将列表列表与字典相结合

Question

感谢所有在这里提供帮助的人。

我有一个列表列表。这些列表包含如下词典：

combined lists = [
[
{'COMPANY': 'company1', 'NUMBER': '111', 'SHIPMENTS': ['1', '2', '3', '4']},
{'COMPANY': 'company2', 'NUMBER': '222', 'SHIPMENTS': ['1']},
{'COMPANY': 'company3', 'NUMBER': '333', 'SHIPMENTS': ['1', '4']},
{'COMPANY': 'company4', 'NUMBER': '444', 'SHIPMENTS': ['2', '5']},
{'COMPANY': 'company5', 'NUMBER': '555', 'SHIPMENTS': ['1', '3', '5', '9']}
], 
[
{'COMPANY': 'company1', 'NUMBER': '111', 'SHIPMENTS': ['5', '6', '7', '8']},
{'COMPANY': 'company3', 'NUMBER': '333', 'SHIPMENTS': ['3', '5']},
{'COMPANY': 'company5', 'NUMBER': '555', 'SHIPMENTS': ['3', '5', '7']},
{'COMPANY': 'company7', 'NUMBER': '777', 'SHIPMENTS': ['2', '4']},
{'COMPANY': 'company9', 'NUMBER': '999', 'SHIPMENTS': ['1', '2', '5', '6', '7']}
], 
]

我根据COMPANY和SHIPMENTS组合这些列表，并且我希望没有重复的SHIPMENTS值。键/值NUMBER无关紧要。

理想情况下，最终输出将是一个字典列表，看起来像这样，其中所有出货量都为公司合并：

final_list = [
{'COMPANY': 'company1', 'SHIPMENTS': ['1', '2', '3', '4', '5', '6', '7', '8']},
{'COMPANY': 'company2', 'SHIPMENTS': ['1']},
{'COMPANY': 'company3', 'SHIPMENTS': ['1', '4', '3', '5']},
{'COMPANY': 'company4', 'SHIPMENTS': ['2', '5']},
{'COMPANY': 'company5', 'SHIPMENTS': ['1', '3', '5', '7', '9']},
{'COMPANY': 'company7', 'SHIPMENTS': ['2', '4']},
{'COMPANY': 'company9', 'SHIPMENTS': ['1', '2', '5', '6', '7']}
]

我知道我没有提供任何我尝试过的东西，但主要是寻找如何获得最终输出。如果这很重要，我正在使用 python3.6

Answer 1

这是一个解决方案，它使用集合来确保没有重复项，但它会丢失发货顺序。

from itertools import chain
combined_lists = [
[
{'COMPANY': 'company1', 'NUMBER': '111', 'SHIPMENTS': ['1', '2', '3', '4']},
{'COMPANY': 'company2', 'NUMBER': '222', 'SHIPMENTS': ['1']},
{'COMPANY': 'company3', 'NUMBER': '333', 'SHIPMENTS': ['1', '4']},
{'COMPANY': 'company4', 'NUMBER': '444', 'SHIPMENTS': ['2', '5']},
{'COMPANY': 'company5', 'NUMBER': '555', 'SHIPMENTS': ['1', '3', '5', '9']}
],
[
{'COMPANY': 'company1', 'NUMBER': '111', 'SHIPMENTS': ['5', '6', '7', '8']},
{'COMPANY': 'company3', 'NUMBER': '333', 'SHIPMENTS': ['3', '5']},
{'COMPANY': 'company5', 'NUMBER': '555', 'SHIPMENTS': ['3', '5', '7']},
{'COMPANY': 'company7', 'NUMBER': '777', 'SHIPMENTS': ['2', '4']},
{'COMPANY': 'company9', 'NUMBER': '999', 'SHIPMENTS': ['1', '2', '5', '6', '7']}
]
]
COMPANY_KEY = 'COMPANY'
SHIPMENTS_KEY = 'SHIPMENTS'
# you're looking to:
# - combine the lists
# - drop the number
# - combine the shipments, removing duplicates
final_dict = {}
for d in chain.from_iterable(combined_lists):
key = d[COMPANY_KEY]
if key in final_dict:
final_dict[key][SHIPMENTS_KEY].update(*d[SHIPMENTS_KEY])
else:
final_dict[key] = {SHIPMENTS_KEY: set(d[SHIPMENTS_KEY])}
print(final_dict)
# if you need a list, not a dict
final_list = [{COMPANY_KEY: key, SHIPMENTS_KEY: value} for key, value in final_dict.items()]
print(final_list)

请注意，如果您只需要一个货件列表，而这确实是字典中唯一的东西，那么更简单的解决方案是：

from collections import defaultdict
better_dict = defaultdict(set)
for d in chain.from_iterable(combined_lists):
better_dict[d[COMPANY_KEY]].update(*d[SHIPMENTS_KEY])
print(better_dict)

Answer 2

这解决了问题，尝试一下并尝试一下，您可以优化代码以获得更好的性能。

def company_exists(company, resulting_list):
for i,dict_ in enumerate(resulting_list):
if company == dict_['COMPANY']:
return i, True
return None, False

def merge_lists(combined_lists):
res = []
for list_ in combined_lists:
for dict_ in list_:
idx, check = company_exists(dict_['COMPANY'], res)
if not check:
res.append(dict_)
else:
res[idx]['SHIPMENTS'].extend(dict_['SHIPMENTS'])
res[idx]['SHIPMENTS'] = list(set(res[idx]['SHIPMENTS']))
return res

希望，它有帮助。

Answer 3

我认为这应该可以解决您的问题

import collections
merged = collections.defaultdict(list)
for x in combined_lists:
for y in x:
merged[y["COMPANY"]] += y["SHIPMENT"]
final_list = []
for x in merged:
final_list.append({"COMPANY": x, "SHIPMENT": merged[x]})