Question

Python3 天哪，对不起以这种或那种方式可能经常重复的问题。对我来说，操纵名单一直就像一个黑匣子的头痛，我现在已经争斗了一个小时没有结果。

我的表格中有一个列表：

[('John', 'first@email.com'), ('John', 'second@email.com'), ('Jack', 'third@email.com')]

我想将其转换为可迭代的字典（然后我可以使用pymongo将其作为文档批量插入），以便它看起来像这样：

 new_posts = [{"sender": "John",
               "email": "first@email.com"},
              {"sender": "John",
               "email": "second@email.com"},
              {"sender": "Jack",
               "email": "third@email.com"}]

如何以易于阅读和有效的方式实现这一目标？

Answer 1

你可以尝试

>>> a = [('John', 'first@email.com'), ('John', 'second@email.com'), ('Jack', 'third@email.com')]
>>> [dict([('sender', x[0]), ('email',x[1])]) for x in a]
[{'sender': 'John', 'email': 'first@email.com'}, {'sender': 'John', 'email': 'second@email.com'}, {'sender': 'Jack', 'email': 'third@email.com'}]

这会将列表转换为dict

Answer 2

data = [('John', 'first@email.com'), ('John', 'second@email.com'), ('Jack', 'third@email.com')]
new_posts = []
for i in range(len(data)):
    new_post = {}
    new_post["sender"] = data[i][0]
    new_post["email"] = data[i][1]
    new_posts.append(new_post)

另一种解决方法..

Answer 3

your_list = [('John', 'first@email.com'), ('John', 'second@email.com'), ('Jack', 'third@email.com')]
your_result = [{'sender': sender, 'email': email} for sender, email in your_list]

Python 3：将2个元素列入字典

3 个答案: