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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.

Seemingly simple text replacement script question

I have a little "problem" that I would like to solve via programming a simple script. I don't have much programming experience and thought I'd ask here for help for what I should look for or know to do this.
Basically I want to take an email address such as placeholder.1234#fakemail.com and replace it into pl*************4#fakemail.com.
I need the script to take the letters after the first two, and before the last, and turn those letters into asterisks, and they have to match the amount of characters.
Just for clarification, I am not asking for someone to write this for me, I just need some guidance for how to go about this. I would also like to do this in Python, as I already have Python setup on my PC.

You can use Python string slicing:
email = "placeholder.1234#fakemail.com"
idx1 = 2
idx2 = email.index("#") - 1
print(email[:idx1] + "*" * (idx2 - idx1) + email[idx2:])
Output:
pl*************4#fakemail.com
Explanation:
Define the string that will contain the email address:
email = "placeholder.1234#fakemail.com"
Define the index of where the asterisks should begin, which is 2:
idx = 2
Define the index of where the asterisks should end, which is the index of where the # symbol is minus 1:
idx2 = email.index("#") - 1
Finally, using the indices defined, you can slice and concatenate the string defined accordingly:
print(email[:idx1] + "*" * (idx2 - idx1) + email[idx2:])

So this email will be a string.
Try use a combination of String indexing and (string replacement or string concatenation).

First, let's think about what data type we would store this in. It should be a String since it contains letters and characters.
We know we want to replace a portion of the String from the THIRD character to the character 2 before "#".
Let's think about this in terms of indexing now. The third character, or our start index for replacement is at index 2. To find the index of the character "#", we can use the: index() function as so:
end = email.index('#')
However, we want to start replacing from 2 before that index, so we can just subtract 2 from it.
Now we have the indexes (start and endpoint) of what we want to replace. We can now use a substring that goes from the starting to ending indexes and use the .replace() function to replace this substring in our email with a bunch of *'s.
To determine how many *'s we need, we can find the difference in indexes and add 1 to get the total number. So we could do something like:
stars = "*" * (end - start + 1)
email = email.replace(email[start:end + 1], stars)
Notice how I did start:end + 1 for the substring. This is because we want to include that ending index, and the substring function on its own will not include the ending index.
I hope this helped answer your question! Please let me know if you need any further clarification or details:)

How to grab a specified number of characters after a part of a specified string?

Let's say I have a string defined like this:
string1 = '23h4b245hjrandomstring345jk3n45jkotherrandomstring'
The goal is to grab the 11 characters (these for example '345jk3n45jk') after a part of the string (this part for example 'randomstring') using a specified search term and the specified number of characters to grab after that search term.
I tried doing something like this:
string2 = substring(string1,'randomstring', 11)
I appreciate any help you guys have to offer!

string2 = string1[string1.find("randomstring")+len("randomstring"):string1.find("randomstring")+len("randomstring")+11]

In one line, using split, and supposing that your randomstring is unique in your string, which seems to be the case as you worded out the question :
string1 = '23h4b245hjrandomstring345jk3n45jkotherrandomstring'
randomstring = 'randomstring'
nb_char_to_take = 11
# split using randomstring as splitter, take part of the string after, i.e the second part of the array, and then the 11 first character
result = string1.split(randomstring)[1][:nb_char_to_take]

You can use a simple regular expression like this
import re
s = "23h4b245hjrandomstring345jk3n45jkotherrandomstring"
result = re.findall("randomstring(.{11})", s)[0]

string1 = '23h4b245hjrandomstring345jk3n45jkotherrandomstring'
string2 = string1[10:22]
print(string2)
randomstring
You could use that. Its called string slicing, you basically count the position of the letters and then the first number before the colon is your starting point the second is your ending point when you enter those position numbers you should get whatever is in-between those position, the last is for a different function I highly suggest you search string slicing on YouTube as my explanation wouldn't really help you, and also search up * Find string method* those should hep you get the idea behind those functions. Sorry couldn't be of much help hope the videos help.

how to get second last and last value in a string after separator in python

In Python, how do you get the last and second last element in string ?
string "client_user_username_type_1234567"
expected output : "type_1234567"

Try this :
>>> s = "client_user_username_type_1234567"
>>> '_'.join(s.split('_')[-2:])
'type_1234567'

You can also use re.findall:
import re
s = "client_user_username_type_1234567"
result = re.findall('[a-zA-Z]+_\d+$', s)[0]
Output:
'type_1234567'

There's no set function that will do this for you, you have to use what Python gives you and for that I present:
split slice and join
"_".join("one_two_three".split("_")[-2:])
In steps:
Split the string by the common separator, "_"
s.split("_")
Slice the list so that you get the last two elements by using a negative index
s.split("_")[-2:]
Now you have a list composed of the last two elements, now you have to merge that list again so it's like the original string, with separator "_".
"_".join("one_two_three".split("_")[-2:])
That's pretty much it. Another way to investigate is through regex.

How to store number from string in python

i have string like this:
string = "The stock item "28031 (111111: Test product)" was added successfully."
I need store from string the first 5 numbers ( for example "28031" ) and save them to another string.
It's because i am selenium tester and every time i am create new stock item he has different first 5 numbers.
Thank you for your help
Filip

m = re.search("\d+", string)
print m.group(0)
prints 28031
It just selects the first group of digits, regardless of the length (2803 would be selected, too)

Firstly I am assuming all these strings have exactly the same format. If so the simplest way to get your stock item number is:
stocknumber = string.split()[3][1:]

After sehe answer I leave mine edited just to show how to match 5 digits
import re
re.search('\d{5}', string).group(0)

EDIT : neurino solution is the smartest!! use it
EDIT : sehe solution is smart and perfect you can add this line to get only the first 5 numbers:
print m.group(0)[0:5]
using [0:5] means to take string elements from 0 to 5 (first 5 elements)
use the str.isdigit built-in function
string = "The stock item 28031 "
Digitstring=''
for i in string:
if i.isdigit():
Digitstring+=i
print Digitstring
Output:
28031
you can count to first x numbers you need and then stop.