python timeit Timer比较两个程序的运行时间

KathyLJQ

于 2019-12-31 10:16:29 发布

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分类专栏： Python

本文链接：https://blog.youkuaiyun.com/KathyLJQ/article/details/103778211

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本文对比了两种用于检查两个字符串是否包含相同字符及其频率的算法。第一种算法使用计数器数组统计每个字符的出现次数，第二种算法则通过字典实现相同的功能。通过运行时测试，比较了这两种算法的效率。

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from timeit import Timer

def checkword01(word1,word2):
    lettercounter1=[0]*26
    lettercounter2=[0]*26
    for i in range(len(word1)):
        pos=ord(word1[i])-ord("a")
        lettercounter1[pos]+=1
    for i in range(len(word2)):
        pos=ord(word2[i])-ord("a")
        lettercounter2[pos]+=1  
    flag=False
    if lettercounter1==lettercounter2:
        flag=True
    return flag

        
def checkword02(word1,word2):
    wordlist01=list(word1)
    wordlist02=list(word2) 
    worddict01={}
    worddict02={}
    for i in wordlist01:
        if i not in worddict01: 
            worddict01[i]=0
        else:
            worddict01[i]+=1
    for i in wordlist02:
        if i not in worddict02: 
            worddict02[i]=0
        else:
            worddict02[i]+=1
    flag=True  
   
    if len(worddict01)!=len( worddict02):        
        flag=False
        return
    for key in worddict01:
        if key not in worddict02:           
            flag=False
            return flag
        else:
            if worddict01[key]!=worddict02[key]:                
                flag=False
                return flag
    return flag          
   

ret01=checkword01("pythoon","typhoon")   
ret02=checkword02("pythoon","typhoon")
print(ret01)
print(ret02)
t1= Timer('checkword01("pythoon","typhoon")', 'from __main__ import checkword01')
t2 = Timer('checkword02("pythoon","typhoon")', 'from __main__ import checkword02')
print(t1.timeit(number=1000))
print(t2.timeit(number=1000))

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