def dsearch(lod, **kw):
    return filter(lambda i: all((i[k] == v for (k, v) in kw.items())), lod)

lod=[{'a':33, 'b':'test2', 'c':'a.ing333'},
     {'a':22, 'b':'ihaha', 'c':'fbgval'},
     {'a':33, 'b':'TEst1', 'c':'s.ing123'},
     {'a':22, 'b':'ihaha', 'c':'dfdvbfjkv'}]



list(dsearch(lod, a=22))

[{'a': 22, 'b': 'ihaha', 'c': 'fbgval'},
 {'a': 22, 'b': 'ihaha', 'c': 'dfdvbfjkv'}]



list(dsearch(lod, a=22, b='ihaha'))

[{'a': 22, 'b': 'ihaha', 'c': 'fbgval'},
 {'a': 22, 'b': 'ihaha', 'c': 'dfdvbfjkv'}]


list(dsearch(lod, a=22, c='fbgval'))

[{'a': 22, 'b': 'ihaha', 'c': 'fbgval'}]

@Frédéric Hamidi的回答很好。在Python 3中。X .next()的语法稍有改变。因此有一个小小的修改:

>>> 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'}

正如@Matt在评论中提到的，你可以添加一个默认值:

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

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}

这里是一个比较，使用遍历列表，使用过滤器+lambda或重构(如果需要或对你的情况有效)你的代码dict of dicts而不是list of dicts

import time

# Build list of dicts
list_of_dicts = list()
for i in range(100000):
    list_of_dicts.append({'id': i, 'name': 'Tom'})

# Build dict of dicts
dict_of_dicts = dict()
for i in range(100000):
    dict_of_dicts[i] = {'name': 'Tom'}


# Find the one with ID of 99

# 1. iterate through the list
lod_ts = time.time()
for elem in list_of_dicts:
    if elem['id'] == 99999:
        break
lod_tf = time.time()
lod_td = lod_tf - lod_ts

# 2. Use filter
f_ts = time.time()
x = filter(lambda k: k['id'] == 99999, list_of_dicts)
f_tf = time.time()
f_td = f_tf- f_ts

# 3. find it in dict of dicts
dod_ts = time.time()
x = dict_of_dicts[99999]
dod_tf = time.time()
dod_td = dod_tf - dod_ts


print 'List of Dictionries took: %s' % lod_td
print 'Using filter took: %s' % f_td
print 'Dict of Dicts took: %s' % dod_td

输出是这样的:

List of Dictionries took: 0.0099310874939
Using filter took: 0.0121960639954
Dict of Dicts took: 4.05311584473e-06

结论: 显然，字典是最有效的搜索方法在这种情况下，你知道，你只能通过id搜索。 有趣的是，使用过滤器是最慢的解决方案。

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)

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

from collections import defaultdict
dicts_by_name=defaultdict(list)
for d in dicts:
    dicts_by_name[d['name']]=d

print dicts_by_name['Tom']

#output
#>>>
#{'age': 10, 'name': 'Tom'}