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
将接受的答案放在函数中,以便于重用
def get_item(collection, key, target):
return next((item for item in collection if item[key] == target), None)
也可以写成
get_item_lambda = lambda collection, key, target : next((item for item in collection if item[key] == target), None)
结果
key = "name"
target = "Pam"
print(get_item(target_list, key, target))
print(get_item_lambda(target_list, key, target))
#{'name': 'Pam', 'age': 7}
#{'name': 'Pam', 'age': 7}
如果键可能不在目标字典中,请使用dict。get和避免KeyError
def get_item(collection, key, target):
return next((item for item in collection if item.get(key, None) == target), None)
get_item_lambda = lambda collection, key, target : next((item for item in collection if item.get(key, None) == target), None)
其他回答
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']
你可以使用列表推导式:
def search(name, people):
return [element for element in people if element['name'] == name]
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'}
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")