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Split according to regex condition

This will be my another question:
string = "Organization: S.P. Dyer Computer Consulting, Cambridge MA"
How can I take all the characters despite it being fullstop, digits, or anything after "Organization: " using regex?
result_organization = re.search("(Organization: )(\w*\.*\w*\.*\w*\s*\w*\s*\w*\s*)", string)
My above code is super long and not wise at all.

I would recommend using find command like this
print(string[string.find("Organization")+14:])

You don't need regex for that, this simple code should give you desired result:
str = "Organization: S.P. Dyer Computer Consulting, Cambridge MA";
if str.startswith("Organization: "):
str = str[14:];
print(str)
You also could use pattern (?<=Organization: ).+
Explanation:
(?<=Organization: ) - positive lookbehind, asserts if what is preceeding is Organization:
.+ - match any character except for newline characters.
Demo

You could use a single capturing group instead of 2 capturing groups.
Instead of specify all the words (\w*\.*\w*\.*\w*\s*\w*\s*\w*\s*) you might choose to match any character except a newline using the dot and then match the 0+ times to match until the end.
But note that that would also match strings like ##$$ ++
^Organization: (.+)
Regex demo | Python demo
For example
import re
string = "Organization: S.P. Dyer Computer Consulting, Cambridge MA"
result_organization = re.search("Organization: (.*)", string)
print(result_organization.group(1))
If you want a somewhat more restrictive pattern you might use a character class and specify what you would allow to match. For example:
^Organization: ([\w.,]+(?: [\w.,]+)*)
Regex demo

Add space between Persian numeric and letter with python re

I want to add space between Persian number and Persian letter like this:
"سعید123" convert to "سعید 123"
Java code of this procedure is like below.
str.replaceAll("(?<=\\p{IsDigit})(?=\\p{IsAlphabetic})", " ").
But I can't find any python solution.

There is a short regex which you may rely on to match boundary between letters and digits (in any language):
\d(?=[^_\d\W])|[^_\d\W](?=\d)
Live demo
Breakdown:
\d Match a digit
(?=[^_\d\W]) Preceding a letter from a language
| Or
[^_\d\W] Match a letter from a language
(?=\d) Preceding a digit
Python:
re.sub(r'\d(?![_\d\W])|[^_\d\W](?!\D)', r'\g<0> ', str, flags = re.UNICODE)
But according to this answer, this is the right way to accomplish this task:
re.sub(r'\d(?=[آابپتثجچحخدذرزژسشصضطظعغفقکگلمنوهی])|[آابپتثجچحخدذرزژسشصضطظعغفقکگلمنوهی](?=\d)', r'\g<0> ', str, flags = re.UNICODE)

I am not sure if this is a correct approach.
import re
k = "سعید123"
m = re.search("(\d+)", k)
if m:
k = " ".join([m.group(), k.replace(m.group(), "")])
print(k)
Output:
123 سعید

You may use
re.sub(r'([^\W\d_])(\d)', r'\1 \2', s, flags=re.U)
Note that in Python 3.x, re.U flag is redundant as the patterns are Unicode aware by default.
See the online Python demo and a regex demo.
Pattern details
([^\W\d_]) - Capturing group 1: any Unicode letter (literally, any char other than a non-word, digit or underscore chars)
(\d) - Capturing group 2: any Unicode digit
The replacement pattern is a combination of the Group 1 and 2 placeholders (referring to corresponding captured values) with a space in between them.
You may use a variation of the regex with a lookahead:
re.sub(r'[^\W\d_](?=\d)', r'\g<0> ', s)
See this regex demo.

python3: regex, find all substrings that starts with and end with certain string

Let's say that I have a string that looks like this:
a = '1253abcd4567efgh8910ijkl'
I want to find all substrings that starts with a digit, and ends with an alphabet.
I tried,
b = re.findall('\d.*\w',a)
but this gives me,
['1253abcd4567efgh8910ijkl']
I want to have something like,
['1234abcd','4567efgh','8910ijkl']
How can I do this? I'm pretty new to regex method, and would really appreciate it if anyone can show how to do this in different method within regex, and explain what's going on.

\w will match any wordcharacter which consists of numbers, alphabets and the underscore sign. You need to use [a-zA-Z] to capture letters only. See this example.
import re
a = '1253abcd4567efgh8910ijkl'
b = re.findall('(\d+[A-Za-z]+)',a)
Output:
['1253abcd', '4567efgh', '8910ijkl']
\d will match digits. \d+ will match one or more consecutive digits. For e.g.
>>> re.findall('(\d+)',a)
['1253', '4567', '8910']
Similarly [a-zA-Z]+ will match one or more alphabets.
>>> re.findall('([a-zA-Z]+)',a)
['abcd', 'efgh', 'ijkl']
Now put them together to match what you exactly want.

