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Regex python - find match items on list that have the same digit between the second character "_" to character "."

I have the following list :
list_paths=imgs/foldeer/img_ABC_21389_1.tif.tif,
imgs/foldeer/img_ABC_15431_10.tif.tif,
imgs/foldeer/img_GHC_561321_2.tif.tif,
imgs_foldeer/img_BCL_871125_21.tif.tif,
...
I want to be able to run a for loop to match string with specific number,which is the number between the second occurance of "_" to the ".tif.tif", for example, when number is 1, the string to be matched is "imgs/foldeer/img_ABC_21389_1.tif.tif" , for number 2, the match string will be "imgs/foldeer/img_GHC_561321_2.tif.tif".
For that, I wanted to use regex expression. Based on this answer, I have tested this regex expression on Regex101:
[^\r\n_]+\.[^\r\n_]+\_([0-9])
But this doesn't match anything, and also doesn't make sure that it will take the exact number, so if number is 1, it might also select items with number 10 .
My end goal is to be able to match items in the list that have the request number between the 2nd occurrence of "_" to the first occirance of ".tif" , using regex expression, looking for help with the regex expression.
EDIT: The output should be the whole path and not only the number.

Your pattern [^\r\n_]+\.[^\r\n_]+\_([0-9]) does not match anything, because you are matching an underscore \_ (note that you don't have to escape it) after matching a dot, and that does not occur in the example data.
Then you want to match a digit, but the available digits only occur before any of the dots.
In your question, the numbers that you are referring to are after the 3rd occurrence of the _
What you could do to get the path(s) is to make the number a variable for the number you want to find:
^\S*?/(?:[^\s_/]+_){3}\d+\.tif\b[^\s/]*$
Explanation
\S*? Match optional non whitespace characters, as few as possible
/ Match literally
(?:[^\s_/]+_){3} Match 3 times (non consecutive) _
\d+ Match 1+ digits
\.tif\b[^\s/]* Match .tif followed by any char except /
$ End of string
See a regex demo and a Python demo.
Example using a list comprehension to return all paths for the given number:
import re
number = 10
pattern = rf"^\S*?/(?:[^\s_/]+_){{3}}{number}\.tif\b[^\s/]*$"
list_paths = [
"imgs/foldeer/img_ABC_21389_1.tif.tif",
"imgs/foldeer/img_ABC_15431_10.tif.tif",
"imgs/foldeer/img_GHC_561321_2.tif.tif",
"imgs_foldeer/img_BCL_871125_21.tif.tif",
"imgs_foldeer/img_BCL_871125_21.png.tif"
]
res = [lp for lp in list_paths if re.search(pattern, lp)]
print(res)
Output
['imgs/foldeer/img_ABC_15431_10.tif.tif']

I'll show you something working and equally ugly as regex which I hate:
data = ["imgs/foldeer/img_ABC_21389_1.tif.tif",
"imgs/foldeer/img_ABC_21389_1.tif.tif",
"imgs/foldeer/img_ABC_15431_10.tif.tif",
"imgs/foldeer/img_GHC_561321_2.tif.tif",
"imgs_foldeer/img_BCL_871125_21.tif.tif"]
numbers = [int(x.split("_",3)[-1].split(".")[0]) for x in data]
First split gives ".tif.tif"
extract the last element
split again by the dot this time, take the first element (thats your number as a string), cast it to int
Please keep in mind it's gonna work only for the format you provided, no flexibility at all in this solution (on the other hand regex doesn't give any neither)

without regex if allowed.
import re
s= 'imgs/foldeer/img_ABC_15431_10.tif.tif'
last =s[s.rindex('_')+1:]
print(re.findall(r'\d+', last)[0])
Gives #
10

[0-9]*(?=\.tif\.tif)
This regex expression uses a lookahead to capture the last set of numbers (what you're looking for)

Try this:
import re
s = '''imgs/foldeer/img_ABC_21389_1.tif.tif
imgs/foldeer/img_ABC_15431_10.tif.tif
imgs/foldeer/img_GHC_561321_2.tif.tif
imgs_foldeer/img_BCL_871125_21.tif.tif'''
number = 1
res1 = re.findall(f".*_{number}\.tif.*", s)
number = 21
res21 = re.findall(f".*_{number}\.tif.*", s)
print(res1)
print(res21)
Results
['imgs/foldeer/img_ABC_21389_1.tif.tif']
['imgs_foldeer/img_BCL_871125_21.tif.tif']

