我有一个Python字典，我想做的是从中获取一些值作为列表，但我不知道实现是否支持。
myDictionary.get('firstKey') # works fine
myDictionary.get('firstKey','secondKey')
# gives me a KeyError -> OK, get is not defined for multiple keys
myDictionary['firstKey','secondKey'] # doesn't work either
有什么办法可以实现这个目标吗？在我的示例中，这看起来很简单，但假设我有一个包含 20 个条目的字典，并且我想要获取 5 个键。除了执行以下操作还有其他方法吗？
myDictionary.get('firstKey')
myDictionary.get('secondKey')
myDictionary.get('thirdKey')
myDictionary.get('fourthKey')
myDictionary.get('fifthKey')
已经存在这样的功能：
from operator import itemgetter
my_dict = {x: x**2 for x in range(10)}
itemgetter(1, 3, 2, 5)(my_dict)
#>>> (1, 9, 4, 25)
itemgetteritemgetteritemgetter(*wanted_keys)(my_dict)
请记住，当仅请求一个键时，
itemgetter使用
forkeys = ['firstKey', 'secondKey', 'thirdKey']
for key in keys:
myDictionary.get(key)
或列表理解：
[myDictionary.get(key) for key in keys]
我建议使用非常有用的
mapmydictionary = {'a': 'apple', 'b': 'bear', 'c': 'castle'}
keys = ['b', 'c']
values = list( map(mydictionary.get, keys) )
# values = ['bear', 'castle']
我在这里没有看到类似的答案 - 值得指出的是，通过使用（列表/生成器）理解，您可以解压这些多个值并将它们分配给单行代码中的多个变量：
first_val, second_val = (myDict.get(key) for key in [first_key, second_key])
我认为列表理解是最干净的方法之一，不需要任何额外的导入：
>>> d={"foo": 1, "bar": 2, "baz": 3}
>>> a = [d.get(k) for k in ["foo", "bar", "baz"]]
>>> a
[1, 2, 3]
或者，如果您希望将值作为单个变量，则使用多重赋值：
>>> a,b,c = [d.get(k) for k in ["foo", "bar", "baz"]]
>>> a,b,c
(1, 2, 3)
%timeititemgetterpydashfrom operator import itemgetter
my_dict = {x: x**2 for x in range(10)}
req_keys = [1, 3, 2, 5]
%timeit itemgetter(1, 3, 2, 5)(my_dict)
257 ns ± 4.61 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
%timeit [my_dict.get(key) for key in req_keys]
604 ns ± 6.94 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
%timeit list( map(my_dict.get, req_keys) )
529 ns ± 34.2 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
!pip install pydash
from pydash import at
%timeit at(my_dict, 1, 3, 2, 5)
22.2 µs ± 572 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit (my_dict.get(key) for key in req_keys)
308 ns ± 6.53 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
s = pd.Series(my_dict)
%timeit s[req_keys]
334 µs ± 58.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
如果您安装了
pandasimport pandas as pd
s = pd.Series(my_dict)
s[['key1', 'key3', 'key2']]
如果你想保留值到键的映射，你应该使用字典理解：
{key: myDictionary[key] for key in [
'firstKey',
'secondKey',
'thirdKey',
'fourthKey',
'fifthKey'
]}
def get_all_values(nested_dictionary):
for key, value in nested_dictionary.items():
if type(value) is dict:
get_all_values(value)
else:
print(key, ":", value)
nested_dictionary = {'ResponseCode': 200, 'Data': {'256': {'StartDate': '2022-02-07', 'EndDate': '2022-02-27', 'IsStoreClose': False, 'StoreTypeMsg': 'Manual Processing Stopped', 'is_sync': False}}}
get_all_values(nested_dictionary)
列表理解方法略有不同。
#doc
[dict[key] for key in (tuple_of_searched_keys)]
#example
my_dict = {x: x**2 for x in range(10)}
print([my_dict[key] for key in (8,9)])
简单方法：
data = { key : base_data.get(key) for key in ['first_name', 'last_name','phone_number']}
如果后备键不是太多，你可以这样做
value = my_dict.get('first_key') or my_dict.get('second_key')
def get_all_values(nested_dictionary):
for key, val in nested_dictionary.items():
data_list = []
if type(val) is dict:
for key1, val1 in val.items():
data_list.append(val1)
return data_list