在一个装饰的装饰之外创建访问函数属性 [英] Accessing function attribute created in a decorator outside that decorator

问题描述

我要计算时代给定函数被调用的次数。

I want to count the number of times a given function has been called.

所以，我做了一个 countcalls 装饰给我的功能，这得到每次通话递增 __ callcount 属性。够简单了。

So, I made a countcalls decorator to give my functions a __callcount attribute which gets incremented on each call. Simple enough.

我的问题是获得 __ callcount 值回出来以后。

My issue is getting the __callcount value back out later.

下面是我的code：

import functools

def countcalls(f):
    f.__callcount = 0

    @functools.wraps(f)
    def _countcalls(*args, **kwds):
        f.__callcount += 1
        print('  Called {0} time(s).'.format(f.__callcount))
        return f(*args, **kwds)
    return _countcalls

@countcalls
def fib(n):
    if n < 0:
        raise ValueError('n must be > 0')
    if n == 0 or n == 1:
        return 1

    return fib(n-1) + fib(n-2)

if __name__ == '__main__':
    print('Calling fib(3)...')
    x = fib(3)
    print('fib(3) = {0}'.format(x))

    print('Calling fib(3) again...')
    x = fib(3)
    print('fib(3) = {0}'.format(x))

    print('fib was called a total of {0} time(s).'.format(fib.__callcount)) 

产生以下输出（Python的3​​.3.0）：

Which generates the following output (Python v3.3.0):

Calling fib(3)...
  Called 1 time(s).
  Called 2 time(s).
  Called 3 time(s).
  Called 4 time(s).
  Called 5 time(s).
fib(3) = 3
Calling fib(3) again...
  Called 6 time(s).
  Called 7 time(s).
  Called 8 time(s).
  Called 9 time(s).
  Called 10 time(s).
fib(3) = 3
fib was called a total of 0 time(s).

为什么 FIB .__ callcount 在最后一行等于 0 ？由于输出显示， __ callcount 被递增， FIB 。

Why does fib.__callcount equal 0 on the last line? As the output shows, __callcount gets incremented, and persists between calls of fib.

我是什么失踪？

推荐答案

那么，这里的原因，一点帮助后。谢谢你们！

Well, here's the reason, after a bit of help. Thanks guys!

问题是功能是不可变的。例如。

The issue is that functions are immutable. e.g.

>>> def f(func):
...     return func()
...
>>> def g():
...     return 'sunflower seeds'
...
>>> id(g)
139636515497336
>>> g = f(g)
>>> id(g)
139636515515112

因此​​，要获得功能˚F我们的定义分配 __ callcount 属性的唯一办法 countcalls 是返回从 callcount 。但是，我们已经返回内部函数 _countcalls 。我们可以同时返回˚F和 _countcalls 但搅乱 @countcalls 修饰语法。

So, the only way to get the function f we assigned the __callcount attribute to in the definition of countcalls is to return that function from callcount. But we're already returning the inner function _countcalls. We can return both f and _countcalls but that messes up the @countcalls decorator syntax.

您仍然可以这样来做，它只是并不像pretty。

You can still do it this way, it's just not as pretty.

import functools

def countcalls(f):
    f.__callcount = 0

    @functools.wraps(f)
    def _countcalls(*args, **kwds):
        f.__callcount += 1
        print('  Called {0} time(s).'.format(f.__callcount))
        return f(*args, **kwds)
    return f, _countcalls

def fib(n):
    if n < 0:
        raise ValueError('n must be > 0')
    if n == 0 or n == 1:
        return 1

    return fib(n-1) + fib(n-2)

if __name__ == '__main__':
    counter, fib = countcalls(fib)

    print('Calling fib(3)...')
    x = fib(3)
    print('fib(3) = {0}'.format(x))

    print('Calling fib(3) again...')
    x = fib(3)
    print('fib(3) = {0}'.format(x))

    print('fib was called a total of {0} time(s).'.format(counter.__callcount))

长话短说，只需使用类从Python的装饰图书馆。 ：D

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