python - 使用字典值对列表进行排序(从最大值到最小值)

Question

这个问题已经有答案了:

How to sort a list of dictionaries by a value of the dictionary in Python? (22 个回答)

已关闭 6 年前。

我有一个数组，里面有很多字典。但是，我想以一种对字典中的特定键具有最大值的方式对字典进行排序。例如，我有一个如下所示的列表

[
    {
        "num_gurus": 40,
        "id": 119749,
        "code": null,
        "name": "ART 198P",
        "short_name": "ART 198P",
        "title": "Directed Group Study",
        "department_long": null,
        "full_name": "Directed Group Study",
        "department_short": "ART"
    },
    {
        "num_gurus": 3,
        "id": 119825,
        "code": null,
        "name": "ASAMST 198P",
        "short_name": "ASAMST 198P",
        "title": "Supervised Group Study",
        "department_long": null,
        "full_name": "Supervised Group Study",
        "department_short": "ASAMST"
    },
    {
        "num_gurus": 200,
        "id": 119904,
        "code": null,
        "name": "AST 636",
        "short_name": "AST 636",
        "title": "Baudelaire: Art Poetry Modernity",
        "department_long": null,
        "full_name": "Baudelaire: Art Poetry Modernity",
        "department_short": "AST"
    }
]

我希望我的输出对我的字典进行排序，其中键属性“num_gurus”的值从最大值到最小值。预期输出为。

[
    {
        "num_gurus": 200,
        "id": 119904,
        "code": null,
        "name": "AST 636",
        "short_name": "AST 636",
        "title": "Baudelaire: Art Poetry Modernity",
        "department_long": null,
        "full_name": "Baudelaire: Art Poetry Modernity",
        "department_short": "AST"
    }
    {
        "num_gurus": 40,
        "id": 119749,
        "code": null,
        "name": "ART 198P",
        "short_name": "ART 198P",
        "title": "Directed Group Study",
        "department_long": null,
        "full_name": "Directed Group Study",
        "department_short": "ART"
    },
    {
        "num_gurus": 3,
        "id": 119825,
        "code": null,
        "name": "ASAMST 198P",
        "short_name": "ASAMST 198P",
        "title": "Supervised Group Study",
        "department_long": null,
        "full_name": "Supervised Group Study",
        "department_short": "ASAMST"
    }

]

到目前为止我已经尝试过了

    for items in load_as_json:
            for key, val in sorted(items['num_gurus'].iteritems(), key=lambda (k,v): (v,k), reverse=True):
                print key,val

This throws me error and doesn't do what I actually want to.
This is the error I got.
  File "utils.py", line 61, in GetPopularCoursesBasedOnGurus
    for key, val in sorted(str(items['num_gurus']).iteritems(), key=lambda (k,v): (v,k)):
AttributeError: 'str' object has no attribute 'iteritems'

Answer 1

试试这个:

my_list.sort(key=lambda my_dict: my_dict["num_gurus"], reverse=True)

这基本上有两件事:

• 关键参数需要一个匿名函数(python 中的 lambda)，然后进行排序原始列表值除以返回的值 lambda 函数。 lambda my_dict: my_dict["num_gurus"] 返回每个字典中的“num_gurus”项，因此列表按这些值排序。
• reverse=True 默认情况下，排序函数从最小到最大排序，因此这只是相反

我还发现这非常“不安全”，因为你在字典中没有“num_gurus”键的保证，或者字典作为键值，因此我个人会用一些异常处理程序来包装它: try\除了

在此处了解更多信息:https://docs.python.org/2.7/tutorial/errors.html ，记住，安全总比后悔好!