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)
其他回答
@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
>>>
你必须遍历列表中的所有元素。没有捷径!
除非在其他地方保存了一个包含指向列表项的名称的字典,但这时必须处理从列表中弹出元素的后果。
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
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'}