从具有固定键值的字符串的列表创建多个字典答案

【问题标题】：Create multiple dictionaries from lists with strings with fixed key values从具有固定键值的字符串的列表创建多个字典
【发布时间】：2021-06-30 18:16:43
【问题描述】：

我有一个用户信息列表，例如

[name1,link1,id1,name2,link2,id2,name3,link3,id3,...]

我想输出一个包含多个字典的列表

[
 {"name":"name1",
  "link":"link1",
  "id":"id1"
 },
 {"name":"name2",
  "link":"link2",
  "id":"id2"
 },
 {"name":"name3",
  "link":"link3",
  "id":"id3"
 }
]

一开始我试过这个

user_info_keys = ["name","link","id","name","link","id","name","link","id"]

user_info_value = ["name1","link1","id1","name2","link2","id2","name3","link3","id3"]

for keys,value in zip(user_info_keys,user_info_value):
    user_info_dict = dict(zip(user_info_keys,user_info_value))

但它只输出

{"name":"name3","link":"link3","id":"id3"}

我应该如何更改代码以获得预期的结果？

【问题讨论】：

你必须为这个列表中的每 3 个元素创建一个新的字典。您发布的代码不断覆盖以前的数据。如果不出意外，请参阅有关如何“分块”列表的帖子。

标签： python python-3.x list dictionary

【解决方案1】：

我可能只是采用一种愚蠢的方法和一个索引来获取每个部分：

user_info_keys = ["name","link","id","name","link","id","name","link","id"]

user_info_value = ["name1","link1","id1","name2","link2","id2","name3","link3","id3"]

outputs = [
    dict(zip(user_info_keys[i:i+3], user_info_value[i:i+3]))
    for i in range(0, len(user_info_keys), 3)
]

请注意，如果每个 dict 没有恰好 3 个键，这将失败。

【讨论】：

【解决方案2】：

你在同一个轨道上我会采取 - 压缩列表一起创建字典。

你只需要把你的列表分成几块，然后对每一块都做。

itertools docs 中有一个 grouper 函数示例可用于此目的。

from itertools import zip_longest

def grouper(iterable, n, fillvalue=None):
    "Collect data into fixed-length chunks or blocks"
    # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return zip_longest(*args, fillvalue=fillvalue)

keys = ["name", "link", "id"]

user_info = ["name1","link1","id1","name2","link2","id2","name3","link3","id3"]

groups = grouper(user_info, n=len(keys), fillvalue='')

users = [dict(zip(keys, values)) for values in groups]

【讨论】：