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……

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")

这里提出的大多数(如果不是全部)实现都有两个缺陷:

他们假设只传递一个键来进行搜索，而对于复杂的字典，有更多的键可能是有趣的 它们假定所有传递用于搜索的键都存在于字典中，因此当KeyError不存在时，它们不会正确处理。

更新后的命题:

def find_first_in_list(objects, **kwargs):
    return next((obj for obj in objects if
                 len(set(obj.keys()).intersection(kwargs.keys())) > 0 and
                 all([obj[k] == v for k, v in kwargs.items() if k in obj.keys()])),
                None)

也许不是最python化的，但至少更安全一点。

用法:

>>> obj1 = find_first_in_list(list_of_dict, name='Pam', age=7)
>>> obj2 = find_first_in_list(list_of_dict, name='Pam', age=27)
>>> obj3 = find_first_in_list(list_of_dict, name='Pam', address='nowhere')
>>> 
>>> print(obj1, obj2, obj3)
{"name": "Pam", "age": 7}, None, {"name": "Pam", "age": 7}

要点。

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}

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}]