python – 将列表转换为字典时的速度问题

2019年8月4日 172次阅读

我有一些关于将列表转换为字典的速度问题,其中以下操作占用总运行时间的约90％：

def list2dict(list_):
    return_dict = {}

    for idx, word in enumerate(list_):
        if word in return_dict:
            raise ValueError("duplicate string found in list: %s" % (word))
        return_dict[word] = idx

    return return_dict

我有麻烦看到它究竟是什么导致了这一点.您是否在代码中看到了明显的瓶颈,或者有关如何加快速度的建议？

谢谢.

最佳答案 编辑：

想象一下,我把它放在首位,因为它更大 – 事实证明,对OP代码的一个小调整会给性能带来很大的提升.

def list2dict(list_):    # OLD
    return_dict = {}
    for idx, word in enumerate(list_):
        if word in return_dict: # this compare is happening every iteration!
            raise ValueError("duplicate string found in list: %s" % (word))
        return_dict[word] = idx
    return return_dict

def list2dictNEW(list_): #NEW HOTNESS
    return_dict = {}
    for idx, word in enumerate(list_):
        return_dict[word] = idx # overwrite if you want to, because...
    if len(return_dict) == len(list_): return return_dict
    # if the lengths aren't the same, something got overwritten so we
    # won't return. If they ARE the same, toss it back with only one
    # compare (rather than n compares in the original
    else: raise ValueError("There were duplicates in list {}".format(list_))

DEMO:
>>> timeit(lambda: list2dictNEW(TEST))
1.9117132451798682
>>> timeit(lambda: list2dict(TEST)):
2.2543816669587216
# gains of a third of a second per million iterations!
# that's a 15.2% speed bost

没有明显的答案,但你可以尝试类似的东西：

def list2dict(list_):
    return_dict = dict()
    for idx,word in enumerate(list_):
        return_dict.setdefault(word,idx)
    return return_dict

您也可以构建一个set并执行list.index,因为您说列表相当小,但我认为这样会比较慢而不是更快.这需要分析才能确定(使用timeit.timeit)

def list2dict(list_):
    set_ = set(list_)
    return {word:list_.index(word) for word in set_}

我冒昧地在一组测试数据上运行一些配置文件.结果如下：

TEST = ['a','b','c','d','e','f','g','h','i','j'] # 10 items

def list2dictA(list_): # build set and index word
    set_ = set(list_)
    return {word:list_.index(word) for word in set_}

def list2dictB(list_): # setdefault over enumerate(list)
    return_dict = dict()
    for idx,word in enumerate(list_):
        return_dict.setdefault(word,idx)
    return return_dict

def list2dictC(list_): # dict comp over enumerate(list)
    return_dict = {word:idx for idx,word in enumerate(list_)}
    if len(return_dict) == len(list_):
        return return_dict
    else:
        raise ValueError("Duplicate string found in list")

def list2dictD(list_): # Original example from Question
    return_dict = {}
    for idx, word in enumerate(list_):
        if word in return_dict:
            raise ValueError("duplicate string found in list: %s" % (word))
        return_dict[word] = idx
    return return_dict

>>> timeit(lambda: list2dictA(TEST))
5.336584700190931
>>> timeit(lambda: list2dictB(TEST))
2.7587691306531
>>> timeit(lambda: list2dictC(TEST))
2.1609074989233292
>>> timeit(lambda: list2dictD(TEST))
2.2543816669587216
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