Cfrgtkky

I have an Phone number here lets say : 98888888xx the last two XX are the numbers which I want to generate a sequence of like 988888811 , 9888888812, 988888813 and so on.

I am trying to learn python programming so can someone help me how would I go on writing a script for the same

You could use a list comprehension and range():

['98888888' + str(number).zfill(2) for number in range(100)]



['9888888800',

'9888888801',

'9888888802',

...

'9888888897',

'9888888898',

'9888888899']

A robust solution is to use recursion to handle many possible occurrences of "x":

import re

s = '98888888xx' 

_len = len(re.sub('d+', '', s))

def combos(d, current = ):

   if len(current) == _len:

     yield current

   else:

     for i in d[0]:

        yield from combos(d[1:], current+[i])



_c = combos([range(1, 10)]*_len)

new_result = [(lambda d:re.sub('x', lambda _:str(next(d)), s))(iter(i)) for i in _c]

Output:

['9888888811', '9888888812', '9888888813', '9888888814', '9888888815', '9888888816', '9888888817', '9888888818', '9888888819', '9888888821', '9888888822', '9888888823', '9888888824', '9888888825', '9888888826', '9888888827', '9888888828', '9888888829', '9888888831', '9888888832', '9888888833', '9888888834', '9888888835', '9888888836', '9888888837', '9888888838', '9888888839', '9888888841', '9888888842', '9888888843', '9888888844', '9888888845', '9888888846', '9888888847', '9888888848', '9888888849', '9888888851', '9888888852', '9888888853', '9888888854', '9888888855', '9888888856', '9888888857', '9888888858', '9888888859', '9888888861', '9888888862', '9888888863', '9888888864', '9888888865', '9888888866', '9888888867', '9888888868', '9888888869', '9888888871', '9888888872', '9888888873', '9888888874', '9888888875', '9888888876', '9888888877', '9888888878', '9888888879', '9888888881', '9888888882', '9888888883', '9888888884', '9888888885', '9888888886', '9888888887', '9888888888', '9888888889', '9888888891', '9888888892', '9888888893', '9888888894', '9888888895', '9888888896', '9888888897', '9888888898', '9888888899']

Note that a simple solution in the form of a nested list comprehension presents itself when there are only two 'x':

d = [s.replace('x', '{}').format(a, b) for a in range(1, 10) for b in range(1, 10)]

However, multiple nested loops is not a clean approach to solving the problem when the input string contains three or more 'x's. Instead, recursion works best:

s = '98888888xxxxx'

_len = len(re.sub('d+', '', s))

_c = combos([range(1, 10)]*_len)

new_result = [(lambda d:re.sub('x', lambda _:str(next(d)), s))(iter(i)) for i in _c]

Output (first twenty strings):

['9888888811111', '9888888811112', '9888888811113', '9888888811114', '9888888811115', '9888888811116', '9888888811117', '9888888811118', '9888888811119', '9888888811121', '9888888811122', '9888888811123', '9888888811124', '9888888811125', '9888888811126', '9888888811127', '9888888811128', '9888888811129', '9888888811131', '9888888811132']