Python：从自身内部获取对函数的引用
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Python: getting a reference to a function from inside itself
提问by Ram Rachum
If I define a function:
如果我定义一个函数：
def f(x):
return x+3
I can later store objects as attributes of the function, like so:
我以后可以将对象存储为函数的属性，如下所示：
f.thing="hello!"
I would like to do this from inside the code of the function itself. Problem is, how do I get a reference to the function from inside itself?
我想从函数本身的代码内部执行此操作。问题是，如何从内部获取对函数的引用？
采纳答案by SurDin
The same way, just use its name.
同样的方法，只需使用它的名字。
>>> def g(x):
... g.r = 4
...
>>> g
<function g at 0x0100AD68>
>>> g(3)
>>> g.r
4
回答by Jeff Ober
If you are trying to do memoization, you can use a dictionary as a default parameter:
如果您尝试进行记忆，可以使用字典作为默认参数：
def f(x, memo={}):
if x not in memo:
memo[x] = x + 3
return memo[x]
回答by Jeff Ober
Or use a closure:
或者使用闭包：
def gen_f():
memo = dict()
def f(x):
try:
return memo[x]
except KeyError:
memo[x] = x + 3
return f
f = gen_f()
f(123)
Somewhat nicer IMHO
稍微好一点恕我直言
