Question

我有一些Python代码远远超过实际应用的代码。

#Generate planets
for t in range(stars*3): #There are 3 planets for every star, but not every star will have 3 planets
    theplanet=Planet()

    if random.randint(0,100) <= 25: #25% of planets have life
        theplanet.tl=random.randint(1,9)    #With a random tech level
    else:
        theplanet.tl=0  #The rest don't

    theplanet.star=stardict[random.choice(list(stardict.keys()))]     #Choose a random star
    theplanet.star.planets+=(theplanet,)    #Append the new planet to the star's list of planets
    theplanet.name=theplanet.star.name+"-"+str(len(theplanet.star.planets)) #Name the planet Starname-X, where X is the length of the star's planets tuple. Since this increases every time a planet is added, it will be 1 for the first planet, 2 for the next, etc...

    if math.floor((t/(stars*3))*100)==(t/(stars*3))*100: print("Generating planets: "+str((t/(stars*3))*100)+"% done.")

我非常确定瓶颈在star=stardict[random.choice(list(等等......行。我在这里猜测，但我认为dicts通过搜索dict中的每个条目并查看哪个条目具有正确的密钥来工作。列表，我再次假设，只是读取从条目号导出的内存位置的信息，对于非常大的（200,000个条目，确切地说）列表/序列要快得多。

将dict的条目转换为列表会使这段代码更快吗？我该怎么做（我以为我看到了它的功能，现在回顾文档......）？有没有其他人注意到这个更快的方式？

Answer 1

每次循环都会创建一个列表，但该列表是不变的。将它移到循环之外。

starlist=list(stardict.keys())
...
    theplanet.star=stardict[random.choice(starlist)]     #Choose a random star

问题几乎肯定不在dict查找中。它们基于非常快的哈希表。

Answer 2

将列表生成list(stardict.keys())移到循环外
尝试分析您的代码（documentation）
假设您正在运行CPython，请检查您的代码是否可以使用Pypy运行。由于其优化的JIT

Answer 3

您只能使用中间值中的键来选择stardict中的随机项。您可以直接使用字典的值列表：

#Generate planets
starlist = stardict.values()
for t in range(stars*3): #There are 3 planets for every star, but not every star will have 3 planets
    theplanet=Planet()

    if random.randint(0,100) <= 25: #25% of planets have life
        theplanet.tl=random.randint(1,9)    #With a random tech level
    else:
        theplanet.tl=0  #The rest don't

    theplanet.star=random.choice(starlist)     #Choose a random star
    theplanet.star.planets+=(theplanet,)    #Append the new planet to the star's list of planets
    theplanet.name=theplanet.star.name+"-"+str(len(theplanet.star.planets)) #Name the planet Starname-X, where X is the length of the star's planets tuple. Since this increases every time a planet is added, it will be 1 for the first planet, 2 for the next, etc...

    if math.floor((t/(stars*3))*100)==(t/(stars*3))*100: print("Generating planets: "+str((t/(stars*3))*100)+"% done.")

通过在Python中使用列表而不是dicts进行优化

3 个答案: