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Regex Ignore Match in the middle

I have string that I want to parse.
The digit (0-99) are appended at the end of the string with '_' delimiter.
The same pattern can also be in the middle of the string with the same delimiter.
For example,
#1 ..._ADDXNT_5_6_7_8_9_10_11_12_1
#2 ...X_VSVFT_0_5_ADL_R_
This is the regex I have
((?<=_)\d{1,2})
It works for #1 and parses it out ['5', '6', '7', '8', '9', '10', '11', '12', '1']
However, it also parses ['0', '5'] which I don't want.
For #2, it shouldn't match.
How can get it to only parse the end appended digits and not in the middle?
for i in ['_ADTCNT_5_6_7_8_9_10_11_12_1', 'X_VDDFEFSET_0_5_ALL_C_']:
match = re.findall(r"(?<=)\d{1,2}", i)
print(match)

You may use this regex with a positive lookahead:
(?<=_)\d{1,2}(?=(?:_\d{1,2})*$)
RegEx Demo
RegEx Details:
(?<=_): Positive lookbehind to assert that we have a _` at previous position
\d{1,2}: Match 1 or 2 digits
(?=(?:_\d{1,2})*$): Positive lookahead to assert that we have 0 or more of _<digits> strings ahead till end

i think this will work :
(?<=_)\d{1,2}(?!.*[A-Za-z])
and this (?!.*[A-Za-z]) mean not followed by alphabetic characters

Another option is to repeatedly match all underscores followed by 1+ digits till the end of the string, and then split on _
(?:_\d{1,2})+$
Regex demo | Python demo
import re
for i in ['_ADTCNT_5_6_7_8_9_10_11_12_1', 'X_VDDFEFSET_0_5_ALL_C_']:
result = re.search(r"(?:_\d{1,2})+$", i)
if result:
print([x for x in result.group().split("_") if x])
Output
['5', '6', '7', '8', '9', '10', '11', '12', '1']

how to find the matching pattern for an input list and then replace the found pattern with the proper pattern conversion using python

note that the final two numbers of this pattern for example FBXASC048 are ment to be ascii code for numbers (0-9)
input example list ['FBXASC048009Car', 'FBXASC053002Toy', 'FBXASC050004Human']
result example ['1009Car', '5002Toy', '2004Human']
what is the proper way to searches for any of these pattern in an input list
num_ascii = ['FBXASC048', 'FBXASC049', 'FBXASC050', 'FBXASC051', 'FBXASC052', 'FBXASC053', 'FBXASC054', 'FBXASC055', 'FBXASC056', 'FBXASC057']
and then replaces the pattern found with one of the items in the conv list but not randomally
because each element in the pattern list equals only one element in the conv_list
conv_list = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
this is the solution in mind:
it has two part
1st part--> is to find for ascii pattern[48, 49, 50, 51, 52, 53, 54, 55, 56,57]
and then replace those with the proper decimal matching (0-9)
so we will get new input list will be called input_modi_list that has ascii replaced with decimal
2nd part-->another process to use fixed pattern to replace using replace function which is this 'FBXASC0'
new_list3
for x in input_modi_list:
y = x.replace('FBXASC0', '')
new_list3.append(new_string)
so new_list3 will have the combined result of the two parts mentioned above.
i don't know if there would be a simplar solution or a better one maybe using regex
also note i don't have any idea on how to replace ascii with decimal for a list of items

