在Python中合并不同的字典

Question

这是一个很长的问题，请耐心等待。我从3个API获得3个词典开始。字典有这样的结构：

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键来做一些处理。我这样做是因为需要完成相当多的处理，并且只需要URL。完成后我有网址与删除的重复和分数是相对于每个URL在列表中的位置的另一个列表的列表：

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] 

我所做的创建从这些2列出了使用新字典功能zip()。

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} 

现在我需要做的就是URL的从ScoredResults与API1，API2或API3链接，这样我有一个新的字典，像这样：

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}, 
...} 

这是对我来说太难了。如果你回顾我的问题历史，我一直在问很多字典问题，但迄今为止还没有实现工作。如果任何人都可以请指出我正确的方向，我将非常感激。

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

在给定的数据...

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

导致成：

{'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

我将改变API的到的东西，是更有意义的你。 URL的字典可能更合适：

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 

Answer 4

会像你的工作？这是相当基本的，通过在URLlist上循环构建您的最终字典。

API1r = API1['results'] 
API2r = API2['hits'] 
API3r = API3['output'] 

FullResults = {} 
for (U, R) in zip(URLlist, Results): 
    FullResults[U] = {} 
    for api in (API1r, API2r, API3r): 
     for v in api: 
      k = dict() 
      k.update(v) 
      if (k.pop('url') == U): 
       FullResults[U].update((k.items()+[('score', R)])) 

注意，作为同url可能存在于你的不同API秒，但与不同的信息，我们需要事先创建在FullResults相应的条目，所以它可能是一个有点棘手简化循环。 LMKHIW。