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/
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