1 #_*_coding:utf-8_*_
 2 __author__ = ‘Linhaifeng‘
 3 import time
 4 
 5 person_list=[‘alex‘,‘wupeiqi‘,‘yuanhao‘,‘linhaifeng‘]
 6 def ask_way(person_list):
 7     print(‘-‘*60)
 8     if len(person_list) == 0:
 9         return ‘没人知道‘
10     person=person_list.pop(0)
11     if person == ‘linhaifeng‘:
12         return ‘%s说:我知道,老男孩就在沙河汇德商厦,下地铁就是‘ %person
13     print(‘hi 美男[%s],敢问路在何方‘ %person)
14     print(‘%s回答道:我不知道,但念你慧眼识猪,你等着,我帮你问问%s...‘ %(person,person_list))
15     time.sleep(3)
16     res=ask_way(person_list)
17     # print(‘%s问的结果是: %res‘ %(person,res))
18     return res
19 
20 
21 
22 res=ask_way(person_list)
23 
24 print(res)

 1 data = http://www.mamicode.com/[1, 3, 6, 7, 9, 12, 14, 16, 17, 18, 20, 21, 22, 23, 30, 32, 33, 35]
 2  
 3  
 4 def binary_search(dataset,find_num):
 5     print(dataset)
 6  
 7     if len(dataset) >1:
 8         mid = int(len(dataset)/2)
 9         if dataset[mid] == find_num:  #find it
10             print("找到数字",dataset[mid])
11         elif dataset[mid] > find_num :# 找的数在mid左面
12             print("\033[31;1m找的数在mid[%s]左面\033[0m" % dataset[mid])
13             return binary_search(dataset[0:mid], find_num)
14         else:# 找的数在mid右面
15             print("\033[32;1m找的数在mid[%s]右面\033[0m" % dataset[mid])
16             return binary_search(dataset[mid+1:],find_num)
17     else:
18         if dataset[0] == find_num:  #find it
19             print("找到数字啦",dataset[0])
20         else:
21             print("没的分了,要找的数字[%s]不在列表里" % find_num)
22  
23  
24 binary_search(data,66)

 1 array=[1,3,4,71,2]
 2 
 3 ret=[]
 4 for i in array:
 5     ret.append(i**2)
 6 print(ret)
 7 
 8 #如果我们有一万个列表,那么你只能把上面的逻辑定义成函数
 9 def map_test(array):
10     ret=[]
11     for i in array:
12         ret.append(i**2)
13     return ret
14 
15 print(map_test(array))
16 
17 #如果我们的需求变了,不是把列表中每个元素都平方,还有加1,减一,那么可以这样
18 def add_num(x):
19     return x+1
20 def map_test(func,array):
21     ret=[]
22     for i in array:
23         ret.append(func(i))
24     return ret
25 
26 print(map_test(add_num,array))
27 #可以使用匿名函数
28 print(map_test(lambda x:x-1,array))
29 
30 
31 #上面就是map函数的功能,map得到的结果是可迭代对象
32 print(map(lambda x:x-1,range(5)))

from functools import reduce
#合并,得一个合并的结果
array_test=[1,2,3,4,5,6,7]
array=range(100)

#报错啊,res没有指定初始值
def reduce_test(func,array):
    l=list(array)
    for i in l:
        res=func(res,i)
    return res

# print(reduce_test(lambda x,y:x+y,array))

#可以从列表左边弹出第一个值
def reduce_test(func,array):
    l=list(array)
    res=l.pop(0)
    for i in l:
        res=func(res,i)
    return res

print(reduce_test(lambda x,y:x+y,array))

#我们应该支持用户自己传入初始值
def reduce_test(func,array,init=None):
    l=list(array)
    if init is None:
        res=l.pop(0)
    else:
        res=init
    for i in l:
        res=func(res,i)
    return res

print(reduce_test(lambda x,y:x+y,array))
print(reduce_test(lambda x,y:x+y,array,50))

 1 #电影院聚集了一群看电影bb的傻逼,让我们找出他们
 2 movie_people=[‘alex‘,‘wupeiqi‘,‘yuanhao‘,‘sb_alex‘,‘sb_wupeiqi‘,‘sb_yuanhao‘]
 3 
 4 def tell_sb(x):
 5     return x.startswith(‘sb‘)
 6 
 7 
 8 def filter_test(func,array):
 9     ret=[]
10     for i in array:
11         if func(i):
12             ret.append(i)
13     return ret
14 
15 print(filter_test(tell_sb,movie_people))
16 
17 
18 #函数filter,返回可迭代对象
19 print(filter(lambda x:x.startswith(‘sb‘),movie_people))

 1 #当然了,map,filter,reduce,可以处理所有数据类型
 2 
 3 name_dic=[
 4     {‘name‘:‘alex‘,‘age‘:1000},
 5     {‘name‘:‘wupeiqi‘,‘age‘:10000},
 6     {‘name‘:‘yuanhao‘,‘age‘:9000},
 7     {‘name‘:‘linhaifeng‘,‘age‘:18},
 8 ]
 9 #利用filter过滤掉千年王八,万年龟,还有一个九千岁
10 def func(x):
11     age_list=[1000,10000,9000]
12     return x[‘age‘] not in age_list
13 
14 
15 res=filter(func,name_dic)
16 for i in res:
17     print(i)
18 
19 res=filter(lambda x:x[‘age‘] == 18,name_dic)
20 for i in res:
21     print(i)
22 
23 
24 #reduce用来计算1到100的和
25 from functools import reduce
26 print(reduce(lambda x,y:x+y,range(100),100))
27 print(reduce(lambda x,y:x+y,range(1,101)))
28 
29 #用map来处理字符串列表啊,把列表中所有人都变成sb,比方alex_sb
30 name=[‘alex‘,‘wupeiqi‘,‘yuanhao‘]
31 
32 res=map(lambda x:x+‘_sb‘,name)
33 for i in res:
34     print(i)

 1 字典的运算：最小值，最大值，排序
 2 salaries={
 3     ‘egon‘:3000,
 4     ‘alex‘:100000000,
 5     ‘wupeiqi‘:10000,
 6     ‘yuanhao‘:2000
 7 }
 8 
 9 迭代字典，取得是key，因而比较的是key的最大和最小值
10 >>> max(salaries)
11 ‘yuanhao‘
12 >>> min(salaries)
13 ‘alex‘
14 
15 可以取values，来比较
16 >>> max(salaries.values())
17 >>> min(salaries.values())
18 但通常我们都是想取出，工资最高的那个人名，即比较的是salaries的值，得到的是键
19 >>> max(salaries,key=lambda k:salary[k])
20 ‘alex‘
21 >>> min(salaries,key=lambda k:salary[k])
22 ‘yuanhao‘
23 
24 
25 
26 也可以通过zip的方式实现
27 salaries_and_names=zip(salaries.values(),salaries.keys()) 
28 
29 先比较值，值相同则比较键
30 >>> max(salaries_and_names)
31 (100000000, ‘alex‘)
32 
33 
34 salaries_and_names是迭代器，因而只能访问一次
35 >>> min(salaries_and_names)
36 Traceback (most recent call last):
37   File "<stdin>", line 1, in <module>
38 ValueError: min() arg is an empty sequence
39 
40 
41 
42 sorted(iterable，key=None,reverse=False)