如何在python中以特定格式返回字典列表[关闭]

Question

关闭。这个问题需要更多focused。它当前不接受答案。

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是否要改善此问题？更新问题，使其仅按editing this post专注于一个问题。

[<10天前关闭。

[{'pf': 'ABC', 'src': 'SCI'}, {'pf': 'ABC', 'src': 'MC'}, {'pf': 'ASR', 'src': 'CC'}, {'pf': 'ASR', 'src': 'CC'}, {'pf': 'ASR', 'src': 'ACS'}, {'pf': 'ABC', 'src': 'CC'}, {'pf': 'NS', 'src': 'ATT'}]

是否有可能获得上述字典的输出？
{
platform:all,
customers:[{
    pf:ABC,
    src:[SCI,MC,CC]
},{
    pf:ASR,
    src:[CC,ACS] #CC - duplicates of ASR are removed
],{
    pf:NS,
    src:[ATT]
}

我尝试使用defaultdict（list），由于我想要的特定格式，它没有用。有帮助吗？

Answer 1

元素的分组可以通过itertools.groupby完成，通过set删除重复项。

例如：

from pprint import pprint
from itertools import groupby

l = [{'pf': 'ABC', 'src': 'SCI'}, {'pf': 'ABC', 'src': 'MC'}, {'pf': 'ASR', 'src': 'CC'}, {'pf': 'ASR', 'src': 'CC'}, {'pf': 'ASR', 'src': 'ACS'}, {'pf': 'ABC', 'src': 'CC'}, {'pf': 'NS', 'src': 'ATT'}]

out = {'customers': [], 'platform': 'all'}
for v, g in groupby(sorted(l, key=lambda k: k['pf']), lambda k: k['pf']):
    out['customers'].append( {'pf': v, 'src': list(set(val['src'] for val in g))} )

pprint(out)

打印：
{'customers': [{'pf': 'ABC', 'src': ['MC', 'SCI', 'CC']},
               {'pf': 'ASR', 'src': ['ACS', 'CC']},
               {'pf': 'NS', 'src': ['ATT']}],
 'platform': 'all'}

Answer 2

如果要使用默认字典，则可以编写如下内容。

from collections import defaultdict
from pprint import pprint


l = [{'pf': 'ABC', 'src': 'SCI'}, {'pf': 'ABC', 'src': 'MC'}, {'pf': 'ASR', 'src': 'CC'}, {'pf': 'ASR', 'src': 'CC'}, {'pf': 'ASR', 'src': 'ACS'}, {'pf': 'ABC', 'src': 'CC'}, {'pf': 'NS', 'src': 'ATT'}]

# use of set so that we have only unique values of 'src' for respective 'pf'
pf_dict = defaultdict(set)

for data_dict in l:
    # with default dict if key does not exist it will create a key in dict with 
    # one element in set, and if key is present it will add new value to already
    # existing set against this key.
    pf_dict[data_dict['pf']].add(data_dict['src'])

customers = []

# converting all sets of 'src' against respective 'pf' to list
for key,value in pf_dict.items():
    customers.append({'pf':key, 'src': list(value)})

result = {
    'platform': 'all',
    'customers': customers
}

pprint(result)

Answer 3

我喜欢安德烈的答案。

自从您提到defaultdict，以下是使用它来避免重复的方法

from collections import defaultdict
from pprint import pprint

ds = [
    {'pf': 'ABC', 'src': 'SCI'},
    {'pf': 'ABC', 'src': 'MC'},
    {'pf': 'ASR', 'src': 'CC'},
    {'pf': 'ASR', 'src': 'CC'},
    {'pf': 'ASR', 'src': 'ACS'},
    {'pf': 'ABC', 'src': 'CC'},
    {'pf': 'NS', 'src': 'ATT'}
]    

pfs = defaultdict(set)
for e in ds:
    pfs[e['pf']].add(e['src'])
pprint(pfs)

out = {'customers': [{'pf': k, 'src': i} for k,i in pfs.items()],
       'platform': 'all'}
pprint(out)

产生
defaultdict(<class 'set'>,
            {'ABC': {'MC', 'SCI', 'CC'},
             'ASR': {'ACS', 'CC'},
             'NS': {'ATT'}})
{'customers': [{'pf': 'ABC', 'src': {'MC', 'SCI', 'CC'}},
               {'pf': 'ASR', 'src': {'ACS', 'CC'}},
               {'pf': 'NS', 'src': {'ATT'}}],
 'platform': 'all'}