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

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]

我测试了各种方法来遍历字典列表并返回键x具有特定值的字典。

结果:

速度:列表推导>生成器表达式>>常规列表迭代>>>过滤器。 所有的比例都与列表中的字典数量线性(10x列表大小-> 10x时间)。 对于大量(数千个)键，每个字典的键不会显著影响速度。请看我计算的图表:https://imgur.com/a/quQzv(方法名称见下文)。

所有测试均使用Python 3.6.4, W7x64完成。

from random import randint
from timeit import timeit


list_dicts = []
for _ in range(1000):     # number of dicts in the list
    dict_tmp = {}
    for i in range(10):   # number of keys for each dict
        dict_tmp[f"key{i}"] = randint(0,50)
    list_dicts.append( dict_tmp )



def a():
    # normal iteration over all elements
    for dict_ in list_dicts:
        if dict_["key3"] == 20:
            pass

def b():
    # use 'generator'
    for dict_ in (x for x in list_dicts if x["key3"] == 20):
        pass

def c():
    # use 'list'
    for dict_ in [x for x in list_dicts if x["key3"] == 20]:
        pass

def d():
    # use 'filter'
    for dict_ in filter(lambda x: x['key3'] == 20, list_dicts):
        pass

结果:

1.7303 # normal list iteration 
1.3849 # generator expression 
1.3158 # list comprehension 
7.7848 # filter

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。

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")

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']