我的第一个想法是，你可能想要考虑为这些字典创建一个字典……举个例子，如果你要多次搜索它。

然而，这可能是一个不成熟的优化。有什么问题:

def get_records(key, store=dict()):
    '''Return a list of all records containing name==key from our store
    '''
    assert key is not None
    return [d for d in store if d['name']==key]

people = [
{'name': "Tom", 'age': 10},
{'name': "Mark", 'age': 5},
{'name': "Pam", 'age': 7}
]

def search(name):
    for p in people:
        if p['name'] == name:
            return p

search("Pam")

为@FrédéricHamidi添加一点点。

如果你不确定字典列表中是否有键，这样做会有帮助:

next((item for item in dicts if item.get("name") and item["name"] == "Pam"), None)

我会像这样创建一个字典的字典:

names = ["Tom", "Mark", "Pam"]
ages = [10, 5, 7]
my_d = {}

for i, j in zip(names, ages):
    my_d[i] = {"name": i, "age": j}

或者，使用与问题中完全相同的信息:

info_list = [{"name": "Tom", "age": 10}, {"name": "Mark", "age": 5}, {"name": "Pam", "age": 7}]
my_d = {}

for d in info_list:
    my_d[d["name"]] = d

然后你可以执行my_d["Pam"]，得到{"name": "Pam"， "age": 7}

我在寻找答案的时候发现了这个帖子 的问题。虽然我知道这回答有点晚，但我想 把它贡献出来，以防对其他人有用:

def find_dict_in_list(dicts, default=None, **kwargs):
    """Find first matching :obj:`dict` in :obj:`list`.

    :param list dicts: List of dictionaries.
    :param dict default: Optional. Default dictionary to return.
        Defaults to `None`.
    :param **kwargs: `key=value` pairs to match in :obj:`dict`.

    :returns: First matching :obj:`dict` from `dicts`.
    :rtype: dict

    """

    rval = default
    for d in dicts:
        is_found = False

        # Search for keys in dict.
        for k, v in kwargs.items():
            if d.get(k, None) == v:
                is_found = True

            else:
                is_found = False
                break

        if is_found:
            rval = d
            break

    return rval


if __name__ == '__main__':
    # Tests
    dicts = []
    keys = 'spam eggs shrubbery knight'.split()

    start = 0
    for _ in range(4):
        dct = {k: v for k, v in zip(keys, range(start, start+4))}
        dicts.append(dct)
        start += 4

    # Find each dict based on 'spam' key only.  
    for x in range(len(dicts)):
        spam = x*4
        assert find_dict_in_list(dicts, spam=spam) == dicts[x]

    # Find each dict based on 'spam' and 'shrubbery' keys.
    for x in range(len(dicts)):
        spam = x*4
        assert find_dict_in_list(dicts, spam=spam, shrubbery=spam+2) == dicts[x]

    # Search for one correct key, one incorrect key:
    for x in range(len(dicts)):
        spam = x*4
        assert find_dict_in_list(dicts, spam=spam, shrubbery=spam+1) is None

    # Search for non-existent dict.
    for x in range(len(dicts)):
        spam = x+100
        assert find_dict_in_list(dicts, spam=spam) is None

你可以使用生成器表达式:

>>> dicts = [
...     { "name": "Tom", "age": 10 },
...     { "name": "Mark", "age": 5 },
...     { "name": "Pam", "age": 7 },
...     { "name": "Dick", "age": 12 }
... ]

>>> next(item for item in dicts if item["name"] == "Pam")
{'age': 7, 'name': 'Pam'}

如果你需要处理不存在的项，那么你可以做用户Matt在他的评论中建议的事情，并使用稍微不同的API提供默认值:

next((item for item in dicts if item["name"] == "Pam"), None)

并且要查找项目的索引，而不是项目本身，您可以枚举()列表:

next((i for i, item in enumerate(dicts) if item["name"] == "Pam"), None)