This is what i have now :
import re
x = "From: Joyce IP: 192.111.1.1 Source: 192.168.1.1"
x = x.replace(' ', '')
m = re.findall('(?<=:)\S+', x)
print m
And I want to have a output like this to make this $ script.py > result.txt:
Joyce 192.111.1.1 192.168.1.1
Instead of finding the matches of the text you want as the result, it may be easier to replace the stuff you don't want:
>>> import re
>>> x = "From: Joyce IP: 192.111.1.1 Source: 192.168.1.1"
>>> re.sub(r'\w+:\s', '', x)
'Joyce 192.111.1.1 192.168.1.1'
However, if you prefer to use re.findall() here is one option that is similar to your current approach:
>>> ' '.join(re.findall(r'(?<=:\s)\S+', x))
'Joyce 192.111.1.1 192.168.1.1'
You need the \s in the negative lookbehind because there is a space after each of the colons in your input string.
a slight change to your code (don't remove the spaces, and include them in the look behind) works perfectly:
import re
x = "From: Joyce IP: 192.111.1.1 Source: 192.168.1.1"
m = re.findall('(?<=:\s)\S+', x)
print " ".join(m)
import re
x = "From: Joyce IP: 192.111.1.1 Source: 192.168.1.1"
reg = r"\d{1,3}(?:[.]\d+){3}"
m = re.findall(reg, x)
for i in m:
print(i)
Result : 192.111.1.1 192.168.1.1
