如果其他元素使用理解匹配，则从两个字典列表中提取元素

Question

如何迭代两个字典列表，通过键匹配列表之间的字典，如果匹配，则将每个字典中的特定键附加到新字典中的键值对中。让我用一个例子来澄清：l1 = [{'id': 52, 'email': '[e-mail]', 'anotherfield': 'some value'},       .....      {'id': 98, 'email': '[e-mail]', 'anotherfield': 'another value'}]l2 = [{'id': 93, 'email': '[e-mail]', 'another key': 'seventeen'},       .....      {'id': 101, 'email': '[e-mail]', 'another key': 'twenty'}]# match the 'email' keys between each list, and if match, create k, v pair from id'sdesired_output = {'52': 93.....'98': 101}通过简单地迭代每个列表，我可以很容易地实现这一点，如下所示：lookup = dict()for l in l1:    for p in l2:        if l['email']==p['email']:            lookup[l['id']]=p['id']            break但是，这有点笨拙，我更喜欢某种理解。我的尝试：lookup = {k['id']: v['id'] for k, v in zip(l1, l2) if k['email'] == v['email']}

Answer 1

试试这个：

from itertools import product

lookup = {k['id']: v['id'] for k, v in product(l1, l2) if k['email'] == v['email']}

Answer 2

不一致列表的解决方法：

l1 = [{"email": "email1", "id": 1}, {"email": "email2", "id": 2}, {"email": "email3", "id": 3}]

l2 = [{"email": "email2", "id": 22}, {"email": "email4", "id": 4}, {"email": "email1", "id": 11}, ]

emails = {}

lookup = {}

for el in l1:

    emails[el["email"]] = el["id"]

for el in l2:

    email = el["email"]

    if email in emails:

        lookup[emails[email]] = el["id"]

# {1: 11, 2: 22}

print(lookup)

# bad solution from question

lookup = {k['id']: v['id'] for k, v in zip(l1, l2) if k['email'] == v['email']}

# {} - empty

print(lookup)

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