Question

我在尝试使用python从json数组中的特定键中获取值列表时遇到麻烦。使用下面的JSON示例，我试图创建一个仅包含name键值的列表。原始JSON：

[
    {
        "id": 1,
        "name": "Bulbasaur",
        "type": [
            "grass",
            "poison"
        ]
    },
    {
        "id": 2,
        "name": "Ivysaur",
        "type": [
            "grass",
            "poison"
        ]
    }
]

预期：

["Bulbasaur", "Ivysaur"]

下面是我的方法的代码：

import json
try:
    with open("./simple.json", 'r') as f:
        contents = json.load(f)
except Exception as e:
    print(e)

print(contents[:]["name"])

我正在尝试一种不需要循环每个索引并附加它们的方法，就像上面的代码一样。使用python的json库可以实现这种方法吗？

Answer 1

您无法执行contents[:]["name"]，因为contents是一个包含整数索引的字典列表，并且您不能使用字符串name来访问元素。

要解决此问题，您需要遍历列表并为每个name获取密钥item的值

import json
contents = []

try:
    with open("./simple.json", 'r') as f:
        contents = json.load(f)
except Exception as e:
    print(e)


li = [item.get('name') for item in contents]
print(li)

输出将为

['Bulbasaur', 'Ivysaur']

Answer 2

尝试使用list comprehensions：

print([d["name"] for d in contents])

Answer 3

这不是该问题的真实答案。真正的答案是使用列表理解。但是，您可以创建一个类，该类允许您专门使用在问题中尝试过的语法。总体思路是将list子类化，以便像[:]这样的切片将特殊视图（另一个类）返回到列表中。然后，该特殊视图将允许同时从所有词典中进行检索和分配。

class DictView:
    """
    A special class for getting and setting multiple dictionaries
    simultaneously. This class is not meant to be instantiated
    in its own, but rather in response to a slice operation on UniformDictList.
    """
    def __init__(parent, slice):
        self.parent = parent
        self.range = range(*slice.indices(len(parent)))

    def keys(self):
        """
        Retreives a set of all the keys that are shared across all
        indexed dictionaries. This method makes `DictView` appear as
        a genuine mapping type to `dict`.
        """
        key_set = set()
        for k in self.range:
            key_set &= self.parent.keys()
        return key_set

    def __getitem__(self, key):
        """
        Retreives a list of values corresponding to all the indexed
        values for `key` in the parent. Any missing key will raise
        a `KeyError`.
        """
        return [self.parent[k][key] for k in self.range]

    def get(self, key, default=None):
        """
        Retreives a list of values corresponding to all the indexed
        values for `key` in the parent. Any missing key will return
        `default`.
        """
        return [self.parent[k].get(key, default) for k in self.range]

    def __setitem__(self, key, value):
        """
        Set all the values in the indexed dictionaries for `key` to `value`.
        """
        for k in self.range:
            self.parent[k][key] = value

    def update(self, *args, **kwargs):
        """
        Update all the indexed dictionaries in the parent with the specified
        values. Arguments are the same as to `dict.update`.
        """
        for k in self.range:
             self.parent[k].update(*args, **kwargs)


class UniformDictList(list):
    def __getitem__(self, key):
        if isinstance(key, slice):
            return DictView(self, key)
        return super().__getitem__(key)

您的原始代码现在可以直接使用，UniformDictList中只需换行即可。

import json
try:
    with open("./simple.json", 'r') as f:
        contents = UniformDictList(json.load(f))
except Exception as e:
    print(e)

print(contents[:]["name"])

如何解析json以获取数组中特定键的所有值？

3 个答案: