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)

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。

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'}

names = [{'name':'Tom', 'age': 10}, {'name': 'Mark', 'age': 5}, {'name': 'Pam', 'age': 7}]
resultlist = [d    for d in names     if d.get('name', '') == 'Pam']
first_result = resultlist[0]

这是一种方法……

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'}]

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'}