Regex: if string ends with numerical values – drop it, OR if string ends with single letter – drop it, OR leave it

Here is a sample of a dataset I have:

I need to transform ‘Project’ column so that:

1. if a row ends with numeric values, they should be dropped
2. if a row ends with single letter, it should be dropped
3. else, leave as it is

Here is how I want it to look like:

I tried to search for some solution, and found this(for first condition with numeric values only):

df['Project']=df.Project[~((df.Project.astype(str).str.match("(.*\d)")) & (df.Project.astype(str).str.len() > 1))] 

It worked, however, I tried to apply it for the second condition as well:

df['Project']=df.Project[~((df.Project.astype(str).str.match("(.*\w)")) & (df.Project.astype(str).str.len() == 1))] 

But, It failed

Can you help me, please?
Thank you!

>Solution :

You can use

df['Project']=df['Project'].str.replace(r'\s+(?:\d+|[A-Za-z])$', '', regex=True)

See the regex demo.

Details:

• \s+ – one or more whitespaces
• (?:\d+|[A-Za-z]) – a non-capturing group matching either
  • \d+ – one or more digits
  • | – or
  • [A-Za-z] – an ASCII letter (use [^\W\d_] to match any Unicode letter)
• $ – end of string.

If your strings have no whitespace before the last number/letter, you may use

df['Project'].str.replace(r'\b(?:\d+|[A-Za-z])$', '', regex=True).str.rstrip()

The \b matches a word boundary making sure there is no letter, digit or _ right before the final digits/letter, and .str.rstrip() will strip the trailing whitespace.