On Wed, 02 Mar 2005 00:58:58 -0600, actuary77
<[EMAIL PROTECTED]> wrote:
> Yes, iteration is 10000000000 times faster than recursion.
>
> The problem I have is:
>
> # need to call this function 50 times with seed of 10
> (afunc(afunc(afunc(... afunc(10))
>
> required_iterations_ 50
> function: afunc(x): x**.5 + 1000 * x + 46.
> initial_seed: 10
>
> Result => afunc(afunc(afunc(afunc(afunc(afunc ... (_aseed)
>
> What is the fastest way to do this without using recursion?
>
> With recursion:
> ================
>
> def rec(afunc,n,x):
> y = afunc(x)
> if n == 0:
> return y
> else:
> return rec(afunc,n-1,y)
>
> def myfunc(x):
> return x**.5 + 1000 * x + 46.
>
> from time import time
>
> _b=time()
> _y = rec(myfunc,50,10)
> _e=time()
> _t=_e-_b
> print "result: %r time: %f\n" % (_y,_t)
>
> output:
> result: 1.00493118428005e+154 time: 0.030000
>
> Is there a much faster way?
Sure.
[EMAIL PROTECTED] tmp]$ python timeit.py -s 'from recursive import
calculate' 'calculate(100, 10)'
1000 loops, best of 3: 295 usec per loop
[EMAIL PROTECTED] tmp]$ python timeit.py -s 'from iterative import
calculate' 'calculate(100, 10)'
10000 loops, best of 3: 134 usec per loop
[EMAIL PROTECTED] tmp]$ cat iterative.py
def calculate(repeats, value):
for x in range(repeats):
value = value ** .5 + 1000*x + 46
return value
[EMAIL PROTECTED] tmp]$ cat recursive.py
def rec(afunc,n,x):
y = afunc(x)
if n == 0:
return y
else:
return rec(afunc,n-1,y)
def myfunc(x):
return x**.5 + 1000 * x + 46.
def calculate(repeats, value):
rec(myfunc, repeats, value)
As you can see, the iterative solution actually results in less code too ;)
Stephen.
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