I think this should do the trick:
import re
input_list = ['FBXASC048009Car', 'FBXASC053002Toy', 'FBXASC050004Human']
pattern = re.compile('FBXASC(\d{3,3})')
def decode(match):
return chr(int(match.group(1)))
result = [re.sub(pattern, decode, item) for item in input_list]
print(result)
Now, there is some explanation due:
1- the pattern object is a regular expression that will match any part of a string that starts with 'FBXASC' and ends with 3 digits (0-9). (the \d means digit, and {3,3} means that it should occur at least 3, and at most 3 times, i.e. exactly 3 times). Also, the parenthesis around \d{3,3} means that the three digits matched will be stored for later use (explained in the next part).
2- The decode function receives a match object, uses .group(1) to extract the first matched group (which in our case are the three digits matched by \d{3,3}), then uses the int function to parse the string into an integer (for example, convert '048' to 48), and finally uses the chr function to find which character has that ASCII-code. (for example chr(48) will return '0', and chr(65) will return 'A')
3- The final part applies the re.sub function to all elements of list which will replace each occurrence of the pattern you described (FBXASC048[3-digits]) with it's corresponding ASCII character.
You can see that this solution is not limited only to your specific examples. Any number can be used as long as it has a corresponding ASCII character recognized by the chr function.
But, if you do want to limit it just to the 48-57 range, you can simply modify the decode function:
def decode(match):
ascii_code = int(match.group(1))
if ascii_code >= 48 and ascii_code <= 57:
return chr(ascii_code)
else:
return match.group(0) # returns the entire string - no modification

This is how I would do it.
make the regex pattern by simply joining the strings with |:
>>> num_ascii = ['FBXASC048', 'FBXASC049', 'FBXASC050', 'FBXASC051', 'FBXASC052', 'FBXASC053', 'FBXASC054', 'FBXASC055', 'FBXASC056', 'FBXASC057']
>>> conv_list = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
>>> regex_pattern = '|'.join(num_ascii)
>>> regex_pattern
'FBXASC048|FBXASC049|FBXASC050|FBXASC051|FBXASC052|FBXASC053|FBXASC054|FBXASC055
|FBXASC056|FBXASC057'
make a look-up dictionary by simply zipping the two lists:
>>> conv_table = dict(zip(num_ascii, conv_list))
>>> conv_table
{'FBXASC048': '0', 'FBXASC049': '1', 'FBXASC050': '2', 'FBXASC051': '3', 'FBXASC
052': '4', 'FBXASC053': '5', 'FBXASC054': '6', 'FBXASC055': '7', 'FBXASC056': '8
', 'FBXASC057': '9'}
iterate over the data and replace the matched string with the corresponding digit:
>>> import re
>>> result = []
>>> for item in ['FBXASC048009Car', 'FBXASC053002Toy', 'FBXASC050004Human']:
... m = re.match(regex_pattern, item)
... matched_string = m[0]
... digit = (conv_table[matched_string])
... print(f'replacing {matched_string} with {digit}')
... result.append(item.replace(matched_string, digit))
...
replacing FBXASC048 with 0
replacing FBXASC053 with 5
replacing FBXASC050 with 2
>>> result
['0009Car', '5002Toy', '2004Human']

How to split algebraic expressions in a string using python?

For example I get following input:
-9x+5x-2-4x+5
And I need to get following list:
['-9x', '5x', '-2', '-4x', '5']
Here is my code, but I don't know how to deal with minuses.
import re
text = '-3x-5x+2=9x-9'
text = re.split(r'\W', text)
print(text)
warning: I cannot use any libraries except re and math.

You could re.findall all groups of characters followed by + or - (or end-of-string $), then strip the + (which, like -, is still part of the following group) from the substrings.
>>> s = "-9x+5x-2-4x+5"
>>> [x.strip("+") for x in re.findall(r".+?(?=[+-]|$)", s)]
['-9x', '5x', '-2', '-4x', '5']
Similarly, for the second string with =, add that to the character group and also strip it off the substrings:
>>> s = '-3x-5x+2=9x-9'
>>> [x.strip("+=") for x in re.findall(r".+?(?=[+=-]|$)", s)]
>>> ['-3x', '-5x', '2', '9x', '-9']
Or apply the original comprehension to the substrings after splitting by =, depending on how the result should look like:
>>> [[x.strip("+") for x in re.findall(r".+?(?=[+-]|$)", s2)] for s2 in s.split("=")]
>>> [['-3x', '-5x', '2'], ['9x', '-9']]
In fact, now that I think of it, you can also just findall that match an optional minus, followed by some digits, and an optional x, with or without splitting by = first:
>>> [re.findall(r"-?\d+x?", s2) for s2 in s.split("=")]
[['-3x', '-5x', '2'], ['9x', '-9']]

One of many possible ways:
import re
term = "-9x+5x-2-4x+5"
rx = re.compile(r'-?\d+[a-z]?')
factors = rx.findall(term)
print(factors)
This yields
['-9x', '5x', '-2', '-4x', '5']

For your example data, you might split on either a plus or equals sign or split when asserting a minus sign on the right which is not at the start of the string.
[+=]|(?=(?<!^)-)
[+=] Match either + or =
| Or
(?=(?<!^)-) Positive lookahead, assert what is on the right is - but not at the start of the string
Regex demo | Python demo
Output for both example strings
['-9x', '5x', '-2', '-4x', '5']
['-3x', '-5x', '2', '9x', '-9']

How to remove alphabets and extract numbers using regex in python?

How to remove alphabets and extract numbers using regex in python?
import re
l=["098765432123 M","123456789012"]
s = re.findall(r"(?<!\d)\d{12}", l)
print(s)
Expected Output:
123456789012

If all you want is to have filtered list, consisting elements with pure digits, use filter with str.isdigit:
list(filter(str.isdigit, l))
Or as #tobias_k suggested, list comprehension is always your friend:
[s for s in l if s.isdigit()]
Output:
['123456789012']

I would suggest to use a negative lookahead assertion, if as stated you want to use regex only.
l=["098765432123 M","123456789012"]
res=[]
for a in l:
s = re.search(r"(?<!\d)\d{12}(?! [a-zA-Z])", a)
if s is not None:
res.append(s.group(0))
The result would then be:
['123456789012']

To keep only digits you can do re.findall('\d',s), but you'll get a list:
s = re.findall('\d', "098765432123 M")
print(s)
> ['0', '9', '8', '7', '6', '5', '4', '3', '2', '1', '2', '3']

So to be clear, you want to ignore the whole string if there is a alphabetic character in it? Or do you still want to extract the numbers of a string with both numbers and alphabetic characters in it?
If you want to find all numbers, and always find the longest number use this:
regex = r"\d+"
matches = re.finditer(regex, test_str, re.MULTILINE)
\d will search for digits, + will find one or more of the defined characters, and will always find the longest consecutive line of these characters.
If you only want to find strings without alphabets:
import re
regex = r"[a-zA-Z]"
test_str = ("098765432123 M", "123456789012")
for x in test_str:
if not re.search(regex, x):
print(x)

Separate each item of a list in an specific way

I have an input, which is a tuple of strings, encoded in a1z26 cipher: numbers from 1 to 26 represent alphabet letters, hyphens represent same word letters and spaces represent an space between words.
For example:
8-9 20-8-5-18-5 should translate to 'hi there'
Let's say that the last example is a tuple in a var called string
string = ('8-9','20-8-5-18-5')
The first thing I find logical is convert the tuple into a list using
string = list(string)
so now
string = ['8-9','20-8-5-18-5']
The problem now is that when I iterate over the list to compare it with a dictionary which has the translated values, double digit numbers are treated as one, so instead of, for example, translating '20' it translate '2' and then '0', resulting in the string saying 'hi bheahe' (2 =b, 1 = a and 8 = h)
so I need a way to convert the list above to the following
list
['8','-','9',' ','20','-','8','-','5','-','18','-','5',]
I've already tried various codes using
list(),
join() and
split()
But it ends up giving me the same problem.
To sum up, I need to make any given list (converted from the input tuple) into a list of characters that takes into account double digit numbers, spaces and hyphens altogether
This is what I've got so far. (The last I wrote) The input is further up in the code (string)
a1z26 = {'1':'A', '2':'B', '3':'C', '4':'D', '5':'E', '6':'F', '7':'G', '8':'H', '9':'I', '10':'J', '11':'K', '12':'L', '13':'M', '14':'N', '15':'O', '16':'P', '17':'Q', '18':'R', '19':'S', '20':'T', '21':'U', '22':'V', '23':'W', '24':'X', '25':'Y', '26':'Z', '-':'', ' ' : ' ', ', ' : ' '}
translation = ""
code = list(string)
numbersarray1 = code
numbersarray2 = ', '.join(numbersarray1)
for char in numbersarray2:
if char in a1z26:
translation += a1z26[char]