Extracting float or int number and substring from a string

I've just learned regex in python3 and was trying to solve a problem.
The problem is something like this:
You have given a string where the first part is a float or integer number and the next part is a substring. You must split the number and the substring and return it as a list. The substring will only contain the alphabet from a-z and A-Z. The values of numbers can be negative.
For example:
Input: 2.5ax
Output:['2.5','ax']
Input: -5bcf
Output:['-5','bcf']
Input:-69.67Gh
Output:['-69.67','Gh']
and so on.
I did several attempts with regex to solve the problem.
1st attempt:
import re
i=input()
print(re.findall(r'^(-?\d+(\.\d+)?)|[a-zA-Z]+$',i))
For the input -2.55xy, the expected output was ['-2.55','xy']
But the output came:
[('-2.55', '.55'), ('', '')]
2nd attempt:
My second attempt was similar to my first attempt just a little different:
import re
i=input()
print(re.findall(r'^(-?(\d+\.\d+)|\d+)|[a-zA-Z]+$',i))
For the same input -2.55xy, the output came as:
[('-2.55', '2.55'), ('', '')]
3rd attempt:
My next attempt was like that:
import re
i=input()
print(re.findall(r'^-?[1-9.]+|[a-z|A-Z]+$',i))
which matched the expected output for -2.55xy and also with the sample examples. But when the input is 2..5 or something like that, it considers that also as a float.
4th attempt:
import re
i=input()
value=re.findall(r"[a-zA-Z]+",i)
print([i.replace(value[0],""),value[0]])
which also matches the expected output but has the same problem as 3rd one that goes with it. Also, it doesn't look like an effective way to do it.
Conclusion:
So I don't know why my 1st and 2nd attempt isn't working. The output comes with a list of tuples which is maybe because of the groups but I don't know the exact reason and don't know how to solve them. Maybe I didn't understand the way the pattern works. Also why the substring didn't show in the output?
In the end, I want to know what's the mistake in my code and how can I write better and more efficient code to solve the problem. Thank you and sorry for my bad English.

The alternation | matches either the left part or the right part.
If the chars a-zA-Z are after the digit, you don't need the alternation | and you can use 2 capture groups to get the matches in that order.
Then using re.findall will return a list of tuples for the capture group values.
(-?\d+(?:\.\d+)?)([a-zA-Z]+)
Explanation
( Capture group 1
-?\d+ Match an optional -
(?:\.\d+)? Optionally match . and 1+ digits using a non capture group (so it is not outputted separately by re.findall)
) Close group 1
( Capture group 2
[a-zA-Z]+ Match 1+ times a char a-z or A-Z
) Close group 2
regex demo
import re
strings = [
"2.5ax",
"-5bcf",
"-69.67Gh",
]
pattern = r"(-?\d+(?:\.\d+)?)([a-zA-Z]+)"
for s in strings:
print(re.findall(pattern, s))
Output
[('2.5', 'ax')]
[('-5', 'bcf')]
[('-69.67', 'Gh')]

lookahead and lookbehind in re.sub simplify things sometimes.
(?<=\d) look behind
(?=[a-zA-Z]) look ahead
that is split between the digit and the letter.
strings = [
"2.5ax",
"-5bcf",
"-69.67Gh",
]
for s in strings:
print(re.split(r'(?<=\d)(?=[a-zA-Z])', s))
['2.5', 'ax']
['-5', 'bcf']
['-69.67', 'Gh']

Python regex negative lookahead matching where it shouldn't

Example first:
import re
details = 'input1 mem001 output1 mem005 data2 mem002 output12 mem006'
input_re = re.compile(r'(?!output[0-9]*) mem([0-9a-f]+)')
print(input_re.findall(details))
# Out: ['001', '005', '002', '006']
I am using negative lookahead to extract the hex part of the mem entries that are not preceded by an output, however as you can see it fails. The desired output should be: ['001', '002'].
What am I missing?

You may use this regex in findall:
\b(?!output\d+)\w+\s+mem([a-zA-F\d]+)
RegEx Demo
RegEx Details:
\b: Word boundary
(?!output\d+): Negative lookahead to assert that we don't have output and 1+ digits ahead
\w+: Match 1+ word characters
\s+: Match 1+ whitespaces
mem([a-zA-F\d]+): Match mem followed by 1+ of any hex character
Code:
import re
s = 'input1 mem001 output1 mem005 data2 mem002 output12 mem006'
print( re.findall(r'\b(?!output\d+)\w+\s+mem([a-zA-F\d]+)', s) )
Output:
['001', '002']

Maybe an easier approach is to split it up in 2 regular expressions ?
First filter out anything that starts with output and is followed by mem like so
output[0-9]* mem([0-9a-f]+)
If you filter this out it would result in
input1 mem001 data2 mem002
When you have filtered them out just search for mem again
mem([0-9a-f]+)
That would result in your desired output
['001', '002']
Maybe not an answer to the original question, but it is a solution to your problem

First of all, let's understand why your original regex doesn't work:
A regex encapsulates two pieces of information: a description of a location within a text, and a description of what to capture from that location. Your original regex tells the regex matcher: "Find a location within the text where the following characters are not 'output'+digits but they are ' mem'+alphanumetics". Think of the logic of that expression: if the matcher finds a location in the text where the following characters are ' mem'+alphanumerics, then, in particular, the following characters are not 'output'+digits. Your look ahead does not add anything to the exoression.
What you really need is to tell the matcher: "Find a location in the text where the following characters are ' mem'+alphanumerics, and the previous characters are not 'output'+digits. So what you really need is a look-behind, not look-ahead.
#ArtyomVancyan proposed a good regex with a look-behind, and it could easily be modified to what you need: instead of a single digit after the 'output', you want potentially more digits, so just put an asterisk (*) after the '\d'.

