[Tutor] Project Euler #8

[Alan Gauld]

On 31/05/13 19:23, Nick Shemonsky wrote:

> or maybe it'd be quicker to compare a to b through each iteration and
> just keep the larger product rather than creating a giant list

You are probably right if it is a giant list.

> str_num = '1234567890'
> n = 5
>
> strings = [str_num[i:i+5] for i in range(0, len(str_num)) if
> len(str_num[i:i+5])==5]
> integers = [int(i) for i in strings[0]]
>
> def product(x):
>      p = 1
>      for n in integers:
>          p*=n
>      return p

You define a parameter x but don't use it?

You could use reduce here:

import operator
...
def product(intlist):
     return reduce(operator.mul, intlist)

HTH
-- 
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/