鸭子将比列表理解或过滤器快得多。它在你的对象上建立一个索引，这样查找就不需要扫描每一个项目。

PIP安装鸭

from ducks import Dex

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

# Build the index
dex = Dex(dicts, {'name': str, 'age': int})

# Find matching objects
dex[{'name': 'Pam', 'age': 7}]

结果:[{'name': 'Pam'， 'age': 7}]

names = [{'name':'Tom', 'age': 10}, {'name': 'Mark', 'age': 5}, {'name': 'Pam', 'age': 7}]
resultlist = [d    for d in names     if d.get('name', '') == 'Pam']
first_result = resultlist[0]

这是一种方法……

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")

你可以通过使用Python中的filter和next方法来实现这一点。

方法过滤给定序列并返回一个迭代器。 Next方法接受迭代器并返回列表中的下一个元素。

所以你可以通过，

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

next(filter(lambda obj: obj.get('name') == 'Pam', my_dict), None)

输出是，

{'name': 'Pam', 'age': 7}

注意:如果没有找到所搜索的名称，上述代码将返回None。

你可以试试这个:

''' lst: list of dictionaries '''
lst = [{"name": "Tom", "age": 10}, {"name": "Mark", "age": 5}, {"name": "Pam", "age": 7}]

search = raw_input("What name: ") #Input name that needs to be searched (say 'Pam')

print [ lst[i] for i in range(len(lst)) if(lst[i]["name"]==search) ][0] #Output
>>> {'age': 7, 'name': 'Pam'} 

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