将列表与词典相结合

Question

谢谢所有在这里帮忙的人

我有一份清单。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']}
        ], 
    ]

我希望根据COMPANY和SHIPMENTS组合这些列表，并且不希望有重复的SHIPMENTS值。NUMBER键/值不相关

理想情况下，最终输出将是一个字典列表，该列表类似于以下内容，其中所有发货都是为公司合并的：

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，但主要是寻找如何接近最终输出。如果有必要的话，我正在使用python3.6

Answer 1

我认为这应该解决你的问题

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]})

Answer 2

这里有一个解决方案，它使用集合来确保没有重复项，但它将丢失装运顺序

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，如果您只需要一份发货清单，而这实际上是字典中唯一的内容，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 3

这就解决了问题，尝试一下，然后玩转它，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，这有帮助