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Convert data into comma separated values

How do i convert data into comma separated values, i want to convert like
I have this data in excel on single cell
"ABCD x3 ABC, BAC x 3"
Want to convert to
ABCD,ABCD,ABCD,ABC,BAC,BAC,BAC
can't find an easy way to do that.
I am trying to solve it in python so i can get a structured data

Hi Zeeshan to try and sort the string into usable data while also multiplying certain parts of the string is kind of tricky for me.
the best solution I can think of is kind of gross but it seems to work. hopefully my comments aren't too confusing <3
import re
data = "ABCD x3 AB BAC x2"
#this will split the string into a list that you can iterate through.
Datalist = re.findall(r'(\w+)', data)
#create a new list for the final result
newlist = []
for object in Datalist:
#for each object in the Datalist list
#if the object starts with 'x'
if re.search("x.*", object):
#convert the multiplier to type(string) and then split the x from the multiplier number string
xvalue = str(object).split('x')
#grab and remove the last item added to the newlist because it hasnt been multiplied.
lastitem = newlist.pop()
#now we can add the last item back in by as many times as the x value
newlist.extend([lastitem] * int(xvalue[1]))
else:
#if the object doesnt start with an x then we can just add it to the list.
newlist.extend([object])
#print result
print(newlist)
#re.search() - looks for a match in a string
#.split() - splits a string into multiple substrings
#.pop() - removes the last item from a list and returns that item.
#.extend() - adds an item to the end of a list
keep in mind that to find the multiplier its looking for x followed by a number (x1). if there is a space for example = (x 1) then it will match x but it wont return a value because there is a space.

there might be multiple ways around this issue and I think the best fix will be to restructure how the data is Formatted into the cell.
here are a couple of ways you can work with the data. it wont directly solve your issue but I hope it will help you think about how you approach it (not being rude I don't actually have a good way to handle your example <3 )
split() will split your string as character 'x' and return a list of substrings you can iterate over.
data = 'ABCD ABCD ABCD ABC BAC BAC BAC'
splitdata = data.split(' ')
print(splitdata)
#prints - ['ABCD', 'ABCD', 'ABCD', 'ABC', 'BAC', 'BAC', 'BAC']
you could also try and match strings from the data
import re
data2 = "ABCD x3 ABC BAC x3"
result = []
for match in re.finditer(r'(\w+) x(\d+)', data2):
substring, count = match.groups()
result.extend([substring] * int(count))
print(result)
use re.finditer to go through the string and match the data with the following format = '(\w+) x(\d+)'
each match then gets added to the list.
'\w' is used to match a character.
'\d' is used to match a digit.
'+' is the quantifier, means one or more.
so we are matching = '(\w+) x(\d+)',
which broken down means we are matching (\w+) one or more characters followed by a 'space' then 'x' followed by (\d+) one or more digits
so because your cell data is essentially a string followed by a multiplier then a string followed by another string and then another multiplier, the data just feels too random for a general solution and i think this requires a direct solution that can only work if you know exactly what data is already in the cell. that's why i think the best way to fix it is to rework the data in the cell first. im in no way an expert and this answer is to help you think of ways around the problem and to add to the discussion :) ,if someone wants to correct me and offer a better solution to this I would love to know myself.

How do you remove first and last char from a string inside of a list depending on specific sets of characters in the list. (Python)

Now, I understand that the title does not exactly correlate to what I am attempting to figure out, but hopefully, you will understand when you see my code.
First of all, there is a list containing a bunch of information, I will include only the first three.
memberList = ['uuid_e04a043abc334bd1a2fbd167bdce1673[MVP+] IgrisGuild Master2020/07/21 '
'02:35:052020/08/09 00:58:55',
'uuid_1f12bce8313040a7978d5c51ceb9d82d[VIP] mistercintPrince2020/08/01 '
'00:31:342020/08/08 23:47:53',
'uuid_405e46954f804487ae9c18689f0c351b[MVP+] zoucePrince2020/08/06 '
'20:11:222020/08/08 22:02:04']
Next, I remove the first 37 and the last 38 characters because they are all not relevant and they are all present at the exact same length in the list.
memberList = [e[37:-38] for e in memberList]
I attempted to do something like this, but I can't quite get it down.
for i in range(len(memberList):
if 'Igris' in memberList:
(remove the first 7 and the last 12 from this specific string inside of the list)
What I want it to end up as is as follows.
print(memberList)
Output:
[Igris, mistercint, zouce]
Thank you for your consideration, this is my first time using this site and I'm very new to coding, so please pardon my incorrect formatting.

import re
members = [re.sub( r"([A-Z])", r" \1", member.split()[1]).split()[0] for member in memberList]
print(members)
...should give you what you want.

