🔒 regex string matching

[fast raptor]

I need to match these strings in this file
I was thinking something like match = re.search("([A-H][LR][01] ){9}[A-H][LR][01]", line), but I was wondering if there was a better way

[teal tangle]

@fast raptor

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[frosty crescent]

why not just ([A-H][LR][01])( \1)*?

or instead of *, put {9} to match 9 more of those

also did you mean to use .search or .findall?

[fast raptor]

does \1 mean the same thing as the first capturing group, i.e, [A-H][LR][01] ?

[fast raptor]

I want it to return the match as a string, like this
"BR0 HR1 CR1 DL0 DR1 ER0 BL1 DL0 ER1 AR1"

[frosty crescent]

yes

[fast raptor]

import re

input_file_path = "/content/Decider_Halt_Hardcoded_Parameters.v.txt"
output_file_path = "output.txt"

with open(input_file_path, "r") as input_file, open(output_file_path, "w") as output_file:
  for line in input_file:
    match = re.search("([A-H][LR][01] ){9}[A-H][LR][01]", line)

[frosty crescent]

python uses \X to refer to group X in the regex

[frosty crescent]

if you want the whole string then i cant really remember if there's a way on the match object to do it

i just do .span(), unpack it, then slice .string

!e

import re

m = re.search('world', "hello world")

s, e = m.span()
print(m.string[s:e])

[teal tangle]

:white_check_mark: Your 3.12 eval job has completed with return code 0.

world

[fast raptor]

I guess I could also do findall for each of the individual 3 letter parts, and then join them with space?

[frosty crescent]

thats probably a bad idea

just use finditer and build the matched string yourself

if you want, you can wrap it in another layer of brackets and then get group 1

[frosty crescent]

i think this might be best for you

[fast raptor]

okay

[frosty crescent]

but with {9} instead of * for a fixed length

[fast raptor]

am I doing this correctly?

import re

input_file_path = "/content/Decider_Halt_Hardcoded_Parameters.v.txt"
output_file_path = "output.txt"

with open(input_file_path, "r") as input_file, open(output_file_path, "w") as output_file:
  for line in input_file:
    match = re.search("([A-H][LR][01])( \1){9}", line)
    if match:
      print(match.group())

I wonder if I can just go with match = re.search("[A-H][LR][01] [A-H][LR][01] [A-H][LR][01] [A-H][LR][01] [A-H][LR][01] [A-H][LR][01] [A-H][LR][01] [A-H][LR][01] [A-H][LR][01] [A-H][LR][01]", line)

[nova void]

why not just r"( \w{3}){10}"

is it really that important that you make it so extraordinarily specific?

[fast raptor]

true

[nova void]

you can also do makeTM (.*),Ha

like that seems good enough

and you can just do re.findall with input_file.read() unless the file is massive (more than 100 mb)

[fast raptor]

the ending suffix may be different, but they all end with a comma

[nova void]

gotcha, then just makeTM (.*),

don't make it more specific than you have to

[fast raptor]

and then I access the match using match.group(1) ?

[nova void]

yep

[fast raptor]

(I'm a newbie)

[nova void]

with re.findall it will only return whats in the group

so this is gonna work how you want it to ```py
for group in re.findall(r"makeTM (.*),", input_file.read()):
print(group.split()) # prints ["BR1", "ER0", "CL1", "AR0", ...] for each line

[fast raptor]

oh that's a good solution, thanks

[nova void]

np

[pallid trellis]

or makeTM ([^,]+)

[nova void]

good point I should mention that you actually have to do r"makeTM (.*?),"

the lazy grabbing is better than [^,] but it is still required

otherwise .* will be too greedy and still gobble up the commas

so

for group in re.findall(r"makeTM (.*?),", input_file.read()):
  print(group.split()) # prints ["BR1", "ER0", "CL1", "AR0", ...] for each line

[fast raptor]

so could I just do this?

with open(input_file_path, "r") as input_file, open(output_file_path, "w") as output_file:
  matches = re.findall(r"makeTM (.*),", input_file.read())
  print(matches)

[nova void]

yes but update the pattern to have (.*?) instead of (.*)

or ([^,]*) or ([^,]+) if you prefer that for some reason

[pallid trellis]

how is it better (objectively)?

[nova void]

intent is clearer and its fewer characters

[fast raptor]

what does the "?" do?

[nova void]

* by default is very greedy and it just keeps matching stuff until it can't anymore

*? means that it will stop matching as soon as it can (it's lazy)

so .*?, means that it's gonna stop matching as soon as it finds a comma

whereas in .*, the star will just continue matching forever even if there is a comma

[fast raptor]

I tried it without the question mark and it seemed to work

[nova void]

great, whatever then

must be because of the line breaks

if you had two matches on one line it wouldn't work

[fast raptor]

what if I just specified the exact number of characters instead of *

[nova void]

that would also succeed its just more work

[pallid trellis]

then you can just slice the string for each line

no regexp required

[fast raptor]

not all the lines have matches though

[teal tangle]

Click here to see this code in our pastebin.

[fast raptor]

this seems to work perfectly

with open(input_file_path, "r") as input_file, open(output_file_path, "w") as output_file:
  matches = re.findall(r"makeTM (.{39}),", input_file.read())
  print(matches)

[pallid trellis]

okay, then it doesn't work

[fast raptor]

I think this is solved now, thanks for the help everyone

!close

[teal tangle]

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