How can I reorganize this list?

Question

I wait from a server to send me a list which includes ip followed by a "," followed by a port followed by ";" and then there's another tuple... and another and x tuples...

the example is:

127.0.0.1,45403;127.0.0.1,47146;127.0.0.1,52888

I want to reorganize it so I have in each loop x iteration

Ipx = 127.0.0.1 
Portx = 45403

In the next iteration of the loop

Ipx = 17.0.0.1
Portx = 47146

etc for every tuple (Ipx and Portx are different varibles)

I have tried

ipx , portx = lista.split(";")
        print ipx
        print portx

but it doesn't work...

You would receive more help on this board if you showed what you've tried. I'd suggest using `split`, like you tagged. As in, `list_of_ips = string_of_ips.split(";")` would get you half way there. — mas, Dec 19 '18 at 19:33
Possible duplicate of [How to split a string into a list?](https://stackoverflow.com/questions/743806/how-to-split-a-string-into-a-list) — mas, Dec 19 '18 at 19:34
@malan is not exactly what I want... I just wanna get each ip and port in different variables. Thank you — Ginés Díaz, Dec 19 '18 at 19:39

score 4 · Accepted Answer · answered Dec 19 '18 at 19:42

4

You need to split by ; and then by , as follows:

lista = "127.0.0.1,45403;127.0.0.1,47146;127.0.0.1,52888"

for address in lista.split(";"):
    ipx, portx = address.split(',')
    print(f'IP: {ipx}, Port: {portx}')

answered Dec 19 '18 at 19:42

Dirk R

602
7
14

Also if you are using an older version of python that does not support f-strings you can substitute the last line with: ```print('IP: {}, Port: {}'.format(ipx, portx))``` – Dirk R Dec 19 '18 at 19:47
thank you very much. Just exactly what I was looking for. – Ginés Díaz Dec 20 '18 at 13:02

score 3 · Answer 2 · answered Dec 19 '18 at 19:43

If your server's response is a string, then you can do this:

inList = '127.0.0.1,45403;127.0.0.1,47146;127.0.0.1,52888'
inList = [[elem for elem in item.split(',')] for item in inList.split(';')]

for ip, port in inList:
  print(ip)
  print(port)

Output:

score 1 · Answer 3 · answered Dec 19 '18 at 19:55

You can create a OrderedDict and keep all ips and port in a list

from collections import OrderedDict
d=OrderedDict()

d.setdefault('ip',[])
d.setdefault('port',[])

inList = '127.0.0.1,45403;127.0.0.1,47146;127.0.0.1,52888'
for i in inList.split(';'):
    temp=i.split(',')
    d['ip'].append(temp[0])
    d['port'].append(temp[1])
print(d)

Output

OrderedDict([('ip', ['127.0.0.1', '127.0.0.1', '127.0.0.1']),
             ('port', ['45403', '47146', '52888'])])

Leandro Eleutério · Answer 4 · 2018-12-19T19:53:21.687

Ok so you are halfway trhu, what you need is:

ipx = lista.split(";")[0].split(“,”)[0]
portx = lista.split(";")[0].split(“,”)[1]
print ipx
print portx

If you want to register more than one IP address I would use a dictionary as:

Ip_port = dict()
For i in all-data:
      a = i.split(";")[0].split(“,”)[0]
     Ip_port[a] = i.split(";")[0].split(“,”)[1]

Imaging that you have all you “lista” in a python list.

How can I reorganize this list?

4 Answers4