Question

我有两个词组列表：prices_distincts，prices。

他们通过hash_brand_artnum连接，两者都按hash_brand_artnum排序我不明白为什么循环工作这么久：

如果prices_distincts的长度为100,000，则适用于30 min
但如果prices_distincts的长度为10,000，则适用于10 sec。

代码：

 prices_distincts = [{'hash_brand_artnum':1202},...,..]
 prices = [{'hash_brand_artnum':1202,'price':12.077},...,...]

 for prices_distinct in prices_distincts:
    for price in list(prices):            
        if prices_distinct['hash_brand_artnum'] == price['hash_brand_artnum']:

            print price['hash_brand_artnum']
            #print prices
            del prices[0]
        else:
            continue

我需要查找价格相同的商品。 price_distincts和价格之间的关系是一对多的。和价格相同的团体价格['hash_brand_artnum']

Answer 1

它的工作时间很长，因为你的算法是O（N ^ 2）和100000 ^ 2 = 10000000000和10000 ^ 2 = 100000000.因此，两个数之间的因子是100，并且因子介于30分钟和10秒~100之间。 / p>

编辑：您的代码和如此少量的数据很难说，我不知道您的任务是什么，但我认为您的词典不是很有用。可以试试这个：

>>> prices_distincts = [{'hash_brand_artnum':1202}, {'hash_brand_artnum':14}]
>>> prices = [{'hash_brand_artnum':1202, 'price':12.077}, {'hash_brand_artnum':14, 'price':15}]
# turning first list of dicts into simple list of numbers
>>> dist = [x['hash_brand_artnum'] for x in prices_distincts]
# turning second list of dicts into dict where number is a key and price is a value
>>> pr = {x['hash_brand_artnum']:x["price"] for x in prices}

不是你可以通过你的号码迭代并获得价格：

>>> for d in dist:
...     print d, pr[d]

Answer 2

正如@RomanPekar所提到的，您的算法运行缓慢，因为它的复杂性为O(n^2)。要修复它，您应该将其写为O(n)算法：

import itertools as it

for price, prices_distinct in it.izip(prices, prices_distincts):
    if prices_distinct['hash_brand_artnum'] == price['hash_brand_artnum']:
        # do stuff

Answer 3

如果price_distincts或多或少价格增长，那么如果你将prices_distincts的大小乘以10，你的原始10秒将乘以10然后再乘以10（循环的第二个），然后乘以~2，因为“列表（价格）”（顺便说一下，应该最终完成循环）：

10秒* 10 * 10 * 2 = 2000秒= 33分钟

这种转换通常很昂贵。

 prices_distincts = [{'hash_brand_artnum':1202},...,..]
 prices = [{'hash_brand_artnum':1202,'price':12.077},...,...]
 list_prices = list(prices)

 for prices_distinct in prices_distincts:
    for price in list_prices:            
        if prices_distinct['hash_brand_artnum'] == price['hash_brand_artnum']:

            print price['hash_brand_artnum']
            #print prices
            del prices[0]
        else:
            continue

循环工作时间太长

3 个答案: