具有两个类似的字典列表
big = [{'id': 1234, 'name': 'ipod'}, {'id': 1235, 'name': 'ipod x'}, {'id': 1236, 'name': 'ipod touch'}]
small = [{'id': 1236, 'url': 'directUrl1'}, {'id': 1235, 'url': 'directUrl2'}]
我想实现一种高效且Python化的方式来查找两个列表(基于id
)之间的交集并创建合并字典的新列表:
res = [{'id': 1236, 'url': 'directUrl1', 'name': 'ipod touch'}, {'id': 1235, 'url': 'directUrl2', 'name': 'ipod x'}]
我目前的做法:
>>> res = []
>>>
>>> for item in [x for x in small if x['id'] in [y['id'] for y in big]]:
... res.append({**item, **[x for x in big if x['id'] == item['id']][0]})
...
>>> res
[{'id': 1236, 'url': 'directUrl1', 'name': 'ipod touch'}, {'id': 1235, 'url': 'directUrl2', 'name': 'ipod x'}]
答案 0 :(得分:2)
Pythonic并不意味着致密;)
怎么样:
new_dictlist = []
for d1 in small:
for d2 in big:
if d1['id'] == d2['id']:
new_dictlist.append({'id':d1['id'], 'url':d1['url'], 'name':d2['name']})
或单线:
[{**x, **y} for x in big for y in small if x['id'] == y['id']]
答案 1 :(得分:0)
您可以为big
创建一个将id
与name
映射的映射。之后,您可以遍历small
列表并构建新的词典列表。
>>> big_map = {d['id']:d['name'] for d in big}
>>> res = [{'id': d['id'], 'url':d['url'], 'name':big_map[d['id']]} for d in small if d['id'] in big_map]
>>> res
[{'id': 1236, 'url': 'directUrl1', 'name': 'ipod touch'}, {'id': 1235, 'url': 'directUrl2', 'name': 'ipod x'}]