Python reformatting strings based on contents - python-3.x

Question

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"])

Answer 1

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

Answer 2

You can use
df = pd.DataFrame({'point_id':['abc123-Target 4-ufs','abc123-target4-ufs','geo.4','j123T4']})
df['point_id'] = df['point_id'].str.replace(r'(?i).*Target\s*(\d+).*', r'target \1', regex=True)
df.loc[df['point_id'].str.contains(r'(?i)\w[.t]\d+$'), 'point_id'] = 'target 4'
# point_id
# 0 target 4
# 1 target 4
# 2 target 4
# 3 target 4
The regex is (?i)Target\s*\d+|\w+[.t]\d+$:
(?i) - case insensitive matching
.* - any 0+ chars other than line break chars, as many as possible
Target\s*(\d+).* - Target, zero or more whitespaces, and one or more digits captured into Group 1
.* - any 0+ chars other than line break chars, as many as possible
The second regex matches
(?i) - case insensitive matching
\w - a word char, then
[.t] - a . or t and then
\d+$ - one or more digits at the end of string.
The second regex is used as a mask, and the values in the point_id column are set to target 4 whenever the pattern matches the regex.
See regex #1 demo and regex #2 demo.