Python Development, memoized functions.

[Jaap Spies]

Alec Mihailovs wrote:
> From: "Antti Karttunen" <antti.karttunen at gmail.com>
> 
>> @M
>> def A000045(n):  if(n < 2): return(n) else: 
>> return(A000045(n-1)+A000045(n-2))
> 
> 
> Another way of doing this is to define A000045_list first (that saves 
> all the calculated values in a list), and then define A000045 as the 
> last element of the list. Such as
> 
> def A000045_list(n):
> result=[0]
> a,b=0,1
> for i in range(n):
>     result.append(b)
>     a,b=b,a+b
> return result
> 
> def A000045(n): return A000045_list(n)[-1]
> 
> That should calculate values faster than the recursive procedure with 
> memoization because that avoids multiple function calls.
> 

Why not use generators? Make a list large enough,

def fib():
     x, y = 0, 1
     while 1:
         x, y = y, x+y
         yield x

f = fib()
a = [f.next() for i in range(1000)]
a.insert(0,0)

def A000045(n):
     return a[n]

def A000045_list(N)
     return a[:N]


Cheers,

Jaap