假设我有这个:
[
{"name": "Tom", "age": 10},
{"name": "Mark", "age": 5},
{"name": "Pam", "age": 7}
]
通过搜索“Pam”作为名称,我想检索相关的字典:{name:“Pam”,年龄:7}
如何做到这一点?
假设我有这个:
[
{"name": "Tom", "age": 10},
{"name": "Mark", "age": 5},
{"name": "Pam", "age": 7}
]
通过搜索“Pam”作为名称,我想检索相关的字典:{name:“Pam”,年龄:7}
如何做到这一点?
当前回答
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
其他回答
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
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))
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