Extract this pattern from a pd column: "2.0.1"

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Hi all,

I load a pd df that has a column containing file name versions and I need to extract the version (removing the letters).
Column:
Filename
filename vs2.0
filename vs2.1.1
filename vs2.2.1
filename vs2.3.3

Desired:
2.0
2.1.1
2.2.1
2.3.3

Thanks!

What I have tried:

I tried:
df['VS'] = df['VS'].str.extract('(\d+\.\d+)', expand=False) 
but I get the first 2 digits of the version, and not the thirds.

Posted 18-Jan-23 6:10am

Mihnea Nichifor

Updated 18-Jan-23 21:35pm

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3 solutions

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OriginalGriff · Answer 1 · 2023-01-18T06:19:00

Solution 1

Try this:

RegEx

(\d+(\.\d+)+)

If you are going to use regular expressions, you need a helper tool. Get a copy of Expresso[^] - it's free, and it examines and generates Regular expressions.

Posted 18-Jan-23 6:19am

OriginalGriff

Richard MacCutchan · Answer 2 · 2023-01-18T06:25:00

Solution 2

Your expression says capture one or more digits, followed by a period. followed by one or more digits. So it will only capture 2.0, 2.1 etc.

Try this:

    (\d+(\.\d+)*)
      ^ ^ ^  ^ ^
      1 2 a  b c
1. One or more digits followed by
2. A group containing
  a. A single period (dot) followed by
  b. One or more digits
  c. Repeated any number of times

Posted 18-Jan-23 6:25am

Richard MacCutchan

Updated 18-Jan-23 6:29am

v2

Comments

Mihnea Nichifor 18-Jan-23 13:46pm

It still gives the same output without the last ".digit"

Richard MacCutchan 19-Jan-23 4:27am

It works for me in Python, and elsewhere, using the standard regex package. Check the pandas documentation to see if they use a different syntax.

Richard MacCutchan 19-Jan-23 5:06am

I just tied this in pandas and managed to get the correct result. Try the following:

df['VS'] = df['VS'].str.extract('(\d+(\.\d+)*)', expand=False)[0]

Without the [0] at the end it returns two fields for some reason, the second being the final dot and digit.

Member 15881193 · Answer 3 · 2023-01-18T19:30:00

Solution 3

Try this:

df['VS'] = df['VS'].str.extract("([\d.\d.\d]+)")

Posted 18-Jan-23 19:30pm

Member 15881193