There's no need to convert the tuple to a list. Tuples are iterable too.
I don't think the list you name is what you actually want. You probably want a 2d iterable (not necessarily a list, as you'll see below we can do this in one pass without generating an intermediary list), where each item corresponds to a word and is a list of the character numbers:
[[8, 9], [20, 8, 5, 18, 5]]
From this, you can convert each number to a letter, join the letters together to form the words, then join the words with spaces.
To do this, you need to pass a parameter to split, to tell it how to split your input string. You can achieve all of this with a one liner:
plaintext = ' '.join(''.join(num_to_letter[int(num)] for num in word.split('-'))
for word in ciphertext.split(' '))
This does exactly the splitting procedure as described above, and then for each number looks into the dict num_to_letter to do the conversion.
Note that you don't even need this dict. You can use the fact that A-Z in unicode is contiguous so to convert 1-26 to A-Z you can do chr(ord('A') + num - 1).

You don't really need hypens, am I right?
I suggest you the following approach:
a = '- -'.join(string).split('-')
Now a is ['8', '9', ' ', '20', '8', '5', '18', '5']
You can then convert each number to the proper character using your dictionary
b = ''.join([a1z26[i] for i in a])
Now b is equal to HI THERE

I think, it's better to apply regular expressions there.
Example:
import re
...
src = ('8-9', '20-8-5-18-5')
res = [match for tmp in src for match in re.findall(r"([0-9]+|[^0-9]+)", tmp + " ")][:-1]
print(res)
Result:
['8', '-', '9', ' ', '20', '-', '8', '-', '5', '-', '18', '-', '5']

using regex here is solution
import re
string = '8-9 20-8-5-18-5'
exp=re.compile(r'[0-9]+|[^0-9]+')
data= exp.findall(string)
print(data)
output
['8', '-', '9', ' ', '20', '-', '8', '-', '5', '-', '18', '-', '5']
if you want to get hi there from the input string , here is a method (i am assuming all character are in uppercase):
import re
string = '8-9 20-8-5-18-5'
exp=re.compile(r'[0-9]+|[^0-9]+')
data= exp.findall(string)
new_str =''
for i in range(len(data)):
if data[i].isdigit():
new_str+=chr(int(data[i])+64)
else:
new_str+=data[i]
result = new_str.replace('-','')
output:
HI THERE

You could also try this itertools solution:
from itertools import chain
from itertools import zip_longest
def separate_list(lst, delim, sep=" "):
result = []
for x in lst:
chars = x.split(delim) # 1
pairs = zip_longest(chars, [delim] * (len(chars) - 1), fillvalue=sep) # 2, 3
result.extend(list(chain.from_iterable(pairs))) # 4
return result[:-1] # 5
print(separate_list(["8-9", "20-8-5-18-5"], delim="-"))
Output:
['8', '-', '9', ' ', '20', '-', '8', '-', '5', '-', '18', '-', '5']
Explanation of above code:
Split each string by delimiter '-'.
Create interspersing delimiters.
Create pairs of characters and separators with itertools.zip_longest.
Extend flattened pairs to result list with itertools.chain.from_iterable.
Remove trailing ' ' from result list added.
You could also create your own intersperse generator function and apply it twice:
from itertools import chain
def intersperse(iterable, delim):
it = iter(iterable)
yield next(it)
for x in it:
yield delim
yield x
def separate_list(lst, delim, sep=" "):
return list(
chain.from_iterable(
intersperse(
(intersperse(x.split(delim), delim=delim) for x in lst), delim=[sep]
)
)
)
print(separate_list(["8-9", "20-8-5-18-5"], delim="-"))
# ['8', '-', '9', ' ', '20', '-', '8', '-', '5', '-', '18', '-', '5']