python练习(7)——构建矩阵，反转矩阵

最新推荐文章于 2024-01-18 16:42:31 发布

王培军

最新推荐文章于 2024-01-18 16:42:31 发布

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分类专栏： Python 算法练习

本文链接：https://blog.youkuaiyun.com/qq_33287645/article/details/79817532

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本文介绍了使用Python进行列表操作及矩阵生成的方法，包括列表扩展、条件生成、矩阵反转等实用技巧，并通过不同方法对比了矩阵转置的效率。

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# 生成[[1,2,3],[4,5,6],[1,2,3],[4,5,6].......]
lst = [[1,2,3],[4,5,6]]
c = [lst.extend(lst) for j in range(3)]   # 2**n
d = [[1,2,3] if not i&1 else [4,5,6] for i in range(6)]  #/2
print(c)
print(d)

##生成 n * m 矩阵 matrix
m,n = 5,5
print([[0]*m]*n)   # 陷阱

tm = [[0]*m for j in range(n)]
print(tm)

# 封装和解构，加方括号，相当于后面加了一个逗号
[*a] = [[1,2,3],[4,5,6]]     # a = [[1,2,3],[4,5,6]]
*a, = [1,2,3],[4,5,6]     # a = [[1,2,3],[4,5,6]]
*a, = ([1,2,3],[4,5,6])    # a = [[1,2,3],[4,5,6]]
[x,*y] = [1,2]    # x,y = (1,[2])

# 反转矩阵效率 matrix
import datetime
square = [[4,5,6] for i in range(30)]
n = 100000
start = datetime.datetime.now()
for i in range(n):
  #  tm = [[0 for col in range(len(square))] for row in range(len(square[0]))]
    tm = [0] * len(square[0])         # 构建大列表空间 乘法效率更高，列表解析式和append相当
    for i in range(len(tm)):
        tm[i] = [0] * len(square)
    for i in range(len(tm)):
        for j in range(len(tm[0])):
            tm[i][j] = square[j][i]
delta = (datetime.datetime.now() - start).total_seconds()
print(delta)
print(tm)

start = datetime.datetime.now()
for i in range(n):
    chsquare = []
    for i in range(len(square[0])):
        chsquare.append([])
        for j in range(len(square)):
            chsquare[i].append(square[j][i])
delta = (datetime.datetime.now()-start).total_seconds()
print(delta)
print(chsquare)

# 效率低
start = datetime.datetime.now()
for i in range(n):
    chsquare = []
    for row in square:
        for i,val in enumerate(row):
            if len(chsquare) < i+1:
                chsquare.append([])
            chsquare[i].append(val)
delta = (datetime.datetime.now()-start).total_seconds()
print(delta)    
print(chsquare)