从 CSV 文件中搜索 IP 地址

Question

I have a CSV file where the columns contain several strings with IP addresses. I successfully ran the regex query but it is adding random characters in the output.

descr = df.loc[:, 'desc']

arr = []


pat = re.compile("(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)")

for i in descr:

     test = pat.findall(i)
     arr.append(test)

df["IPA"] = arr

It gives IP address output, but I want the output as 10.35.50.4 etc [(10, 35, 50, 4)] and [(10, 35, 50, 3)] .

Answer 1

You need to transform your groups (Anything inside a parentheses) into non-capturing groups. You can do that by adding a "?:" right after opening the parentheses.

pat = re.compile("(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)")

The reason is written on the definition of the function "findall":

Return all non-overlapping matches of pattern in string, as a list of strings. The string is scanned left-to-right, and matches are returned in the order found. If one or more groups are present in the pattern, return a list of groups; this will be a list of tuples if the pattern has more than one group . Empty matches are included in the result.

Therefore it was returning all the groups, numbers inside your address, separately.