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 doesn't
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 Saturday, 21 April 2012 09:25:40 UTC+1, Steven D#39;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 of
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 looked up
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 than once with the same arguments, I've written a
decorator to
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 than
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 function
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
On Fri, Apr 20, 2012 at 9:57 AM, Rotwang sg...@hotmail.co.uk 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;
On Fri, Apr 20, 2012 at 6:07 PM, Ian Kelly ian.g.ke...@gmail.com 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
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