算法之二维数组转置

现在有个二维数组：data=[0,1,2,3],[0,1,2,3],[0,1,2,3],[0,1,2,3],每一行都是[0,1,2,3]，要求顺时针旋转90度，变成每一列都是[0,1,2,3]。

#首先生成一维数组的语句是：
row=[i for i in range(6)]
print(row)
#运行结果：
#[0, 1, 2, 3, 4, 5]
#生成二维数组的语句是：
data=[[i for i in range(4)] for k in range(4)]
for j in data:
    print(j)
'''
相当于是把一维数组的整体作为上面的i
运行结果是：
[0, 1, 2, 3]
[0, 1, 2, 3]
[0, 1, 2, 3]
[0, 1, 2, 3]
'''

 数组扩充行和列：

def trans(m):
    a = [[] for i in m[0]]
    for i in m:
        for j in range(len(i)):
            a[j].append(i[j])
    return a
m = [[1, 2], [3, 4], [5, 6]]    # 想像第一个列表是原始的，后面的是往里添加的
print(trans(m))    # result:[[1, 3, 5], [ 2, 4, 6]]
#来自于https://www.cnblogs.com/anpengapple/p/5427367.html

矩阵转置（含过程）：

m = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
def trans(m):
    for i in range(len(m)):
        for j in range(i):
            m[i][j], m[j][i] = m[j][i], m[i][j]
            print(m)
trans(m)
'''
运行结果：
[[1, 4, 3], [2, 5, 6], [7, 8, 9]]
[[1, 4, 7], [2, 5, 6], [3, 8, 9]]
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]
'''

矩阵转置（不含中间过程）：

m = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
def trans(m):
    for i in range(len(m)):
        for j in range(i):
            m[i][j], m[j][i] = m[j][i], m[i][j]
    return m
print(trans(m))
#注意：只适合方阵

 矩阵的运算参考：https://www.cnblogs.com/chamie/p/4870078.html