枚举函数：
enumerate(iterable[, start]) – > iterator for index, value of iterable
返回一个可迭代对象，将原有可迭代对象的元素和从start开始的数字配对。
练习题：
解答：
#方法1，列表扫描追加法
import datetime
matrix = [[1,2,3], [4,5,6]]
tm = []
count = 0
for row in matrix:
for i,col in enumerate(row):
if len(tm) < i + 1: # row有m列，tm就要就要有m行
tm.append([])
tm[i].append(col)
count += 1
print(matrix)
print(tm)
print(count)
#方法2，直接开辟目标空间，矩阵元素交换法（避免引用类型）
matrix = [[1,2,3], [4,5,6]]
tm = [[0 for col in range(len(matrix))] for row in range(len(matrix[0]))]
count = 0
# tm = []
# for row in range(len(matrix[0])):
# tm.append([])
# for col in range(len(matrix)):
# tm[row].append(0)
for i,row in enumerate(tm):
for j,col in enumerate(row):
tm[i][j] = matrix[j][i] #matrix元素搬到tm
count += 1
print(matrix)
print(tm)
print(count)
效率测试：
datetime 或者 %%timeit(Ipython中)
import datetime
matrix = [[1,2,3], [4,5,6], [7,8,9]]
start = datetime.datetime.now()
for c in range(100000):
tm = [] #目标矩阵
for row in matrix:
for i, item in enumerate(row):
if len(tm) < i + 1:
tm.append([])
tm[i].append(item)
delta = (datetime.datetime.now() – start).total_seconds()
print(delta)
print(matrix)
print(tm)
start = datetime.datetime.now()
for c in range(100000):
tm = [0]* len(matrix[0])
for i in range(len(tm)):
tm[i] = [0] * len(matrix)
for i, row in enumerate(tm):
for j, col in enumerate(row):
tm[i][j] = matrix[j][i]
delta = (datetime.datetime.now() – start).total_seconds()
print(delta)
print(matrix)
print(tm)
matrix = [[1,2,3], [4,5,6], [1,2,3], [4, 5, 6],[1,2,3], [4,5,6], [1,2,3], [4, 5, 6],[1,2,3], [4,5,6], [1,2,3], [4, 5, 6],[1,2,3], [4,5,6], [1,2,3], [4, 5, 6],[1,2,3], [4,5,6], [1,2,3], [4, 5, 6],[1,2,3], [4,5,6], [1,2,3], [4, 5, 6],[1,2,3], [4,5,6], [1,2,3], [4, 5, 6],[1,2,3], [4,5,6], [1,2,3], [4, 5, 6],[1,2,3], [4,5,6], [1,2,3], [4, 5, 6]]
4*4开始，先开辟空间效率更高啦！
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