转置矩阵（Python）

题目：

#方法1
lis = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
count = 0
for i in range(len(lis)):
    for j in range(len(lis[i])):
        if i > j:
            lis[i][j], lis[j][i] = lis[j][i], lis[i][j]
            count += 1
print(*lis, sep = '\n')
print(count)

结果：

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

#方法二
lis = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
count = 0
for i, _ in enumerate(lis):
    for j, _ in enumerate(lis[i]):
        if i > j:
            lis[i][j], lis[j][i] = lis[j][i], lis[i][j]
            count += 1
print(*lis, sep = '\n')
print(count)
    

结果：

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

#方法三
lis = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
count = 0
for i in range(len(lis)):
    for j in range(i):
        lis[i][j], lis[j][i] = lis[j][i], lis[i][j]
        count += 1
print(*lis, sep = '\n')
print(count)

结果：

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

二乘三转置矩阵：

#方法四
lis = [[1, 2, 3], [4, 5, 6]]
list_ = []

for i, row in enumerate(lis):
    for j, col in enumerate(row):
        if i == 0:
            list_.append([])
        list_[j].append(col)
print(*list_, sep='\n')
或者
lis = [[1, 2, 3], [4, 5, 6]]
list_ = []

for i, row in enumerate(lis):
    for j, col in enumerate(row):
        if len(list_) < j + 1:
            list_.append([])
        list_[j].append(col)
print(*list_, sep='\n')

结果：

[1, 4]
[2, 5]
[3, 6]

方法五
lis = [[1, 2, 3], [4, 5, 6]]
list_ = [[None for i in range(len(lis))] for i in range(len(lis[0]))]
for i in range(len(lis)):
    for j in range(len(lis[0])):
        list_[j][i] = lis[i][j]
print(*list_, sep='\n')```
结果：
```python
[1, 4]
[2, 5]
[3, 6]