On 21/04/2012 09:36, Steven D'Aprano wrote:
[...]
Here is how I would write the above.
import functools
def memoise(func):
"""Decorator to memoise a function."""
cache = {}
@functools.wraps(func)
def inner(*args, **kwargs):
# Make sure keyword args are always look
On Saturday, 21 April 2012 09:25:40 UTC+1, Steven D'Aprano wrote:
> On Fri, 20 Apr 2012 09:10:15 -0700, Jon Clements wrote:
>
> >> But I don't know how. I know that I can see the default arguments of
> >> the original function using func.__defaults__, but without knowing the
> >> number and names
On Fri, 20 Apr 2012 16:57:06 +0100, Rotwang wrote:
> def memo(func):
> def memofunc(*args, **kwargs):
> twargs = tuple(kwargs.items())
> if (args, twargs) in memofunc.d:
> return copy(memofunc.d[(args, twargs)])
> memofunc.d[(args, twargs)] = func(*args
On Fri, 20 Apr 2012 09:10:15 -0700, Jon Clements wrote:
>> But I don't know how. I know that I can see the default arguments of
>> the original function using func.__defaults__, but without knowing the
>> number and names of func's positional arguments (which I don't know how
>> to find out) this
On Fri, Apr 20, 2012 at 6:07 PM, Ian Kelly wrote:
> (args, varargs, varkw, defaults) = inspect.getargspec(func)
> if varargs:
> args.append(varargs)
> if varkw:
> args.append("tuple(sorted(%s.items()))" % varkw)
Note that in Python 3, this would need to become something lik
On Fri, Apr 20, 2012 at 9:57 AM, Rotwang wrote:
> As far as I know, the decorated function will always return the same value
> as the original function. The problem is that the dictionary key stored
> depends on how the function was called, even if two calls should be
> equivalent; hence the origi
On 20/04/2012 17:10, Jon Clements wrote:
On Friday, 20 April 2012 16:57:06 UTC+1, Rotwang wrote:
Hi all, here's a problem I don't know how to solve. I'm using Python 2.7.2.
I'm doing some stuff in Python which means I have cause to call
functions that take a while to return. Since I often want
Rotwang wrote:
> I've written a
> decorator to eliminate repeated calls by storing a dictionary whose
> items are arguments and their results:
> The problem is that the dictionary key
> stored depends on how the function was called, even if two calls should
> be equivalent; hence the original fun
On Friday, 20 April 2012 16:57:06 UTC+1, Rotwang wrote:
> Hi all, here's a problem I don't know how to solve. I'm using Python 2.7.2.
>
> I'm doing some stuff in Python which means I have cause to call
> functions that take a while to return. Since I often want to call such a
> function more th
