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
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]
其他回答
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'}
@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
>>>
鸭子将比列表理解或过滤器快得多。它在你的对象上建立一个索引,这样查找就不需要扫描每一个项目。
PIP安装鸭
from ducks import Dex
dicts = [
{"name": "Tom", "age": 10},
{"name": "Mark", "age": 5},
{"name": "Pam", "age": 7}
]
# Build the index
dex = Dex(dicts, {'name': str, 'age': int})
# Find matching objects
dex[{'name': 'Pam', 'age': 7}]
结果:[{'name': 'Pam', 'age': 7}]
简单地使用列表理解:
[i for i in dct if i['name'] == 'Pam'][0]
示例代码:
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'}
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']