, 6, 7, 8]
          [9, 10, 11, 12]
          [1, 2, 3, 4]

down:     [1, 2, 3, 4]
          [5, 6, 7, 8]
          [9, 10, 11, 12]

left:     [2, 3, 4, 1]
          [6, 7, 8, 5]
          [10, 11, 12, 9]

right:    [1, 2, 3, 4]
          [5, 6, 7, 8]
          [9, 10, 11, 12]

Answer 3

使用numpy可能是这样的：

def shift(x, direction='up'):
    if direction == 'up':
        temp = range(x.shape[0])
        indicies = temp[1:] + [temp[0]]
        return x[indicies]
    elif direction == 'left':
        temp = range(x.shape[1])
        indicies = temp[1:] + [temp[0]]
        return x[:, indicies]
    else:
        print 'Error direction not known'

结果：

>>> shift(l, direction='up')
array([[2, 5, 2],
       [3, 3, 3],
       [1, 1, 1]])
>>> shift(l, direction='left')
array([[1, 1, 1],
       [5, 2, 2],
       [3, 3, 3]])
>>> shift(l, direction='to the moon')
Error direction not known

Answer 4

这是一个通用版本，您可以在任意次数的所有四个方向上旋转

l = [[1,1,1],
     [2,5,2],
     [3,3,3]]

def shift(direction, count, myList):
    myLen = len(myList)
    if direction == "up":
        return [myList[i % myLen] for i in range(count, count + myLen)]
    elif direction == "down":
        return [myList[-i] for i in range(count, count - myLen, -1)]
    elif direction == "left":
        tlist = zip(*myList)
        return map(list, zip(*[tlist[i % myLen] for i in range(count, count + myLen)]))
    elif direction == "right":
        tlist = zip(*myList)
        return map(list, zip(*[tlist[-i] for i in range(count, count - myLen, -1)]))

print shift("up", 1, l)
print shift("up", 2, l)
print shift("down", 2, l)
print shift("down", 1, l)
print shift("left", 1, l)
print shift("right", 1, l)

<强>输出

[[2, 5, 2], [3, 3, 3], [1, 1, 1]]
[[3, 3, 3], [1, 1, 1], [2, 5, 2]]
[[2, 5, 2], [3, 3, 3], [1, 1, 1]]
[[3, 3, 3], [1, 1, 1], [2, 5, 2]]
[[1, 1, 1], [5, 2, 2], [3, 3, 3]]
[[1, 1, 1], [2, 2, 5], [3, 3, 3]]

在两个方向上移动2D矩阵的有效方法？

4 个答案: