你试过熊猫套餐吗?它非常适合这类搜索任务，也进行了优化。

import pandas as pd

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

# Create a data frame, keys are used as column headers.
# Dict items with the same key are entered into the same respective column.
df = pd.DataFrame(listOfDicts)

# The pandas dataframe allows you to pick out specific values like so:

df2 = df[ (df['name'] == 'Pam') & (df['age'] == 7) ]

# Alternate syntax, same thing

df2 = df[ (df.name == 'Pam') & (df.age == 7) ]

我在下面添加了一些基准测试，以说明熊猫在更大范围内(即10万+条目)的更快运行时间:

setup_large = 'dicts = [];\
[dicts.extend(({ "name": "Tom", "age": 10 },{ "name": "Mark", "age": 5 },\
{ "name": "Pam", "age": 7 },{ "name": "Dick", "age": 12 })) for _ in range(25000)];\
from operator import itemgetter;import pandas as pd;\
df = pd.DataFrame(dicts);'

setup_small = 'dicts = [];\
dicts.extend(({ "name": "Tom", "age": 10 },{ "name": "Mark", "age": 5 },\
{ "name": "Pam", "age": 7 },{ "name": "Dick", "age": 12 }));\
from operator import itemgetter;import pandas as pd;\
df = pd.DataFrame(dicts);'

method1 = '[item for item in dicts if item["name"] == "Pam"]'
method2 = 'df[df["name"] == "Pam"]'

import timeit
t = timeit.Timer(method1, setup_small)
print('Small Method LC: ' + str(t.timeit(100)))
t = timeit.Timer(method2, setup_small)
print('Small Method Pandas: ' + str(t.timeit(100)))

t = timeit.Timer(method1, setup_large)
print('Large Method LC: ' + str(t.timeit(100)))
t = timeit.Timer(method2, setup_large)
print('Large Method Pandas: ' + str(t.timeit(100)))

#Small Method LC: 0.000191926956177
#Small Method Pandas: 0.044392824173
#Large Method LC: 1.98827004433
#Large Method Pandas: 0.324505090714

在我看来，这是最python的方式:

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

filter(lambda person: person['name'] == 'Pam', people)

result(在Python 2中作为列表返回):

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

注意:在python3中，返回一个filter对象。所以python3的解决方案是:

list(filter(lambda person: person['name'] == 'Pam', people))

简单地使用列表理解:

[i for i in dct if i['name'] == 'Pam'][0]

示例代码:

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

print([i for i in dct if i['name'] == 'Pam'][0])

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

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]

你可以使用列表推导式:

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

使用列表推导式的一个简单方法是，如果l是列表

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

然后

[d['age'] for d in l if d['name']=='Tom']