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
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。
这里提出的大多数(如果不是全部)实现都有两个缺陷:
他们假设只传递一个键来进行搜索,而对于复杂的字典,有更多的键可能是有趣的 它们假定所有传递用于搜索的键都存在于字典中,因此当KeyError不存在时,它们不会正确处理。
更新后的命题:
def find_first_in_list(objects, **kwargs):
return next((obj for obj in objects if
len(set(obj.keys()).intersection(kwargs.keys())) > 0 and
all([obj[k] == v for k, v in kwargs.items() if k in obj.keys()])),
None)
也许不是最python化的,但至少更安全一点。
用法:
>>> obj1 = find_first_in_list(list_of_dict, name='Pam', age=7)
>>> obj2 = find_first_in_list(list_of_dict, name='Pam', age=27)
>>> obj3 = find_first_in_list(list_of_dict, name='Pam', address='nowhere')
>>>
>>> print(obj1, obj2, obj3)
{"name": "Pam", "age": 7}, None, {"name": "Pam", "age": 7}
要点。
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
def dsearch(lod, **kw):
return filter(lambda i: all((i[k] == v for (k, v) in kw.items())), lod)
lod=[{'a':33, 'b':'test2', 'c':'a.ing333'},
{'a':22, 'b':'ihaha', 'c':'fbgval'},
{'a':33, 'b':'TEst1', 'c':'s.ing123'},
{'a':22, 'b':'ihaha', 'c':'dfdvbfjkv'}]
list(dsearch(lod, a=22))
[{'a': 22, 'b': 'ihaha', 'c': 'fbgval'},
{'a': 22, 'b': 'ihaha', 'c': 'dfdvbfjkv'}]
list(dsearch(lod, a=22, b='ihaha'))
[{'a': 22, 'b': 'ihaha', 'c': 'fbgval'},
{'a': 22, 'b': 'ihaha', 'c': 'dfdvbfjkv'}]
list(dsearch(lod, a=22, c='fbgval'))
[{'a': 22, 'b': 'ihaha', 'c': 'fbgval'}]