Question

这是一个很长的问题所以请耐心等待。我从3个API获得3个dicts开始。 dicts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密钥以进行一些处理。我之所以这样做，是因为需要进行大量处理，只需要URL。完成后，我有一个删除了重复项的URL列表，以及另一个与列表中每个URL位置相关的分数列表：

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()函数从这两个列表中创建一个新词典。

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ScoredResults的网址与API1，API2或API3相关联，这样我就有了这样的新词典：

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。如果你回顾我的问题历史，我一直在问很多字典问题，但到目前为止还没有实现。如果有人能指出我正确的方向，我将非常感激。

Answer 1

我会将API转换为对您更有意义的东西。网址的字典可能更合适：

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 2

使用给定的数据......

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

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 4

这样的事情对你有用吗？它是相当基础的，通过循环URLlist构建你的最终字典。

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)]))

请注意，由于您的不同url中可能存在相同的API但信息不同，我们需要事先在FullResults中创建相应的条目，因此它可能是简化循环有点棘手。 LMKHIW。

在Python中合并不同的词典

4 个答案: