On Mon, Feb 21, 2005 at 01:14:00PM -0800, Paul Rubin wrote:
> "markscottwright" <[EMAIL PROTECTED]> writes:
> > But when I try to do it iteratively, it just hangs when I try to
> > evaluate the results (for count > 1):
> >
> > def repeated2(f, count):
> > newfun = f
> > for i in range(count-1):
> > newfun = lambda x: newfun(f(x))
> > return newfun
> >
> > For the life of me, I can't figure out why. It seems like for count =
> > 2, for example, the results from repeated2 should be lambda x: f(f(x)),
> > but it doesn't seem to be.
>
> It's Python's scoping madness. Try:
>
> def repeated2(f, count):
> newfun = lambda x: x # identity
> for i in range(count):
> newfun = lambda x, g=newfun: g(f(x))
> return newfun
Ahh, but not sufficienty evil or pernicious.
def evil(f, count):
def apply_evil(accum, func):
return func(accum)
def pernicious(x):
return reduce(apply_evil, [f]*count, x)
return pernicious
def f(x):
return x+x
print evil(f, 3)(2)
More seriously I'd go without the recursion and just make a wrapper
that applies the function count times in a wrapper.
def benign(f, count):
def wrap(x):
result = f(x)
for (i) in range(count-1):
result = f(result)
return result
return wrap
print benign(f, 3)(2)
-Jack
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