字典搜索的Python列表

假设我有这个：

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

并通过搜索“ Pam”作为名称，我想检索相关的字典： {name: "Pam", age: 7}

如何实现呢？

您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'}

如果您需要处理不存在的项目，则可以执行用户Matt 在其注释中建议的操作，并使用略有不同的API提供默认值：

next((item for item in dicts if item["name"] == "Pam"), None)

为了找到项目的索引，而不是项目本身，可以枚举（）列表：

next((i for i, item in enumerate(dicts) if item["name"] == "Pam"), None)

在我看来，这是最蟒蛇的方式：

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

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

结果（在Python 2中作为列表返回）：

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

注意：在Python 3中，将返回一个过滤器对象。因此，python3解决方案将是：

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

@FrédéricHamidi的回答很好。在Python 3.x中，语法.next()略有变化。因此稍作修改：

>>> 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'}

如@Matt的评论中所述，您可以这样添加默认值：

>>> 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
>>>

您可以使用列表推导：

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

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")

我测试了各种方法来浏览字典列表，然后返回键x具有特定值的字典。

结果：

• 速度：列表理解>生成器表达式>>普通列表迭代>>>过滤器。
• 全部缩放与列表中的字典数量成线性关系（10倍列表大小-> 10倍时间）。
• 对于大量（数千）键，每个词典的键不会显着影响速度。请查看我计算出的以下图表：https : //imgur.com/a/quQzv（方法名称请参见下文）。

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

向@FrédéricHamidi添加一点点。

如果您不确定某个键是否在字典列表中，可以使用以下方法：

next((item for item in dicts if item.get("name") and item["name"] == "Pam"), None)

您是否尝试过熊猫包装？它非常适合此类搜索任务，并且也进行了优化。

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，以大范围地（即100k +项）说明熊猫的运行时间：

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

这是在字典列表中搜索值的一般方法：

def search_dictionaries(key, value, list_of_dictionaries):
    return [element for element in list_of_dictionaries if element[key] == value]

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

只需使用列表推导：

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

样例代码：

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'}

您可以通过在Python中使用filter和next方法来实现。

filter方法过滤给定的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。

因此，您可以通过以下方式找到元素

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：None如果找不到我们正在搜索的名称，上述代码将返回以防万一。

我的第一个想法是，您可能要考虑创建一个包含这些词典的字典...例如，如果您要搜索的词典次数不止一次。

但是，这可能是过早的优化。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]

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'}

使用列表推导的一种简单方法是，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']

您可以尝试以下方法：

''' 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'} 

这是一个使用迭代遍历列表的比较，使用filter + lambda或重构（如果需要或对您的情况有效）的代码将您的代码用于命令，而不是命令列表

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： 在这些情况下，显然拥有字典词典是最有效的搜索方式，在这种情况下，您知道您将仅通过id进行搜索。有趣的是，使用过滤器是最慢的解决方案。

您必须遍历列表的所有元素。没有捷径！

除非在其他地方保留了指向列表项的名称字典，否则您必须注意从列表中弹出元素的后果。

我在寻找同一问题的答案时找到了这个线程。虽然我意识到这是一个迟来的答案，但我认为我会做出贡献，以防它对其他人有用：

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

这里提出的大多数（如果不是全部）实现都有两个缺陷：

• 他们假定只传递一个键来进行搜索，而对于复杂的字典有更多键可能很有趣
• 他们假定传递给搜索的所有键都存在于字典中，因此当键错误不存在时，它们将无法正确处理。

更新的主张：

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的，但至少具有更多的故障保护功能。

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}

该要点。