How to use python re module to filter the int number by digital

I want to use python re module to filter the int number by the number digital.

    1
  700
76093
71365
35837
75671
 ^^                 
 ||--------------------- this position should not be 6,7,8,9,0 
 |---------------------- this position should not be 5,6,7

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Code:

int_list=[1,700,76093,71365,35837,75671]
str_list = [str(x).zfill(5) for x in int_list]
reexp = r"\d[0-4,8-9][1-5]\d\d"
import re
p = re.compile(reexp)
result = [int("".join(str(y) for y in x)) for x in str_list if p.match(x)]

I have 2 questions:

1.Is it possible to generate the reexp string from below code:

thousand_position = set([1,2,3,4,5,1,1,1,1,1,1,1,1,1,1])
hundred_position  = set([1,2,3,4,8,9,0,1,2,3,2,3,1,2])

2.how to make the reexp be more simple avoid below 0-prefixed bug?

00700
00500          <--- this will also drops into the reexp, it is a 
                     bug because it has no kilo number
10700

reexp = r"\d[0-4,8-9][1-5]\d\d"

Thanks for your time

B.Rgs

PS: thanks for suggstion for the math solution below, I know it may be easy and faster, but I want the re based version to balance other thoughts.

---

Are you sure you want to be using the re module? You can get at what you're trying to do with some simple math operations.

def valid_number(n):
  return 0 < n%1000/100 < 6 and not 5 >= n%10000/1000 >= 7

int_list = [1,700,76093,71365,35837,75671,]
result   = [x for x in int_list if valid_number(x)]

or alternatively:

result    = filter(valid_number, int_list)

---

Ok, first, I'm going to post some code that actually does what you describe initially:

>>> int_list=[1, 700, 76093, 71365, 35837, 75671]
>>> str_list = [str(i).zfill(5) for i in int_list]
>>> filtered =  [s for s in str_list if re.match('\d[0-4,8-9][1-5]\d\d', s)]
>>> filtered
['71365']

Edit: Ok, I think I understand your question now. Instead of using zfill, you could use rjust, which will insert spaces instead of zeros.

>>> int_list=[1,700,76093,71365,35837,75671,500]
>>> str_list = [str(i).rjust(5) for i in int_list]
>>> re_str = '\d' + str(list(set([0, 1, 3, 4, 8, 9]))) + str(list(set([1, 2, 3, 4, 5]))) + '\d\d'
>>> filtered =  [s for s in str_list if re.match(re_str, s)]
>>> filtered
['71365']

I think doing this mathematically as yan suggests will be faster in the end, but perhaps you have your reasons for using regular expressions.