Python reformatting strings based on contents

In a pandas dataframe I have rows with contents in the following format:

1) abc123-Target 4-ufs
2) abc123-target4-ufs
3) geo.4
4) j123T4

All of these should be simply: target 4

So far my cleaning procedure is as follows:

df["point_id"] = df["point_id"].str.lower()
df["point_id"] = df['point_id'].str.replace('^.*?(?=target)', '')

This returns:

1) target 4-ufs
2) target4-ufs
3) geo.14
4) geo.2
5) j123T4

What I believe I need is:

a. Remove anything after the last number in the string, this solves 1
b. If 'target' does not have a space after it add a space, this with the above solves 2
c. If the string ends in a point and a number of any length remove everything before the point (incl. point) and replace with 'target ', this solves 3 and 4
d. If the string ends with a 't' followed by a number of any length remove everything before 't' and replace with 'target ', this solves 5

I'm looking at regex and re but the following is not having effect (add space before the last number)

df["point_id"] = re.sub(r'\D+$', '', df["point_id"])

Solution

Reading the rules, you might use 2 capture groups and check for the group values:

\btarget\s*(\d+)|.*[t.](\d+)$

\btarget\s*(\d+) Match target, optional whitespace chars and capture 1+ digits in group 1
| Or
.*[t.] Match 0+ characters followed by either t or a .
(\d+)$ Capture 1+ digits in group 2 at the end of the string

Regex demo | Python demo

Python example:

import re
import pandas as pd

pattern = r"\btarget\s*(\d+)|.*[t.](\d+)$"
strings = [
    "abc123-Target 4-ufs",
    "abc123-target4-ufs",
    "geo.4",
    "j123T4"
]

df = pd.DataFrame(strings, columns=["point_id"])

def change(s):
    m = re.search(pattern, s, re.IGNORECASE)
    return "target " + (m.group(2) if m.group(2) else m.group(1))

df["point_id"] = df["point_id"].apply(change)
print(df)

Output

   point_id
0  target 4
1  target 4
2  target 4
3  target 4