Python基础教程第2版 6:抽象
http://blog.chinaunix.net/uid-21142030-id-5676171.html
抽象与结构,事件流程
创建函数是组织程序的关键.def
使用函数改变数据结构(比如list,dict)是将程序抽象化的好方法.
函数定义中的参数是形参
函数调用中的参数是实参
函数通过参数获取值.
函数内部作用域修改参数值不会影响外部值.
抽象的要点是隐藏更新时的烦琐细节.
函数只能修改参数对象本身.
一个将姓名查询添加程序
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- #!/usr/bin/env python
- #-*- coding:utf-8 -*-
- '''
- 斐波那契数列:初始为0,1,列表中的一个数字是前两个元素的和
- '''
- num=int(raw_input('请输入要计算的斐波那契数列的数目:'))
- def fibs(num):
- '文档字符串 fibs.__doc__查看'
- result=[0,1]
- for i in range(num-2):
- result.append(result[-2]+result[-1])
- return result
- print fibs(num)
- #初始化数据库
- def initdata(data):
- data['first']={}
- data['middle']={}
- data['last']={}
- storage={}
- initdata(storage)
- me='eric lei wang'
- you='talen tf HAO'
- her='mei liu'
- #查询方法
- def lookup(database,lable,name):
- result=database[lable].get(name)
- return result
- #录入数据库
- def store(database,username):#把用户存入数据库中
- #1.分割用户名
- names=username.split()
- lables=['first','middle','last']
- if len(names)==2:
- names.insert(1,'')
- for lable,name in zip(lables,names):
- #判断是否已经存在
- people=lookup(database,lable,name)
- if people:
- #如果列表中有值,直接添加
- people.append(username)
- else:
- #如果列表没有值,新创建,注意,这里生成的是列表.
- database[lable][name]=[username]
- store(storage,me)
- store(storage,me)
- store(storage,you)
- store(storage,her)
- print storage
位置参数与关键字参数是可以联合使用的.除非完全清楚程序的功能和参数的意义,否则应该避免混合使用.
使用*收集任意多的参数,返回数组.
使用**收集任意多的关键字参数,返回字典.
修改一下上面的程序,支持多个名字同时录入
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- #!/usr/bin/env python
- #-*- coding:utf-8 -*-
- '''
- 斐波那契数列:初始为0,1,列表中的一个数字是前两个元素的和
- '''
- num=int(raw_input('请输入要计算的斐波那契数列的数目:'))
- def fibs(num):
- '文档字符串 fibs.__doc__查看'
- result=[0,1]
- for i in range(num-2):
- result.append(result[-2]+result[-1])
- return result
- print fibs(num)
- #初始化数据库
- def initdata(data):
- data['first']={}
- data['middle']={}
- data['last']={}
- storage={}
- initdata(storage)
- me='eric lei wang'
- you='talen tf HAO'
- her='mei liu'
- #查询方法
- def lookup(database,lable,name):
- result=database[lable].get(name)
- return result
- #录入数据库
- def store(database,*usernames):#把用户存入数据库中
- #1.分割用户名
- for username in usernames:
- names=username.split()
- lables=['first','middle','last']
- if len(names)==2:
- names.insert(1,'')
- for lable,name in zip(lables,names):
- #判断是否已经存在
- people=lookup(database,lable,name)
- if people:
- #如果列表中有值,直接添加
- people.append(username)
- else:
- #如果列表没有值,新创建,注意,这里生成的是列表.
- database[lable][name]=[username]
- store(storage,me,you)
- store(storage,me)
- store(storage,you)
- store(storage,her)
- print storage
点击(此处)折叠或打开
- 请输入要计算的斐波那契数列的数目:11
- [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
- {'middle': {'tf': ['talen tf HAO', 'talen tf HAO'], 'lei': ['eric lei wang', 'eric lei wang'], '': ['mei liu']}, 'last': {'liu': ['mei liu'], 'wang': ['eric lei wang', 'eric lei wang'], 'HAO': ['talen tf HAO', 'talen tf HAO']}, 'first': {'mei': ['mei liu'], 'eric': ['eric lei wang', 'eric lei wang'], 'talen': ['talen tf HAO', 'talen tf HAO']}}
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- 参数练习
- def story(**kwds):
- return 'Once upon a time,there was a %(job)s called %(name)s.' % kwds
- def power(x,y,*others):
- if others:
- print 'Received redundant parameters:',others
- return pow(x,y)
- def interval(start, stop=None,step=1):
- '打印指定范围的数字'
- if stop is None:
- start,stop=0,start
- result=[]
- while start stop:
- result.append(start)
- start+=step
- return result
- print story(job='ttserver',name='workhard')
- print story(name='mile',job='assssfsffa')
- params={'job':'job1','name':'nameserver'}
- print story(**params)
- print power(2,3)
- print power(y=4,x=5)
- params=(5,)*2
- print power(*params)
- print power(3,3,'talen')
- print interval(10)
- print interval(1,5)
- print interval(3,12,4)
- print power(*interval(3,7))
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- Once upon a time,there was a ttserver called workhard.
- Once upon a time,there was a assssfsffa called mile.
- Once upon a time,there was a job1 called nameserver.
- 8
- 625
- 3125
- Received redundant parameters: ('talen',)
- 27
- [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
- [1, 2, 3, 4]
- [3, 7, 11]
- Received redundant parameters: (5, 6)
- 81
-
def A(a):
def B(b):
a+b
return B
A(a)(b)或
C=A(a)
C(b)
C是A的返回函数即C=B
递归
函数调用自身的行为就是递归
递归必须以最小可能性为结束
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- #!/usr/bin/env python
- #-*- coding:utf-8 -*-
- def factorial(n):
- result=n
- for i in range(1,n):
- result*=i
- print result
- return result
- print('最终结果是%s' %factorial(5))
-
- def factorial2(n):
- if n == 1 :
- return 1
- else:
- return n*factorial2(n-1)
- print('阶乘的结果是%s' % factorial2(6))
-
- def power(x,n):
- result=x
- if n == 0:
- return 1
- else:
- return result * power(x,n-1)
- print(power(3,3))
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- 5
- 10
- 30
- 120
- 最终结果是120
- 阶乘的结果是720
- 27
如果不自己写,标准库有bisect模块可以有效实现二元查找.
宇宙中的粒子数约为10的87次方,只要290个问题就能分辨它们.
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- #二元运算
-
- #元素:序列,数,位置,范围最小值位置,最大值位置
- def search(sequence,number,lower=0,upper=None):
- #先判断是否在序列中
- if number in sequence:
- if upper is None:
- upper=len(sequence)-1
- if lower==upper:
- return upper
- else:
- #定义中间数,不能
- middle=(lower+upper)//2
- #判断在哪一边
- if number > sequence[middle]:
- return search(sequence,number,middle+1,upper)
- else:
- return search(sequence,number,lower,middle)
- else:
- return '没有这个数字'
- seq=[34,67,8,123,4,100,95]
- seq.sort()
- print seq
- print('数字位于%s' % search(seq,34))
- print('数字位于%s' % search(seq,55))
- print('数字位于%s' % search(seq,95))
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- [4, 8, 34, 67, 95, 100, 123]
- 数字位于2
- 数字位于没有这个数字
- 数字位于4
