Question

我需要根据特定键合并两个字典列表，否则需要追加

我使用以下代码生成了字典列表，并且按我的主要问题可以预期的工作是当我尝试合并两个不同的列表，如果找到相同的键，那么我需要根据字典键附加该值。

feed_mapping = {'BC': 11, 'HA':12, 'AB':16,'GR':18}
L = ["AB-24519152","BC-24519152"]

years_dict = dict()
l1 = []
for line in L:
    feed_id = feed_mapping[line.split('-')[0]]
    if feed_id in years_dict:
        # append the new number to the existing array at this slot
        years_dict[feed_id].append(line)
    else:
        # create a new array in this slot
        years_dict[feed_id] = [line]


for element,values in years_dict.items():
    l1.append({"id":values,"feed":element})


print(l1)


l1=  [{'id': ['AB-23473427','AB-99999'], 'feed': 16}, {'id': ['HA-4848231'], 'feed': 12}]
l2 = [{'id': ['AB-23473427'], 'feed': 14}, {'id': ['HA-4848231'], 'feed': 12}]

预期输出：

[{'id': ['AB-23473427','AB-99999'], 'feed': 16}, {'id': ['HA-4848231','HA-4848231'], 'feed': 12},{'id': ['AB-23473427'], 'feed': 14}]

Answer 1

您应该利用更适合此任务的数据结构，它实际上是您自己的years_dict。

from collections import defaultdict
l1=  [{'id': ['AB-23473427','AB-99999'], 'feed': 16}, {'id': ['HA-4848231'], 'feed': 12}]
l2 = [{'id': ['AB-23473427'], 'feed': 14}, {'id': ['HA-4848231'], 'feed': 12}]

temp = defaultdict(list) 

for i in l1:
    temp[i['feed']].extend(i['id'])

for i in l2:
    temp[i['feed']].extend(i['id'])

#Notice that temp is essentially same as years_dict, but with information of both lists
result = [{"id":v,"feed":k} for k,v in temp.items()]
#Output:
[{'id': ['AB-23473427', 'AB-99999'], 'feed': 16},
 {'id': ['HA-4848231', 'HA-4848231'], 'feed': 12},
 {'id': ['AB-23473427'], 'feed': 14}]

因此，尽管此代码有效，但我的建议是保留所有列表的years_dict，并只在最后将其转换为最终输出。

Answer 2

l1=  [{'id': ['AB-23473427','AB-99999'], 'feed': 16}, {'id': ['HA-4848231'], 'feed': 12}]
l2 = [{'id': ['AB-23473427'], 'feed': 14}, {'id': ['HA-4848231'], 'feed': 12}]

for elm2 in l2:
    for elm1 in l1:
        if elm2['feed'] == elm1['feed']:
            elm1['id'].extend(elm2['id'])
            break
    else:
        l1.append(elm2)

尝试一下。这会将l1设置为[{'id': ['AB-23473427', 'AB-99999'], 'feed': 16}, {'id': ['HA-4848231', 'HA-4848231'], 'feed': 12}, {'id': ['AB-23473427'], 'feed': 14}]

根据特定键合并2个字典列表

2 个答案: