I can’t parse the result of the program, got confused with the algorithm?

Question

Python

I am not very strong in programming, especially in python.
My task:
There is a txt file that contains the result of the masscan operation. It is
required to display each ip with all its open ports found:
res.txt
Code:

# Masscan 1.0.3 scan initiated Sun Dec 23 23:00:31 2018
# Ports scanned: TCP(1;80-80) UDP(0;) SCTP(0;) PROTOCOLS(0;)
Host: 192.168.1.1 () Ports: 80/open/tcp////
Host: 192.168.1.1 () Ports: 801/open/tcp////
Host: 192.168.1.2 () Ports: 801/open/tcp////
Host: 192.168 .1.2 () Ports: 445/open/tcp////
Host: 192.168.1.3 () Ports: 80/open/tcp////
Host: 192.168.1.3 () Ports: 8080/open/tcp/// /
Host: 192.168.1.3 () Ports: 21/open/tcp////
# Masscan done at Sun Dec 23 23:00:45 2018

My task is to get the data in this format.

192.168.1.1 80, 801
192.168.1.2 801, 445
192.168.1.3 80, 8080, 21

#!/usr/bin/env python
import sys, re, os

ports = []
ip = None

with open('mres.txt','r') as f:

    for elem in f.readlines():

        if not elem.startswith('#'):
              if ip != None:
                  print('1')
                  if elem.split(' ')[1] == ip:
                      ports.append(elem.split(' ')[3].split('/')[0])
                      continue
                  else:
                      print(ports)
                      print('nmap '+ip +' ports: '+str(ports))
                      ports=[]
              else:
                  print('2')
                  #print('Unic: '+str(ip) + ' port: '+ str(elem.split(' ')[3].split('/')[0]))
                  ip = elem.split(' ')[1]
                  ports.append(elem.split(' ')[3].split('/')[0])

Answer 1

Sketched a bit, try it.

import re
from collections import defaultdict

ips = defaultdict(list)
regular = re.compile(r'Host: ([\d\.]+).+?Ports: (\d+)/')

with open('res.txt', 'r') as f:
    for line in f:
        line = line.strip()
        if not line.startswith('#'):
            ip, port = regular.search(line).groups()
            ips[ip].append(port)

for k, v in ips.items():        # Выведет:
    print(k, ', '.join(v))      # 192.168.1.1 80, 801
                                # 192.168.1.2 801, 445
                                # 192.168.1.3 80, 8080, 21

with open('outputfile.txt', 'w') as f: # Запишет тоже самое
    for k, v in ips.items():
        f.write('{} {}\n'.format(k, ', '.join(v)))