Subset a large number into smaller series of numbers using python
I have a big number lets say of around hundred digits. I want to subset that big number into consecutive number of 5 digits and find the product of those 5 digits. For example my first 5 digit number would be 73167. I need to check the product of the individual numbers in 73167 and so on.
The sample number is as follows:
73167176531330624919225119674426574742355349194934969835203127745063262395783180169848018694788518438586156078911294949545950173795833195285320880551112540698747158523863050715693290963295227443043557
I have a problem subsetting the small numbers out of the big number.
My basic starting code is :
b = 73167176531330624919225119674426574742355349194934969835203127745063262395783180169848018694788518438586156078911294949545950173795833195285320880551112540698747158523863050715693290963295227443043557
jd = str(b)
for ch in jd:
number = ch
print (number)
Any help is highly appreciated.
Edit: I believe that grouper is overkill in this solution, look at the solution by @Haidro https://stackoverflow.com/a/16078696/1219006
Using the grouper recipe from itertools
I'm assuming b is a string to begin with because it would be a crazy waste of memory to make a number that big.
from itertools import izip_longest
from operator import mul
def grouper(n, iterable, fillvalue=None):
"Collect data into fixed-length chunks or blocks"
# grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx
args = [iter(iterable)] * n
return izip_longest(fillvalue=fillvalue, *args)
for g in grouper(5, b, fillvalue=''):
# And to work out the product of the digits
num = ''.join(g)
prod = reduce(mul, map(int, num))
尝试这个：
from operator import mul
def chunker(seq, size):
return (seq[pos:pos + size] for pos in xrange(0, len(seq), size))
for i in chunker(str(myint),5): # Where myint is that big number
reduce(mul, map(int, i))
一行为你：
import re; re.findall("d{5}", number)