我有一个简单的memoizer装饰:memoize的装饰未能memoize的(不使用修饰语法时)
def funcmemo(f):
memo = {}
@wraps(f)
def wrapper(*args):
if args in memo:
return memo[args]
else:
temp = f(*args)
print "memoizing: ", args, temp
memo[args] = temp
return temp
return wrapper
现在,当我通过 “@” 符号使用,
@funcmemo
def fib(n):
print "fib called with:", n
if n < 2: return n
return fib(n-2) + fib(n-1)
res = fib(3)
print "result:", res
它工作正常,如在打印输出中看到:
fib called with: 3
fib called with: 1
memoizing: (1,) 1
fib called with: 2
fib called with: 0
memoizing: (0,) 0
memoizing: (2,) 1
memoizing: (3,) 2
result: 2
然而,当我这样做:
def fib(n):
print "fib called with:", n
if n < 2: return n
return fib(n-2) + fib(n-1)
memfib = funcmemo(fib)
res = memfib(3)
print "result:", res
显然未修饰的FIB被调用,只有最后的返回值“达到”高速缓存(显然产生了巨大的经济放缓):
fib called with: 3
fib called with: 1
fib called with: 2
fib called with: 0
fib called with: 1
memoizing: (3,) 2
result: 2
奇怪的是,这一个正常工作:
def fib(n):
print "fib called with:", n
if n < 2: return n
return fib(n-2) + fib(n-1)
fib = funcmemo(fib)
res = fib(3)
print "result:", res
而且,同样的事情发生与基于类的版本:
class Classmemo(object):
def __init__ (self, f):
self.f = f
self.mem = {}
def __call__ (self, *args):
if args in self.mem:
return self.mem[args]
else:
tmp = self.f(*args)
print "memoizing: ", args, temp
self.mem[args] = tmp
return tmp
使用“匿名”装饰功能时,就像
res = Classmemo(fib)(3)
我会很高兴得到启发关于这背后的原因也发生
的问题。
是的,你是对的。但你的问题在哪里? – Alfe 2012-03-22 09:54:43