Question

>>> def rotate_matrix( k: List[List[int]]):
    """
    For example, if I have:
    m = [[1,2,3],
         [2,3,3],
         [5,4,3]]
    rotate_matrix(m) should give me [[3,3,3],[2,3,4],[1,2,5]]. 
    """

编辑：最好没有numpy。

Answer 1

您可以使用numpy函数rot90

import numpy as np
m = np.array([[1,2,3],
         [2,3,3],
         [5,4,3]])

def rotate_matrix(mat):
    return np.rot90(mat)

Answer 2

在纯python中，这是逆时针旋转矩阵作为一行（即，不带numpy）：

new_matrix = [[m[j][i] for j in range(len(m))] for i in range(len(m[0])-1,-1,-1)]

如果要在函数中执行此操作，则

def rotate_matrix( m ):
    return [[m[j][i] for j in range(len(m))] for i in range(len(m[0])-1,-1,-1)]

以及任一方式，

的结果

m = [ [1,2,3], [2,3,3], [5,4,3]]

是

[[3, 3, 3], [2, 3, 4], [1, 2, 5]]

此外，如果您想要常规的转置，那么简单的一行纯python版本就是

[[m[j][i] for j in range(len(m))] for i in range(len(m[0]))]

Answer 3

看看这段代码。这是使用scratch将任何矩阵旋转90度的方法

[这是将矩阵旋转 90 度的代码][1]

    n = input("enter input:")
    l =[]

    while (n !=""):
    
        l.append(n.split(" "))
        n = input("enter input:")
        l1=[]
    for i in range(len(l[0])):
        l1.append([])
    for j in range (len(l1)):
        for p in range(len(l)):
           l1[j].append(l[p][-(j+1)])

    for s in l1:
         print(*s)
'''


  [1]: https://i.stack.imgur.com/c8jHC.png

Answer 4

def rotate_90_degree_anticlckwise(matrix):
    new_matrix = []
    for i in range(len(matrix[0]), 0, -1):
        new_matrix.append(list(map(lambda x: x[i-1], matrix)))

    return new_matrix


def rotate_90_degree_clckwise(matrix):
    new_matrix = []
    for i in range(len(matrix[0])):
        li = list(map(lambda x: x[i], matrix))
        li.reverse()
        new_matrix.append(li)

    return new_matrix

def rotate_90_degree_clckwise_by_zip_method(matrix):
    return list(list(x)[::-1] for x in zip(*matrix))

Answer 5

只需垂直翻转矩阵，然后将右上三角形与左下三角形切换

def rotate_matrix_ccw(mat):
    if mat is None:
        return None
    n = len(mat)
    if n == 1:
        return mat
    for i in range(n):
        if len(mat[i]) != n:
            raise Exception("Matrix must be square")
    # flip the matrix vertically
    for j in range(n // 2):
        for i in range(n):
            mat[i][j], mat[i][n - 1 - j] = mat[i][n - 1 - j], mat[i][j]
    # switch the upper-right triangle with the lower-left triangle
    for i in range(n):
        for j in range(i):
            mat[i][j], mat[j][i] = mat[j][i], mat[i][j]
    return mat

Answer 6

#solution1

print(list(list(x) for x in zip(*m))[::-1])

#solution2

print([[x[i] for x in m] for i in range(len(m))][::-1])

#solution3

r = []
i = 0
for row in m:
  listToAdd = [item[i] for item in m]
  r.append(listToAdd)
  i +=1
print(r[::-1])

在Python中，如何将矩阵逆时针旋转90度？

6 个答案: