Replace ip partially with x in python-CodePudding

I have several ip addresses like

162.1.10.15
160.15.20.222
145.155.222.1

I am trying to replace the ip's like below.

162.x.xx.xx
160.xx.xx.xxx
145.xxx.xxx.x

How to achieve this in python.

CodePudding user response：

We can use re.sub with a callback function here:

def repl(m):
    return m.group(1)   '.'   re.sub(r'.', 'x', m.group(2))   '.'   re.sub(r'.', 'x', m.group(3))   '.'   re.sub(r'.', 'x', m.group(4))

inp = "160.15.20.222"
output = re.sub(r'\b(\d )\.(\d )\.(\d )\.(\d )\b', repl, inp)
print(output)  # 160.xx.xx.xxx

In the callback, the idea is to use re.sub to surgically replace each digit by x. This keeps the same width of each original number.

CodePudding user response：

Here’s a slightly simpler solution

import re

txt = "192.1.2.3"

x = txt.split(".", 1) # ['192', '1.2.3']
y = x[0]   "."   re.sub(r"\d", "x", x[1])

print(y) # 192.x.x.x

CodePudding user response：

There are multiple ways to go about this. Regex is the most versatile and fancy way to write string manipulation codes. But you can also do it by same old for-loops with split and join functions.

ip = "162.1.10.15"

#Splitting the IPv4 address using '.' as the delimiter
ip = ip.split(".")

#Converting the substrings to x's except 1st string
for i,val in enumerate(ip[1:]):
    cnt = 0
    for x in val:
        cnt  = 1
    ip[i 1] = "x" * cnt

#Combining the substrings back to ip
ip = ".".join(ip)
print(ip)

I highly recommend checking Regex but this is also a valid way to go about this task.

Hope you find this useful!

CodePudding user response：

This is not the optimize solution but it works for me .

import re
Ip_string = "160.15.20.222"
Ip_string = Ip_string.split('.')
Ip_String_x =""
flag = False
for num in Ip_string:
  if flag:
    num = re.sub('\d','x',num)
    Ip_String_x = Ip_String_x   '.'  num
  else:
    flag = True
    Ip_String_x = num

CodePudding user response：

Pass an array of IPs to this function:

def replace_ips(ip_list):
    r_list=[]
    for i in ip_list:
        first,*other=i.split(".",3)
        r_item=[]
        r_item.append(first)
        for i2 in other:
            r_item.append("x"*len(i2))
        r_list.append(".".join(r_item))
    return r_list

In case of your example:

print(replace_ips(["162.1.10.15","160.15.20.222","145.155.222.1"]))#==> expected output: ["162.x.xx.xx","160.xx.xx.xxx","145.xxx.xxx.x"]

CodePudding user response：

Solution 1

Other answers are good, and this single regex works, too:

import re


strings = [
    '162.1.10.15',
    '160.15.20.222',
    '145.155.222.1',
]

for string in strings:
    print(re.sub(r'(?:(?<=\.)|(?<=\.\d)|(?<=\.\d\d))\d', 'x', string))

output:

162.x.xx.xx
160.xx.xx.xxx
145.xxx.xxx.x

Explanation

(?<=\.) means following by single dot.
(?<=\.\d) means follwing by single dot and single digit.
(?<=\.\d\d) means following by single dot and double digit.
\d means a digit.
So, all digits that following by single dot and none/single/double digits are replaced with 'x'
(?<=\.\d{0,2}) or similar patterns are not allowed since look-behind ((?<=...)) should has fixed-width.

Solution 2

Without re module and regex,

for string in strings:
    first, *rest = string.split('.')
    print('.'.join([first, *map(lambda x: 'x' * len(x), rest)]))

above code has same result.