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
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'}
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'}
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")
@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}]
你可以试试这个:
''' 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'}