带字典值的排序列表（从大到小）

Question

我有一个包含大量字典的数组。但是，我希望对字典进行排序，使字典中的特定键具有最大值。 例如，我有一个如下的列表

[
    {
        "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"
    }
]

我希望我的输出对字典进行排序，其中键属性'num\u gurus'的值从最大到最小。预期产出将是。你知道吗

[
    {
        "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"
    }

]

我已经试过了

    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

要将排序后的列表存储为新列表，可以使用^{}作为：

sorted(my_list, key=lambda x: x['num_gurus'], reverse=True)
# returns sorted list

其中my_list是list个dict对象。你知道吗

否则，如果要对原始列表的内容进行排序，即my_list，则使用list.sort()作为：

my_list.sort(key=lambda x: x["num_gurus"], reverse=True)
# sorts the original list

检查文档位置：How to do Sorting in list

Answer 2

试试这个：

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

这基本上是两件事：

• key paramater需要一个匿名函数（python中的lambda）然后排序 原始列表值由 lambda函数。lambda my_dict: my_dict["num_gurus"]返回每个字典中的“num\u gurus”项，因此列表按这些值排序。你知道吗
• reverse=True默认情况下，排序函数从最小值排序到最大值，因此 这正好相反

另外，我发现这非常“不安全”，因为您在字典中没有“num\u gurus”键的担保，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\except

在这里阅读更多：https://docs.python.org/2.7/tutorial/errors.html，记住安全总比抱歉好！你知道吗