在Python中合并不同的词典

Question

这是一个很长的问题，请耐心听我说。我从3个API中获得的3个dict开始。格言的结构如下：

API1 = {'results':[{'url':'www.site.com','title':'A great site','snippet':'This is a great site'},
{'url':'www.othersite.com','title':'Another site','snippet':'This is another site'},
{'url':'www.wiki.com','title':'A wiki site','snippet':'This is a wiki site'}]}

API2 = {'hits':[{'url':'www.dol.com','title':'The DOL site','snippet':'This is the dol site'},
{'url':'www.othersite.com','title':'Another site','snippet':'This is another site'},
{'url':'www.whatever.com','title':'Whatever site','snippet':'This is a site about whatever'}]}

API3 = {'output':[{'url':'www.dol.com','title':'The DOL site','snippet':'This is the dol site'},
{'url':'www.whatever.com','title':'Whatever site','snippet':'This is a site about whatever'},
{'url':'www.wiki.com','title':'A wiki site','snippet':'This is a wiki site'}]}

我从API1、API2和API3中提取URL键来进行一些处理。我这样做是因为有相当多的处理工作要做，只有网址是需要的。完成后，我有一个删除重复的URL列表和另一个相对于列表中每个URL位置的分数列表：

URLlist = ['www.site.com','www.wiki.com','www.othersite.com','www.dol.com','www.whatever.com']

Results = [1.2, 6.5, 3.5, 2.1, 4.0]

我所做的是使用zip()函数从这两个列表创建一个新字典。你知道吗

ScoredResults = dict(zip(URLlist,Results))

{'www.site.com':1.2,'www.wiki.com':6.5, 'www.othersite.com':3.5, 'www.dol.com':2.1, 'www.whatever.com':4.0}

现在我需要做的是将URL的fromScoredResults与API1、API2或API3链接起来，这样我就有了一个新的字典：

Full Results = 
{'www.site.com':{'title':'A great site','snippet':'This is a great site','score':1.2},
 'www.othersite.com':{'title':'Another site','snippet':'This is another site','score':3.5},
...}

这对我来说太难了。如果你回顾一下我的问题历史，我已经问了很多字典问题，但到目前为止还没有实现。如果有人能给我指出正确的方向，我将不胜感激。你知道吗

Answer 1

快速尝试：

from itertools import chain

full_result = {}

for blah in chain.from_iterable(d.itervalues() for d in (API1, API2, API3)):
    for d in blah:
        full_result[d['url']] = {
            'title': d['title'],
            'snippet': d['snippet'],
            'score': ScoredResults[d['url']]
        }

print full_result

Answer 2

用给定的数据

Full_Results = {d['url']: {'title': d['title'], 'snippet': d['snippet'], 'score': ScoredResults[d['url']]} for d in API1['results']+API2['hits']+API3['output']}

导致：

{'www.dol.com': {'score': 2.1,
  'snippet': 'This is the dol site',
  'title': 'The DOL site'},
 'www.othersite.com': {'score': 3.5,
  'snippet': 'This is another site',
  'title': 'Another site'},
 'www.site.com': {'score': 1.2,
  'snippet': 'This is a great site',
  'title': 'A great site'},
 'www.whatever.com': {'score': 4.0,
  'snippet': 'This is a site about whatever',
  'title': 'Whatever site'},
 'www.wiki.com': {'score': 6.5,
  'snippet': 'This is a wiki site',
  'title': 'A wiki site'}}

Answer 3

我会把API转换成对你更有意义的东西。URL的dict可能更合适：

def transform_API(API):
    list_of_dict=API.get('results',API.get('hits',API.get('output')))
    if(list_of_dict is None):
       raise KeyError("results, hits or output not in API")
    d={}
    for dct in list_of_dict:
        d[dct['url']]=dct
        dct.pop('url')
    return d

API1=transform_API(API1)
API2=transform_API(API2)
API3=transform_API(API3)

master={}
for d in (API1,API2,API3):
    master.update(d)

urls=list(master.keys())
scores=get_scores_from_urls(urls)

for k,score in zip(urls,scores):
    master[k]['score']=score