本文介绍了字典列表列表中的键列表?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

假设我有一个列表的字典列表。



我很确定Python有一个很好的和简短的方式(不用写2个循环)来检索与该关键字相关联的值。 (每个字典都有一个键)。



我该如何做?



编辑:有一个示例(JSON表示)输入

  [
[
{
lat:45.1845931,
lgt :5.7316984,
name:Chavant,
sens:,
line:[],
stationID:
$,
{
lat:45.1845898,
lgt:5.731746,
name:Chavant,
sens ,
line:[],
stationID:
}
],
[
{
:45.1868233,
lgt:5.7565727,
name:Neyrpic - Belledonne,
sens:,
line:[],
stationID:
},
{
lat:45.1867322,
lgt:5.7568569,
name:Neyrpic - Belledonne,
sens:,
line:[],
stationID:
}
]
]

作为输出,我想要一个名单列表。



PS:ODBL下的数据。

解决方案

如果您需要全部平面列表中的名称:

  response =#你的大列表
[d.get('name')for lst如果要使用内部列表获取结果,则为
  [[d.get('name')for d in lst] for lst in response] 


Suppose I have a list of a list of dictionaries.

I'm pretty sure Python has a nice and short way (without writing 2 for loops) to retrieve the value associated with the key. (Every dictionary has a key).

How can I do this?

Edit : There is a sample (JSON representation) of the input

[
    [
        {
            "lat": 45.1845931,
            "lgt": 5.7316984,
            "name": "Chavant",
            "sens": "",
            "line" : [],
            "stationID" : ""
        },
        {
            "lat": 45.1845898,
            "lgt": 5.731746,
            "name": "Chavant",
            "sens": "",
            "line" : [],
            "stationID" : ""
        }
    ],
    [
        {
            "lat": 45.1868233,
            "lgt": 5.7565727,
            "name": "Neyrpic - Belledonne",
            "sens": "",
            "line" : [],
            "stationID" : ""
        },
        {
            "lat": 45.1867322,
            "lgt": 5.7568569,
            "name": "Neyrpic - Belledonne",
            "sens": "",
            "line" : [],
            "stationID" : ""
        }
    ]
]

As output I'd like to have a list of names.

PS: Data under ODBL.

解决方案

If you need all names in flat list:

response = # your big list
[d.get('name') for lst in response for d in lst]

if you want to get result with inner list:

[[d.get('name') for d in lst] for lst in response]

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07-31 03:35