如何从字典键“Message”中提取第一个 IP 地址，并使用提取的 IP 地址添加一个名为“IP”的新键？

Question

So I have a list with dictionaries inside which I generated from a log file. I want to create a new key called "IP" with the extracted IP address from "Message"

This is an example of the list of dictionaries.

[{'Date': 'Jun 29', 'Time': '03:22:22', 'PID': '13251', 'Message': 'Authentication failed from 163.27.187.39 (163.27.187.39): Permission denied in replay cache code', 'Access Type': 'Success'}
...
{'Date': 'Jun 29', 'Time': '03:22:22', 'PID': '13263', 'Message': 'connection from 61.74.96.178 () at Wed Jun 29 03:22:22 2005', 'Access Type': 'Success'}]

I thought of using regex but i get an error saying my dictionary changed sized during iteration.

for Dict in data:
    for k,v in Dict.items():
        if k == 'Message':
            re.findall(r"[0-9]+(?:\.[0-9]+){3}\s", v)
            Dict["host/IP address"] = re.findall(r"[0-9]+(?:\.[0-9]+){3}\s", v)
        else:
            Dict["host/IP address"] = "" 
    print(Dict)

Answer 1

You don't need to iterate over the dict, as you know the keys, just create the new one from the IP of key Message .

• I remove the \s in the regex because you don't want the space
• r"[0-9]{1,3}(?:\.[0-9]{1,3}){3} better regex that check digit group are length [1-3]
• use [0] at the end to get the first IP from Message , if you want a list of all the IP found, remove it

---

In case there is no IP, you should first compute the IPs, then add the mapping regarding the result of findall

for value in data:
    ips = re.findall(r"[0-9]{1,3}(?:\.[0-9]{1,3}){3}", value['Message'])
    if ips:
        value["host/IP address"] = ips[0]

If you want to put empty string in case of no IP

for value in data:
    ips = re.findall(r"[0-9]{1,3}(?:\.[0-9]{1,3}){3}", value['Message'])
    value["host/IP address"] = ips[0] if ips else ""

---

• with [0] : 'host/IP address': '163.27.187.39'

• without [0] : 'host/IP address': ['163.27.187.39', '163.27.187.39']