Python Algorithms – chapter2 基础知识

2018-02-11 19:41:22来源:cnblogs.com作者:huangqiancun人点击

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一、渐进记法

三个重要的记号

Ο、Ω、Θ,Ο记法表示渐进上界,Ω记法表示渐进下界,Θ记法同时提供了函数的上下界

几种常见的渐进运行时间实例

complexity

三种重要情况

最好的情况,最坏的情况,平均情况

最坏的情况通常是最有用的情况,可以对算法效率做出最佳保证

实证式算法评估

Tip1:If possible, don’t worry about it.

Tip2:用timeit模块进行计时

import timeittimeit.timeit("x = 2+2") #0.003288868749876883timeit.timeit("x = sum(range(10))") #0.003288868749897271

Tip3:用profiler找出瓶颈

使用cProfiler获取运行情况的内容,打印出程序中各函数的计时结果,如果python版本中没有cProfiler可以使用profiler代替

import cProfilecProfile.run("helloworld()")

Tip4:绘制出结果

可以使用matplotlib绘制出结果,可参考http://www.cnblogs.com/huangqiancun/p/8379502.html

Tip5:在根据计时比对结果做出判断时要小心仔细

Tip6:通过相关实验对渐进时间做出判断时要小心仔细

二、图与树

graphrep

1 图的实现

邻接表

邻接集

a, b, c, d, e, f, g, h = range(8)N = [    {b, c, d, e, f},    # a    {c, e},             # b    {d},                # c    {e},                # d    {f},                # e    {c, g, h},          # f    {f, h},             # g    {f, g}              # h]
b in N[a] # Truelen(N[f]) # 3

邻接列表

a, b, c, d, e, f, g, h = range(8)N = [    [b,c,d,e,f], #a    [c,e], #b    [d], #c    [e], #d    [f], #e    [c,g,h], #f    [f,h], #g    [f,g] #h    ]

加权邻接字典

a, b, c, d, e, f, g, h = range(8)N = [    {b:2, c:1, d:3, e:9, f:4},    # a    {c:4, e:3},                   # b    {d:8},                        # c    {e:7},                        # d    {f:5},                        # e    {c:2, g:2, h:2},              # f    {f:1, h:6},                   # g    {f:9, g:8}                    # h]
b in N[a] #  Truelen(N[f]) #  3N[a][b] #  2

邻接集的字典表示法

N = {    'a': set('bcdef'),    'b': set('ce'),    'c': set('d'),    'd': set('e'),    'e': set('f'),    'f': set('cgh'),    'g': set('fh'),    'h': set('fg')}

邻接矩阵

a, b, c, d, e, f, g, h = range(8)N = [[0,1,1,1,1,1,0,0], # a     [0,0,1,0,1,0,0,0], # b     [0,0,0,1,0,0,0,0], # c     [0,0,0,0,1,0,0,0], # d     [0,0,0,0,0,1,0,0], # e     [0,0,1,0,0,0,1,1], # f     [0,0,0,0,0,1,0,1], # g     [0,0,0,0,0,1,1,0]] # hN[a][b] # Neighborhood membership -> 1sum(N[f]) # Degree -> 3

对不存在的边赋予无限大权值的加权矩阵

a, b, c, d, e, f, g, h = range(8)_ = float('inf')W = [[0,2,1,3,9,4,_,_], # a     [_,0,4,_,3,_,_,_], # b     [_,_,0,8,_,_,_,_], # c     [_,_,_,0,7,_,_,_], # d     [_,_,_,_,0,5,_,_], # e     [_,_,2,_,_,0,2,2], # f     [_,_,_,_,_,1,0,6], # g     [_,_,_,_,_,9,8,0]] # hW[a][b] < inf # Truesum(1 for w in W[a] if w < inf) - 1  # 5

注意:在对度值求和时务必要记得从中减1,因为我们不想把对角线也计算在内

Numpy库中的专用数组

N = [[0]*10 for i in range(10)]
import numpy as npN = np.zeros([10,10])

更多内容可参考http://www.cnblogs.com/huangqiancun/p/8379241.html

2 树的实现

treerep

T = [["a", "b"], ["c"], ["d", ["e","f"]]]T[0][1] # 'b'T[2][1][0] # 'e'

二叉树类

class Tree:    def __init__(self, left, right):        self.left = left        self.right = rightt = Tree(Tree("a", "b"), Tree("c", "d"))t.right.left  # 'c'

多路搜索树类(左孩子,右兄弟)

class Tree:    def __init__(self, kids, next=None):        self.kids = self.val = kids        self.next = nextreturn Treet = Tree(Tree("a", Tree("b", Tree("c", Tree("d")))))t.kids.next.next.val  # 'c'

Bunch模式

bunch类

class Bunch(dict):    def __init__(self, *args, **kwds):        super(Bunch, self).__init__(*args, **kwds)        self.__dict__ = self
x = Bunch(name = "Jayne Cobb", position = "Public Relations")x.name #'Jayne Cobb'
T = Buncht = T(left = T(left = "a",right = "b"), right = T(left = "c"))t.left # {'right': 'b', 'left': 'a'}t.left.right #' b'"left" in t.right # True

三、黑盒子

1 隐性平方级操作

from random import randrangeL = [randrange(10000) for i in range(1000)]42 in L # FalseS = set(L)42 in S #False

看起来使用set毫无意义,但是成员查询在list中是线性级的,在set中则是常数级的

lists = [[1,2], [3,4,5], [6]]sum(lists, []) #[1, 2, 3, 4, 5, 6]res = []for lst in lists:    res.extend(lst)# [1, 2, 3, 4, 5, 6]

sum函数是平方级的运行时间,第二个为更好的选择,当list的长度很短时,他们之间没有太大差距,但一旦超出某个长度,sum版本就会彻底完败

2 浮点运算的麻烦

sum(0.1 for i in range(10)) == 1.0 #False
def almost_equal(x, y, places=7):    return round(abs(x-y), places) == 0almost_equal(sum(0.1 for i in range(10)), 1.0) # True
from decimal import *sum(Decimal("0.1") for i in range(10)) == Decimal("1.0") #True



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