Regex to pull particular value from the field in Python

Question

I have below type of components in jira.I am trying to create the regex to fetch the value from these components .This value could be anything e.g 1.1 or 1.a or a.a or nothing for any components but remember I don't need anything like R or (U1) just need 1.0,1.1,1.a,a.1,1.x

so I need only the first character or number before period(.) and first character or number after period(.) or if there is no period(.) then blank

                                            need

PCN 9.4U1 (Act)----------------------------1.4  
PCN 9.5 (Act)------------------------------1.5  
PCN 9.6 (Act)------------------------------1.6
R AA 7.5U5 (Arch)--------------------------2.5              
R AA 7.6U2 (Rel)--------------------------2.6               
R AA 37.7R (Arch)---------------------------2.7             
R TEST 1.x (Fut)-------------------------2.x
R testp U2---------------------------------------no value

I am using the below regex to get the value

Fixversionmat = re.findall(r"(\d+\.\d+)", jsonToPython['name'])

but this gives me result only when value is something like that 1.1 and in other cases it fails.

If it can be "anything" then how is a regexp supposed to match it? You need to specify what it can be precisely. — Barmar, Apr 04 '18 at 22:13
@Barmar sorry for this..anything means here it could be number 1.1 or mix of characters 1.a or a.4 or nothing — unknown, Apr 04 '18 at 22:15
@Barmar In most of the cases yes but sometime doesn't contain anything for e.g R olcPa GA — unknown, Apr 04 '18 at 22:17
So in those cases the regexp won't match, and you can test for that separately. — Barmar, Apr 04 '18 at 22:19
You can't match *nothing* unless there's a pattern for something before or after it. — Barmar, Apr 04 '18 at 22:20
@Barmar it makes sense but for now I can exclude this one case — unknown, Apr 04 '18 at 22:22
You don't even need regexes on this example. The alphanumeric value is always the rightmost non-parenthesized string which contains any digit. So just throw away everything from '(' on, split on space, iterate right-to-left, and return the first string containing a number. Else return `None`. — smci, Apr 04 '18 at 22:38
I'm downvoting this question because despite numerous requests you refuse to describe the pattern of your version numbers clearly. Of course, if you knew what the pattern was you probably could write the regexp yourself. — Barmar, Apr 04 '18 at 23:05

Barmar · Answer 1 · 2018-04-04T22:33:18.543

3

\S+\.\S+ will two strings of alphanumeric characters separated by ..

Fixversionmatch = re.findall(r"\w+\.\w+", jsonToPython['name'])

edited Apr 04 '18 at 22:33

answered Apr 04 '18 at 22:29

Barmar

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This will not work "\S+\.\S+" because it will give u 1.1(U1) as well and i don't want that just need 1.1 or 1.a or a.a – unknown Apr 04 '18 at 22:32
Where is that in the sample input? I told you you need to be very precise about what the pattern is. – Barmar Apr 04 '18 at 22:33
Update the question to clearly state what characters can be in the version. One or two examples is not enough for me to figure out the full pattern. – Barmar Apr 04 '18 at 22:35
this one is r"\w+\.\w+" not working and giving 3.8U3 – unknown Apr 04 '18 at 22:45
But you said it can be `1.a`. Why is the `a` OK there, but not `U3`? – Barmar Apr 04 '18 at 22:46
Stop giving examples, and give a precise *description* of how these version numbers can be. I can't figure out the pattern from examples and exceptions. – Barmar Apr 04 '18 at 22:47
Are lowercase letters allowed, but not uppercase? – Barmar Apr 04 '18 at 22:47
Yes I said I need only 1.a or 1.x or 1.1 but not 1.1(U1) or 3.8(U2) I gave few examples in my question – unknown Apr 04 '18 at 22:48
It is possible both lowercase and upercase – unknown Apr 04 '18 at 22:48
so basically if there is 1.xR,1.1R then I need 1.x or 1.1 – unknown Apr 04 '18 at 22:51
I give up. How many times do I have to ask for a precise description, not examples? – Barmar Apr 04 '18 at 22:53
So far you've said it can't be R1 or U1, what about X1 or Z1? – Barmar Apr 04 '18 at 22:54
What about 1.R? – Barmar Apr 04 '18 at 22:54
check my updated question what I need for each component.Hope this clears – unknown Apr 04 '18 at 22:59
I think your answer is illustrative enough, I agree with it, you answered first and dynamic questions are difficult. So I am upvoting this and deleting mine, which actually was essentially the same. – progmatico Apr 04 '18 at 23:00
@Sushant Do you really not understand the concept of a *description* instead of *examples*? – Barmar Apr 04 '18 at 23:03
@Sushant Use `[a-z\d]` to match lowercase letters and digits. Does that help? – Barmar Apr 04 '18 at 23:03
Thanks for your input and patience I upvoted the answer – unknown Apr 04 '18 at 23:07

Patrick Artner · Answer 2 · 2018-04-04T22:38:26.423

You get most of your examples with:

(\d[^ \n]*| [a-zA-Z]+?\d[^ \n]*).*$

Link: https://regexr.com/3ncbs

It does not catch version numbers that consist of letters only - it catches mixed ones.

t = """CTX     
CTX 4.0R (Released)     
CTX 4.1 (Released)
CTX 4.2 (Released)
CTX 4.2R2 (FRtRre)  
CTX 4.3 (Released)  
CTX 4.4 (Released)  
CTX 4.4R1 (Active)  
CTX 4.5 (Active)    
CTX 4.6 (Active)
R PX 3.5R3 (Archived)           
R PX 3.5R4 (Archived)               
R PX 3.5R5 (Archived)               
R PX 3.6R2 (Released)               
R PX 3.6R3 (Rnreleased)             
R PX 3.6R4 (Released)               
R PX 3.6R5 (Active)             
R PX 3.7R (Archived)                
R PX 3.7R1 (Released)           
R PX 3.7R2 (Active)             
R PX 3.8R (Released)            
R PX 3.8R1 (Released)               
R PX 3.8R2 (Released)           
R PX 3.8R3 (Released)               
R PX 3.8R4 (Active)
R LPTT GA   
R LPTT R1   
R LPTT R2
R Cianara 4.1R2 (Early Access)
R Cianara 4.x (FRtRre)
R NRnPA R2"""

import re

vers = re.findall(r'(\d[^ \n]*| [a-zA-Z]+?\d[^ \n]*).*$',t,re.MULTILINE)
print(vers)

Output:

['4.0R', '4.1', '4.2', '4.2R2', '4.3', '4.4', '4.4R1', '4.5', 
 '4.6', '3.5R3', '3.5R4', '3.5R5', '3.6R2', '3.6R3', '3.6R4', 
 '3.6R5', '3.7R', '3.7R1', '3.7R2', '3.8R', '3.8R1', '3.8R2', 
 '3.8R3', '3.8R4', ' R1', ' R2', '4.1R2', '4.x', ' R2']

It does not find "empty" ones.

It doesn't have to start with a number, he said it can be `a.1` or `a.a`. — Barmar, Apr 04 '18 at 22:25
@Barmar you are right, I got a bit closer, still not getting the a.1 completely though... — Patrick Artner, Apr 04 '18 at 22:32

Regex to pull particular value from the field in Python

2 Answers2