[英]Combining list of lists with dictionaries
感謝所有在這里提供幫助的人。
我有一個列表列表。 這些列表包含如下字典:
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']}
],
]
我根據COMPANY
和SHIPMENTS
組合這些列表,並且我希望沒有重復的SHIPMENTS
值。 NUMBER
鍵/值無關緊要。
最終的 output 理想情況下是一個看起來像這樣的字典列表,其中公司的出貨量全部合並:
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']}
]
我知道我沒有提供任何我嘗試過的東西,但主要是尋找如何接近最終的 output。 如果這很重要,我正在使用 python3.6
這是一個解決方案,它使用集合來確保沒有重復,但它會丟失發貨順序。
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)
請注意,如果您只需要一份貨運清單,而這確實是您字典中唯一的內容,那么更簡單的解決方案是:
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)
我認為這應該可以解決您的問題
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]})
這樣就解決了問題,試一試,你可以優化代碼以獲得更好的性能。
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,它有幫助。
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