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
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]
其他回答
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)
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。
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]
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))