临时解压字典

Question

说，我有这样的dic

my_dictionary = {'a':1,'c':5,'b':20,'d':7}

现在，我想用我的 dic 做这个：

if my_dictionary['a'] == 1 and my_dictionary['d'] == 7:
    print my_dictionary['c']

这看起来很荒谬，因为我输入了 my_dictionary 3 次！

那么有没有语法可以让我做这样的事情：

within my_dictionary:
    if a == 1 and d == 7:
        print c

如果我的 dic 中没有更多内容（在本例中为 b），这实际上会起作用：

def f(a,d,c):
    if a == 1 and d == 7:
        print c 

f(**my_dictionary)

Answer 1

您可以将函数更改为

def f(a,d,c,**args):
    if a == 1 and d == 7:
        print c

即使字典中有其他项目，它也能正常工作。

Answer 2

您可以使用 operator.itemgetter 来最小化多重索引：

>>> if operator.itemgetter('a','d')(my_dictionary)==(1,7):
...      print operator.itemgetter('c')(my_dictionary)

并且您可以在函数中使用它:

>>> def get_item(*args):
...   return operator.itemgetter(*args)(my_dictionary)
... 
>>> 
>>> if get_item('a','d')==(1,7):
...      print get_item('c')
... 
5

Answer 3

回答

So is there any syntax which would allow me to do something like this:
within my_dictionary:
if a == 1 and d == 7:
    print c

您可以子class dict 使其具有with 魔术方法。为此，class 必须具有 __enter__ 和 __exit__ 方法。然后，您可以将密钥导出到 with 语句的本地范围，并使用 exit 方法清除它们。

使用 this answer 我能够创建一个这样做的子class:

import inspect
import ctypes

locals_to_fast = ctypes.pythonapi.PyFrame_LocalsToFast
locals_to_fast.restype = None
locals_to_fast.argtypes = [ctypes.py_object, ctypes.c_int]

class WithDict(dict):
    def __enter__(self):
        frame = self.get_frame()
        for k,v in self.iteritems():
            frame.f_locals[str(k)] = v
            locals_to_fast(frame, 1)

    def __exit__(self, exc_type, exc_value, traceback):
        frame = self.get_frame()
        for k in self.keys():
            del frame.f_locals[str(k)]

    def get_frame(self):
        return inspect.getouterframes(inspect.currentframe())[2][0]

使用您的原始示例的测试用例

my_dictionary = WithDict({'a':1,'c':5,'b':20,'d':7})
with my_dictionary:
    if a == 1 and d == 7:
        print c

打印5

with 语句完成后变量被删除

临时解压字典

Temporarily unpack dictionary

python

dictionary

unpack