Python 将外矩阵旋转n步

我有一个矩阵

matrix1 = [1,2
           3,4]

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

我想把这个矩阵旋转一步，比如r=1，那么输出就会是

output_matrix = [3,1
                 4,2]

output_matrix = [4,1,2
                 7,5,3
                 8,9,6]

我如何才能做到这一点，旋转的步骤将是动态的

我找到了这个解决方案，但这是针对固定旋转的，即步骤=1

def rotateMatrix(mat):

    if not len(mat):
        return

    top = 0
    bottom = len(mat)-1

    left = 0
    right = len(mat[0])-1

    while left < right and top < bottom:

        # Store the first element of next row,
        # this element will replace first element of
        # current row
        prev = mat[top+1][left]

        # Move elements of top row one step right
        for i in range(left, right+1):
            curr = mat[top][i]
            mat[top][i] = prev
            prev = curr

        top += 1

        # Move elements of rightmost column one step downwards
        for i in range(top, bottom+1):
            curr = mat[i][right]
            mat[i][right] = prev
            prev = curr

        right -= 1

        # Move elements of bottom row one step left
        for i in range(right, left-1, -1):
            curr = mat[bottom][i]
            mat[bottom][i] = prev
            prev = curr

        bottom -= 1

        # Move elements of leftmost column one step upwards
        for i in range(bottom, top-1, -1):
            curr = mat[i][left]
            mat[i][left] = prev
            prev = curr

        left += 1

    return mat


matrix =[[1,2,3],[4,5,6], [7,8,9]]

matrix = rotateMatrix(matrix)
# # Print modified matrix
print(matrix)

---

你可以结合旋转和转置来得到你想要的

import numpy as np

m = np.array([[1,2],[3,4]], int)
m1 = np.rot90(m)
m2 = m1.transpose()
print (m2)

这会给你

[[2 1]
 [4 3]]

随心所欲地改变

实际上你可以选择这样的元素

import numpy as np
m = np.array([[1,2],[3,4]], int)
# create tuple
new = (m[0][1],m[1][0])
# print list
print (list(new))

[2, 3]

---

这就是我将如何实施它，但我认为仍有一些改进的余地。支持负r。希望这有帮助。

为什么不多次调用旋转矩阵来实现您想要的结果呢？我可以，但这不是一个通用的解决方案。我们不能不使用任何包来实现它，我正在寻找不使用任何库的解决方案。您真的得到了他们的矩阵2示例正确吗？我对此表示怀疑。Stefan Pochmann是正确的。该解决方案不适用于3*3矩阵。是的，Steven是正确的，但举例来说，SVD可以在他的情况下提供帮助。他提到他不需要任何软件包。

def rotate(r,matrix):
        '''for rotating the values on the outer ring of a matrix of size HxW'''
        height = len(matrix)
        width = len(matrix[0])
        matrixMap = mapMatrix(height,width)
        r %= len(matrixMap)
        rotatedMap = matrixMap[-r:]+matrixMap[:-r]
        newMatrix = {el:matrix[el[0]][el[1]] for el in matrixMap}
        for i,el in enumerate(rotatedMap):
            matrix[matrixMap[i][0]][matrixMap[i][1]] = newMatrix[el]
        return matrix

def mapMatrix(h,w):
        matrix = []
        for i in range(w):
            matrix.append((0,i))
        for i in range(1,h):
            matrix.append((i,w-1))
        for i in range(w-2,-1,-1):
            matrix.append((h-1,i))
        for i in range(h-2,0,-1):
            matrix.append((i,0))
        return matrix