Pythonic way to find the last position in a string matching a negative regex

In Python, I try to find the last position in an arbitrary string that does match a given pattern, which is specified as negative character set regex pattern. For example, with the string uiae1iuae200, and the pattern of not being a number (regex pattern in Python for this would be [^0-9]), I would need '8' (the last 'e' before the '200') as result.
What is the most pythonic way to achieve this?
As it's a little tricky to quickly find method documentation and the best suited method for something in the Python docs (due to method docs being somewhere in the middle of the corresponding page, like re.search() in the re page), the best way I quickly found myself is using re.search() - but the current form simply must be a suboptimal way of doing it:
import re
string = 'uiae1iuae200' # the string to investigate
len(string) - re.search(r'[^0-9]', string[::-1]).start()
I am not satisfied with this for two reasons:
- a) I need to reverse string before using it with [::-1], and
- b) I also need to reverse the resulting position (subtracting it from len(string) because of having reversed the string before.
There needs to be better ways for this, likely even with the result of re.search().
I am aware of re.search(...).end() over .start(), but re.search() seems to split the results into groups, for which I did not quickly find a not-cumbersome way to apply it to the last matched group. Without specifying the group, .start(), .end(), etc, seem to always match the first group, which does not have the position information about the last match. However, selecting the group seems to at first require the return value to temporarily be saved in a variable (which prevents neat one-liners), as I would need to access both the information about selecting the last group and then to select .end() from this group.
What's your pythonic solution to this? I would value being pythonic more than having the most optimized runtime.
Update
The solution should be functional also in corner cases, like 123 (no position that matches the regex), empty string, etc. It should not crash e.g. because of selecting the last index of an empty list. However, as even my ugly answer above in the question would need more than one line for this, I guess a one-liner might be impossible for this (simply because one needs to check the return value of re.search() or re.finditer() before handling it). I'll accept pythonic multi-line solutions to this answer for this reason.

You can use re.finditer to extract start positions of all matches and return the last one from list. Try this Python code:
import re
print([m.start(0) for m in re.finditer(r'\D', 'uiae1iuae200')][-1])
Prints:
8
Edit:
For making the solution a bit more elegant to behave properly in for all kind of inputs, here is the updated code. Now the solution goes in two lines as the check has to be performed if list is empty then it will print -1 else the index value:
import re
arr = ['', '123', 'uiae1iuae200', 'uiae1iuae200aaaaaaaa']
for s in arr:
lst = [m.start() for m in re.finditer(r'\D', s)]
print(s, '-->', lst[-1] if len(lst) > 0 else None)
Prints the following, where if no such index is found then prints None instead of index:
--> None
123 --> None
uiae1iuae200 --> 8
uiae1iuae200aaaaaaaa --> 19
Edit 2:
As OP stated in his post, \d was only an example we started with, due to which I came up with a solution to work with any general regex. But, if this problem has to be really done with \d only, then I can give a better solution which would not require list comprehension at all and can be easily written by using a better regex to find the last occurrence of non-digit character and print its position. We can use .*(\D) regex to find the last occurrence of non-digit and easily print its index using following Python code:
import re
arr = ['', '123', 'uiae1iuae200', 'uiae1iuae200aaaaaaaa']
for s in arr:
m = re.match(r'.*(\D)', s)
print(s, '-->', m.start(1) if m else None)
Prints the string and their corresponding index of non-digit char and None if not found any:
--> None
123 --> None
uiae1iuae200 --> 8
uiae1iuae200aaaaaaaa --> 19
And as you can see, this code doesn't need to use any list comprehension and is better as it can just find the index by just one regex call to match.
But in case OP indeed meant it to be written using any general regex pattern, then my above code using comprehension will be needed. I can even write it as a function that can take the regex (like \d or even a complex one) as an argument and will dynamically generate a negative of passed regex and use that in the code. Let me know if this indeed is needed.

To me it sems that you just want the last position which matches a given pattern (in this case the not a number pattern).
This is as pythonic as it gets:
import re
string = 'uiae1iuae200'
pattern = r'[^0-9]'
match = re.match(fr'.*({pattern})', string)
print(match.end(1) - 1 if match else None)
Output:
8
Or the exact same as a function and with more test cases:
import re
def last_match(pattern, string):
match = re.match(fr'.*({pattern})', string)
return match.end(1) - 1 if match else None
cases = [(r'[^0-9]', 'uiae1iuae200'), (r'[^0-9]', '123a'), (r'[^0-9]', '123'), (r'[^abc]', 'abcabc1abc'), (r'[^1]', '11eea11')]
for pattern, string in cases:
print(f'{pattern}, {string}: {last_match(pattern, string)}')
Output:
[^0-9], uiae1iuae200: 8
[^0-9], 123a: 3
[^0-9], 123: None
[^abc], abcabc1abc: 6
[^1], 11eea11: 4

This does not look Pythonic because it's not a one-liner, and it uses range(len(foo)), but it's pretty straightforward and probably not too inefficient.
def last_match(pattern, string):
for i in range(1, len(string) + 1):
substring = string[-i:]
if re.match(pattern, substring):
return len(string) - i
The idea is to iterate over the suffixes of string from the shortest to the longest, and to check if it matches pattern.
Since we're checking from the end, we know for sure that the first substring we meet that matches the pattern is the last.