From the Python manual on regular expressions, it tells us that \w:
matches any alphanumeric character and the underscore; this is equivalent to the set [a-zA-Z0-9_]
So you are actually over capturing what you need. Refine your regular expression a bit:
>>> re.findall(r'(\d+[a-z]+)', a, re.I)
['1253abcd', '4567efgh', '8910ijkl']
The re.I makes your expression case insensitive, so it will match upper and lower case letters as well:
>>> re.findall(r'(\d+[a-z]+)', '12124adbad13434AGDFDF434348888AAA')
['12124adbad']
>>> re.findall(r'(\d+[a-z]+)', '12124adbad13434AGDFDF434348888AAA', re.I)
['12124adbad', '13434AGDFDF', '434348888AAA']

\w matches string with any alphanumeric character. And you have used \w with *. So your code will provide a string which is starting with a digit and contains alphanumeric characters of any length.
Solution:
>>>b=re.findall('\d*[A-Za-z]*', a)
>>>b
['1253abcd', '4567efgh', '8910ijkl', '']
you will get '' (an empty string) at the end of the list to display no match. You can remove it using
b.pop(-1)

Regular expressions: replace comma in string, Python

Somehow puzzled by the way regular expressions work in python, I am looking to replace all commas inside strings that are preceded by a letter and followed either by a letter or a whitespace. For example:
2015,1674,240/09,PEOPLE V. MICHAEL JORDAN,15,15
2015,2135,602832/09,DOYLE V ICON, LLC,15,15
The first line has effectively 6 columns, while the second line has 7 columns. Thus I am trying to replace the comma between (N, L) in the second line by a whitespace (N L) as so:
2015,2135,602832/09,DOYLE V ICON LLC,15,15
This is what I have tried so far, without success however:
new_text = re.sub(r'([\w],[\s\w|\w])', "", text)
Any ideas where I am wrong?
Help would be much appreciated!

The pattern you use, ([\w],[\s\w|\w]), is consuming a word char (= an alphanumeric or an underscore, [\w]) before a ,, then matches the comma, and then matches (and again, consumes) 1 character - a whitespace, a word character, or a literal | (as inside the character class, the pipe character is considered a literal pipe symbol, not alternation operator).
So, the main problem is that \w matches both letters and digits.
You can actually leverage lookarounds:
(?<=[a-zA-Z]),(?=[a-zA-Z\s])
See the regex demo
The (?<=[a-zA-Z]) is a positive lookbehind that requires a letter to be right before the , and (?=[a-zA-Z\s]) is a positive lookahead that requires a letter or whitespace to be present right after the comma.
Here is a Python demo:
import re
p = re.compile(r'(?<=[a-zA-Z]),(?=[a-zA-Z\s])')
test_str = "2015,1674,240/09,PEOPLE V. MICHAEL JORDAN,15,15\n2015,2135,602832/09,DOYLE V ICON, LLC,15,15"
result = p.sub("", test_str)
print(result)
If you still want to use \w, you can exclude digits and underscore from it using an opposite class \W inside a negated character class:
(?<=[^\W\d_]),(?=[^\W\d_]|\s)
See another regex demo

\w matches a-z,A-Z and 0-9, so your regex will replace all commas. You could try the following regex, and replace with \1\2.
([a-zA-Z]),(\s|[a-zA-Z])
Here is the DEMO.

Searching an input string for occurences of integers and characters using a single regular expression in Python

I have an input string which is considered valid only if it contains:
At least one character in [a-z]
At least one integer in [0-9], and
At least one character in [A-Z]
There is no constraint on the order of occurrence of any of the above. How can I write a single regular expression that validates my input string ?

Try this
^(?=.*[a-z])(?=.*[A-Z])(?=.*[0-9]).*$
See it here online on Regexr
The ^ and $ are anchors which bind the pattern to the start and the end of the string.
The (?=...) are lookahead assertions. they check if the pattern after the = is ahead but they don't match it. So to match something there needs to be a real pattern also. Here it is the .* at the end.
The .* would match the empty string also, but as soon as one of the lookaheads fail, the complete expression will fail.
For those who are concerned about the readability and maintainability, use the re.X modifier to allow pretty and commented regexes:
reg = re.compile(r'''
^ # Match the start of the string
(?=.*[a-z]) # Check if there is a lowercase letter in the string
(?=.*[A-Z]) # Check if there is a uppercase letter in the string
(?=.*[0-9]) # Check if there is a digit in the string
.* # Match the string
$ # Match the end of the string
'''
, re.X) # eXtented option whitespace is not part of he pattern for better readability

Do you need regular expression?
import string
if any(c in string.uppercase for c in t) and any(c in string.lowercase for c in t) and any(c in string.digits for c in t):
or an improved version of #YuvalAdam's improvement:
if all(any(c in x for c in t) for x in (string.uppercase, string.lowercase, string.digits)):