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Extract string with specific format

I'm novice to Python and I am trying to extract a string from another string with specific format, for example:
I have original string: -
--#$_ABC1234-XX12X
I need to extract exactly the string ABC1234 (must include three first characters and followed by four digits).

You can use the curly brace repetition qualifiers {} to match exactly three alphabetic characters and exactly four numeric characters:
>>> from re import search
>>>
>>> string = '---#$_ABC1234-XX12X'
>>> match = search('[a-zA-Z]{3}\d{4}', string)
>>> match
<_sre.SRE_Match object; span=(6, 13), match='ABC1234'>
>>> match.group(0) # Use this to get the string that was matched.
'ABC1234'
Explanation of regex:
[a-zA-Z]: Match any letter upper case of lower case...
{3}: Exactly three times. And...
\d: Any digit character...
{4} Exactly four times.

You can make use of re module in Python
matcher = re.search((?P<matched_string>[a-zA-Z]{3}\d{4}))
needed_string = matcher.groupdict()['matched_string']
needed_string will be your desired output.
For the re module refer to: https://docs.python.org/3.4/library/re.html

If you now the exact coordinates of the string you can use something like this:
>>> var = "--#$_ABC1234-XX12X"
>>> newstring = var[5:12]
>>> newstring
'ABC1234'
a python string has a slice method.

How to print substring using RegEx in Python?

This is two texts:
1) 'provider:sipoutilp1.ym.ms'
2) 'provider:sipoutqtm.ym.ms'
I would like to print ilp when reaches to the fist line and qtm when reaches to the second line.
This is my solution but it is not working.
RE_PROVIDER = re.compile(r'(?P<provider>\((ilp+|qtm+)')
or in the line below,
182938,DOMINICAN REPUBLIC-MOBILE
to DOMINICAN REPUBLIC , can I use the same approach re.compile?
Thank you for any help.

Your regex is not correct because you have a open parenthesis before your keywords, since there is no such character in your lines.
As a more general way you can get capture the alphabetical character after sipout or provider:sipout.
>>> s1 = 'provider:sipoutilp1.ym.ms'
>>> s2 = 'provider:sipoutqtm.ym.ms'
>>> RE_PROVIDER = re.compile(r'(?P<provider>(?<=sipout)(ilp|qtm))')
>>> RE_PROVIDER.search(s1).groupdict()
{'provider': 'ilp'}
>>> RE_PROVIDER.search(s2).groupdict()
{'provider': 'qtm'}
(?<=sipout) is a positive look-behind which will makes the regex engine match the patter which is precede with sipout.
After edit:
If you want to match multiple strings with different structure, you have to use a optional preceding patterns for matching your keywords, and due to this point that you cannot use unfixed length patterns within look-behind you cannot use it for this aim. So instead you can use a capture group trick.
You can define the optional preceding patterns within a none capture group and your keyword within a capture group then after match get the second matched gorup (group(1), group(0) is the whole of your match).
>>> RE_PROVIDER = re.compile(r'(?:sipout|\d+,)(?P<provider>(ilp|qtm|[A-Z\s]+))')
>>> RE_PROVIDER.search(s1).groupdict()
{'provider': 'ilp'}
>>> RE_PROVIDER.search(s2).groupdict()
{'provider': 'qtm'}
>>> s3 = "182938,DOMINICAN REPUBLIC-MOBILE"
>>> RE_PROVIDER.search(s3).groupdict()
{'provider': 'DOMINICAN REPUBLIC'}
Note that gorupdict doesn't works in this case because it will returns

Why does the Python regex ".*PATTERN*" match "XXPATTERXX"?

Suppose I want to find "PATTERN" in a string, where "PATTERN" could be anywhere in the string. My first try was *PATTERN*, but this generates an error saying that there is "nothing to repeat", which I can accept so I tried .*PATTERN*. This regex does however not give the expected result, see below
import re
p = re.compile(".*PATTERN*")
s = "XXPATTERXX"
if p.match(s):
print s + " match with '.*PATTERN*'"
The result is
XXPATTERXX match with '.*PATTERN*'
Why does "PATTER" match?
Note: I know that I could use .*PATTERN.* to get the expected result, but I am curious to find out why the asterisk on it self fails to get the results.

Your pattern matches 0 or more N characters at the end, but doesn't say anything about what comes after those N characters.
You could add $ to the pattern to anchor to the end of the input string to disallow the XX:
>>> import re
>>> re.compile(".*PATTERN*$")
<_sre.SRE_Pattern object at 0x10029fb90>
>>> import re
>>> p = re.compile(".*PATTERN*$")
>>> p.match("XXPATTERXX") is None
True
>>> p.match("XXPATTER") is None
False
>>> p.match("XXPATTER")
<_sre.SRE_Match object at 0x1004627e8>
You may want to look into the different types of anchor. \b may also fit your needs; it matches word boundaries (so between a \w and \W class character, or between \W and \w), or you could use negative look-ahead and look-behinds to disallow other characters around your PATTERN string.