Python re.sub() is not replacing every match

I'm using Python 3 and I have two strings: abbcabb and abca. I want to remove every double occurrence of a single character. For example:
abbcabb should give c and abca should give bc.
I've tried the following regex (here):
(.)(.*?)\1
But, it gives wrong output for first string. Also, when I tried another one (here):
(.)(.*?)*?\1
But, this one again gives wrong output. What's going wrong here?
The python code is a print statement:
print(re.sub(r'(.)(.*?)\1', '\g<2>', s)) # s is the string

It can be solved without regular expression, like below
>>>''.join([i for i in s1 if s1.count(i) == 1])
'bc'
>>>''.join([i for i in s if s.count(i) == 1])
'c'

re.sub() doesn't perform overlapping replacements. After it replaces the first match, it starts looking after the end of the match. So when you perform the replacement on
abbcabb
it first replaces abbca with bbc. Then it replaces bb with an empty string. It doesn't go back and look for another match in bbc.
If you want that, you need to write your own loop.
while True:
newS = re.sub(r'(.)(.*?)\1', r'\g<2>', s)
if newS == s:
break
s = newS
print(newS)
DEMO

Regular expressions doesn't seem to be the ideal solution
they don't handle overlapping so it it needs a loop (like in this answer) and it creates strings over and over (performance suffers)
they're overkill here, we just need to count the characters
I like this answer, but using count repeatedly in a list comprehension loops over all elements each time.
It can be solved without regular expression and without O(n**2) complexity, only O(n) using collections.Counter
first count the characters of the string very easily & quickly
then filter the string testing if the count matches using the counter we just created.
like this:
import collections
s = "abbcabb"
cnt = collections.Counter(s)
s = "".join([c for c in s if cnt[c]==1])
(as a bonus, you can change the count to keep characters which have 2, 3, whatever occurrences)

EDIT: based on the comment exchange - if you're just concerned with the parity of the letter counts, then you don't want regex and instead want an approach like #jon's recommendation. (If you don't care about order, then a more performant approach with very long strings might use something like collections.Counter instead.)
My best guess as to what you're trying to match is: "one or more characters - call this subpattern A - followed by a different set of one or more characters - call this subpattern B - followed by subpattern A again".
You can use + as a shortcut for "one or more" (instead of specifying it once and then using * for the rest of the matches), but either way you need to get the subpatterns right. Let's try:
>>> import re
>>> pattern = re.compile(r'(.+?)(.+?)\1')
>>> pattern.sub('\g<2>', 'abbcabbabca')
'bbcbaca'
Hmm. That didn't work. Why? Because with the first pattern not being greedy, our "subpattern A" can just match the first a in the string - it does appear later, after all. So if we use a greedy match, Python will backtrack until it finds as long of a pattern for subpattern A that still allows for the A-B-A pattern to appear:
>>> pattern = re.compile(r'(.+)(.+?)\1')
>>> pattern.sub('\g<2>', 'abbcabbabca')
'cbc'
Looks good to me.

The site explains it well, hover and use the explanation section.
(.)(.*?)\1 Does not remove or match every double occurance. It matches 1 character, followed by anything in the middle sandwiched till that same character is encountered again.
so, for abbcabb the "sandwiched" portion should be bbc between two a
EDIT:
You can try something like this instead without regexes:
string = "abbcabb"
result = []
for i in string:
if i not in result:
result.append(i)
else:
result.remove(i)
print(''.join(result))
Note that this produces the "last" odd occurrence of a string and not first.
For "first" known occurance, you should use a counter as suggested in this answer . Just change the condition to check for odd counts. pseudo code(count[letter] %2 == 1)

The difference between ( [^,]*) and (.*,) in regular expression? Using python

When I tried to transform the string into a dict-like form, I met this problem
s = '&a: 12, &b:13, &c:14, &d: 15' # the string I want to convert
Before converting it, I tried to find all the matched results at first so I used
dict_form = re.compile(r'(&[a-zA-Z]*:)(.*),')
result = dict_form.findall(s)
print(result) # [('&a:', ' 12, &b:13, &c:14')]
It's quite unexpected, and a little bit messy
But when I tried another way to match the string:
dict_form1 = re.compile(r'(&[a-zA-Z]*:)([^,]*)')
result = dict_form1.findall(s)
print(result) # [('&a:', ' 12'), ('&b:', '13'), ('&c:', '14'), ('&d:', ' 15')]
This time, I get a better one with key and item separately stored in a tuple.
The only difference I made was (.), into [^,]
The first one I thought was to find anything until it matches a comma
The second one I thought was to find anything but comma
What's the difference?

In the first instance:
dict_form = re.compile(r'(&[a-zA-Z]*:)(.*),')
the (.*) operator is greedy. This means it will match everything up to the last comma, which is why you see the match extend up to &c:14.
In the second instance, by excluding the comma, you are forcing the match to be bound by a comma-- it's like saying "match everything until we hit a comma". This will cause the matching behavior you were expecting in the first place.

as have been said the .* will be greedy and try to match as much as possible, to make it non-greedy use the question mark (?) as in .*?. In your code:
dict_form = re.compile(r'(&[a-zA-Z]*:)(.*?),')
result = dict_form.findall(s)
print(result)
Another maybe easier solution is to just use string splits instead of regex:
result = [_s.split(':') for _s in s.split(',')]