Extract date from inside a string with Python

I have the following string, while the first letters can differ and can also be sometimes two, sometimes three or four.
PR191030.213101.ABD
I want to extract the 191030 and convert that to a valid date.
filename_without_ending.split(".")[0][-6:]
PZA191030_392001_USB
Sometimes it looks liket his
This solution is not valid since this is also might differ from time to time. The only REAL pattern is really the first six numbers.
How do I do this?
Thank you!

You could get the first 6 digits using a pattern an a capturing group
^[A-Z]{2,4}(\d{6})\.
^ Start of string
[A-Z]{2,4} Match 2, 3 or 4 uppercase chars
( Capture group 1
\d{6} Match 6 digits
)\. Close group and match trailing dot
Regex demo | Python demo
For example
import re
regex = r"^[A-Z]{2,4}(\d{6})\."
test_str = "PR191030.213101.ABD"
matches = re.search(regex, test_str)
if matches:
print(matches.group(1))
Output
191030

You can do:
a = 'PR191030.213101.ABD'
int(''.join([c for c in a if c.isdigit()][:6]))
Output:
191030

This can also be done by:
filename_without_ending.split(".")[0][2::]
This splits the string from the 3rd letter to the end.

Since first letters can differ we have to ignore alphabets and extract digits.
So using re module (for regular expressions) apply regex pattern on string. It will give matching pattern out of string.
'\d' is used to match [0-9]digits and + operator used for matching 1 digit atleast(1/more).
findall() will find all the occurences of matching pattern in a given string while #search() is used to find matching 1st occurence only.
import re
str="PR191030.213101.ABD"
print(re.findall(r"\d+",str)[0])
print(re.search(r"\d+",str).group())

Extract Only Digits from Dollar Figures

What I'm trying to do is extract only the digits from dollar figures.
Format of Input
...
$1,289,868
$62,000
$421
...
Desired Output
...
1289868
62000
421
...
The regular expression that I was using to extract only the digits and commas is:
r'\d+(,\d+){0,}'
which of course outputs...
...
1,289,868
62,000
421
...
What I'd like to do is convert the output to an integer (int(...)), but obviously this won't work with the commas. I'm sure I could figure this out on my own, but I'm running really short on time right now.
I know I can simply use r'\d+', but this obviously separates each chunk into separate matches...

You can't match discontinuous texts within one match operation. You can't put a regex into re.findall against 1,345,456 to receive 1345456. You will need to first match the strings you need, and then post-process them within code.
A regex you may use to extract the numbers themselves
re.findall(r'\$(\d{1,3}(?:,\d{3})*)', s)
See this regex demo.
Alternatively, you may use a bit more general regex to be used with re.findall:
r'\$(\d+(?:,\d+)*)'
See this regex demo.
Note that re.findall will only return the captured part of the string (the one matched with the (...) part in the regex).
Details
\$ - a dollar sign
(\d{1,3}(?:,\d{3})*) - Capturing group 1:
\d{1,3} - 1 to 3 digits (if \d+ is used, 1 or more digits)
(?:,\d{3})* - 0 or more sequences of
, - a comma
\d{3} - 3 digits (or if \d+ is used, 1 or more digits).
Python code sample (with removing commas):
import re
s = """$1,289,868
$62,000
$421"""
result = [x.replace(",", "") for x in re.findall(r'\$(\d{1,3}(?:,\d{3})*)', s)]
print(result) # => ['1289868', '62000', '421']

Using re.sub
Ex:
import re
s = """$1,289,868
$62,000
$421"""
print([int(i) for i in re.sub(r'[^0-9\s]', "", s).splitlines()])
Output:
[1289868, 62000, 421]

You don't need regex for this.
int(''.join(filter(str.isdigit, "$1,000,000")))
works just fine.
If you did want to use regex for some reason:
int(''.join(re.findall(r"\d", "$1,000,000")))

If you know how to extract the numbers with comma groupings, the easiest thing to do is just transform that into something int can handle:
for match in matches:
i = int(match.replace(',', ''))
For example, if match is '1,289,868', then match.replace(',', '') is '1289868', and obviously int(<that>) is 1289868.

You dont need regex for this. Just string operations should be enough
>>> string = '$1,289,868\n$62,000\n$421'
>>> [w.lstrip('$').replace(',', '') for w in string.splitlines()]
['1289868', '62000', '421']
Or alternatively, you can use locale.atoi to convert string of digits with commas to int
>>> import locale
>>> locale.setlocale(locale.LC_ALL, 'en_US.UTF8')
>>> list(map(lambda x: locale.atoi(x.lstrip('$')), string.splitlines()))
[1289868, 62000, 421]