算法：几个重要函数应用于一个题目

今天欣赏一个包含众多知识点的解法。题目是这样的：

给定两个字符串s1和s2，统计它们各自包含的各个小写字母的个数，然后互相比较。如果对某个字母，s1中出现的次数多，就打印“1:aaaa” (表示字母a在s1中出现了4次)；如果s2中出现的次数多，就打印“2:aaaa”；如果次数相同，就打印“=:aaaa”。最后输出整个的比较结果，输出时先按次数排序，再按1-2-=排序，最后按字母排序。

举个例子：

s1 = "my&friend&Paul has heavy hats! &"

s2 = "my friend John has many many friends &"

mix(s1, s2) --> "2:nnnnn/1:aaaa/1:hhh/2:mmm/2:yyy/2:dd/2:ff/2:ii/2:rr/=:ee/=:ss/1:l/1:t/1:u/1:v/2:o"

分析：

这个题目其实挺简单的。统计字母个数？遍历+字典轻松搞定。比较两个字典？逐个即可。最后输出的先按...排序，再按...排序，最后按...排序？有点麻烦，不过用itemgetter也可以做到。

本人参考两种解法，并比较两种解法的性能：

#-*- coding:utf-8 -*-
__author__ = 'Administrator'

from collections import Counter
import re
import time
#s1 = "my&friend&Paul has heavy hats! &"
#s2 = "my friend John has many many friends &"
#str_mix(s1, s2) --> "2:nnnnn/1:aaaa/1:hhh/2:mmm/2:yyy/2:dd/2:ff/2:ii/2:rr/=:ee/=:ss/1:l/1:t/1:u/1:v/2:o"
def str_min(s1,s2):
    c1=Counter(s1)
    c2=Counter(s2)
    d=[]
    key1=c1.keys();
    key2=c2.keys();

    for key in set(list(key1)+list(key2)):
        if re.match("[a-z]",key):
            if c1[key]>c2[key]:
                d.append(("1",key,c1[key]))
            elif c1[key]<c2[key]:
                d.append(("2",key,c2[key]))
            else:
                d.append(("=",key,c1[key]))

    rec=["{0}:{1}".format(v,s*n) for v,s,n in d]
    return rec

def str_min2(s1,s2):
    c1=Counter(filter(str.islower,s1))
    c2=Counter(filter(str.islower,s2))
    d=[]
    for key in set(list(c1.keys())+list(c2.keys())):
        n1=c1.get(key,0)
        n2=c2.get(key,0)
        if n1>0 or n2>0:
            if n1>n2:
                d.append(("1",key,n1))
            elif n1<n2:
                d.append(("2",key,n2))
            else:
                d.append(("=",key,n1))
    rec=["{}:{}".format(v,s*n) for v,s,n in d]
    return rec

if __name__=="__main__":
    s1 = "my&friend&Paul has heavy hats! &"
    s2 = "my friend John has many many friends &"
    t0=time.time()
    for i in range(100000):
         rec=str_min(s1,s2)
    print('/'.join(sorted(rec,key=lambda s:(-len(s),s))))
    t1=time.time()-t0
    print(t1)
    print("-"*100)
    t0=time.time()
    for i in range(100000):
         rec=str_min2(s1,s2)
    print('/'.join(sorted(rec,key=lambda s:(-len(s),s))))
    t1=time.time()-t0
    print(t1)
    
    

输出：

2:nnnnn/1:aaaa/1:hhh/2:mmm/2:yyy/2:dd/2:ff/2:ii/2:rr/=:ee/=:ss/1:l/1:t/1:u/1:v/2:o
14.07480502128601
----------------------------------------------------------------------------------------------------
2:nnnnn/1:aaaa/1:hhh/2:mmm/2:yyy/2:dd/2:ff/2:ii/2:rr/=:ee/=:ss/1:l/1:t/1:u/1:v/2:o
8.899508953094482