Python Regex matching already matched sub-string

I'm fairly new to Python Regex and I'm not able to understand the following:
I'm trying to find one small letter surrounded by three capital letters.
My first problem is that the below regex is giving only one match instead of the two matches that are present ['AbAD', 'DaDD']
>>> import re
>>>
>>> # String
... str = 'AbADaDD'
>>>
>>> pat = '[A-Z][a-z][A-Z][A-Z]'
>>> regex = re.compile(pat)
>>>
>>> print regex.findall(str)
['AbAD']
I guess the above is due to the fact that the last D in the first regex is not available for matching any more? Is there any way to turn off this kind of matching.
The second issue is the following regex:
>>> import re
>>>
>>> # String
... str = 'AbADaDD'
>>>
>>> pat = '[^A-Z][A-Z][a-z][A-Z][A-Z][^A-Z]'
>>> regex = re.compile(pat)
>>>
>>> print regex.findall(str)
[]
Basically what I want is that there shouldn't be more than three capital letters surrounding a small letter, and therefore I placed a negative match around them. But ['AbAD'] should be matched, but it is not getting matched. Any ideas?

It's mainly because of the overlapping of matches. Just put your regex inside a lookahead inorder to handle this type of overlapping matches.
(?=([A-Z][a-z][A-Z][A-Z]))
Code:
>>> s = 'AbADaDD'
>>> re.findall(r'(?=([A-Z][a-z][A-Z][A-Z]))', s)
['AbAD', 'DaDD']
DEMO
For the 2nd one, you should use negative lookahead and lookbehind assertion like below,
(?=(?<![A-Z])([A-Z][a-z][A-Z][A-Z])(?![A-Z]))
Code:
>>> re.findall(r'(?=(?<![A-Z])([A-Z][a-z][A-Z][A-Z])(?![A-Z]))', s)
['AbAD']
DEMO
The problem with your second regex is, [^A-Z] consumes a character (there isn't a character other than uppercase letter exists before first A) but the negative look-behind (?<![A-Z]) also do the same but it won't consume any character . It asserts that the match would be preceded by any but not of an uppercase letter. That;s why you won't get any match.

The problem with you regex is tha it is eating up the string as it progresses leaving nothing for second match.Use lookahead to make sure it does not eat up the string.
pat = '(?=([A-Z][a-z][A-Z][A-Z]))'
For your second regex again do the same.
print re.findall(r"(?=([A-Z][a-z][A-Z][A-Z](?=[^A-Z])))",s)
.For more insights see
1)After first match the string left is aDD as the first part has matched.
2)aDD does not satisfy pat = '[A-Z][a-z][A-Z][A-Z]'.So it is not a part of your match.

1st issue,
You should use this pattern,
r'([A-Z]{1}[a-z]{1}[A-Z]{1})'
Example
>>> import re
>>> str = 'AbADaDD'
>>> re.findall(r'([A-Z]{1}[a-z]{1}[A-Z]{1})', str)
['AbA', 'DaD']
2nd issue
You should use,
(?=(?<![A-Z])([A-Z]{1}[a-z]{1}[A-Z]{1}[A-Z]{1})(?![A-Z]))
Example
>>> import re
>>> str = 'AbADaDD'
>>> re.findall(r'(?=(?<![A-Z])([A-Z]{1}[a-z]{1}[A-Z]{1}[A-Z]{1})(?![A-Z]))', str)
['AbAD']

How to print regex match results in python 3?

I was in IDLE, and decided to use regex to sort out a string. But when I typed in what the online tutorial told me to, all it would do was print:
<_sre.SRE_Match object at 0x00000000031D7E68>
Full program:
import re
reg = re.compile("[a-z]+8?")
str = "ccc8"
print(reg.match(str))
result:
<_sre.SRE_Match object at 0x00000000031D7ED0>
Could anybody tell me how to actually print the result?

You need to include .group() after to the match function so that it would print the matched string otherwise it shows only whether a match happened or not. To print the chars which are captured by the capturing groups, you need to pass the corresponding group index to the .group() function.
>>> import re
>>> reg = re.compile("[a-z]+8?")
>>> str = "ccc8"
>>> print(reg.match(str).group())
ccc8
Regex with capturing group.
>>> reg = re.compile("([a-z]+)8?")
>>> print(reg.match(str).group(1))
ccc
re.match(pattern, string, flags=0)
If zero or more characters at the beginning of string match the regular expression pattern, return a corresponding MatchObject instance. Return None if the string does not match the pattern; note that this is different from a zero-length match.
Note that even in MULTILINE mode, re.match() will only match at the beginning of the string and not at the beginning of each line.

If you need to get the whole match value, you should use
m = reg.match(r"[a-z]+8?", text)
if m: # Always check if a match occurred to avoid NoneType issues
print(m.group()) # Print the match string
If you need to extract a part of the regex match, you need to use capturing groups in your regular expression. Enclose those patterns with a pair of unescaped parentheses.
To only print captured group results, use Match.groups:
Return a tuple containing all the subgroups of the match, from 1 up to however many groups are in the pattern. The default argument is used for groups that did not participate in the match; it defaults to None.
So, to get ccc and 8 and display only those, you may use
import re
reg = re.compile("([a-z]+)(8?)")
s = "ccc8"
m = reg.match(s)
if m:
print(m.groups()) # => ('ccc', '8')
See the Python demo