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Python: Extract values after decimal using regex

I am given a string which is number example "44.87" or "44.8796". I want to extract everything after decimal (.). I tried to use regex in Python code but was not successful. I am new to Python 3.
import re
s = "44.123"
re.findall(".","44.86")

Something like s.split('.')[1] should work

If you would like to use regex try:
import re
s = "44.123"
regex_pattern = "(?<=\.).*"
matched_string = re.findall(regex_pattern, s)
?<= a negative look behind that returns everything after specified character
\. is an escaped period
.* means "match all items after the period
This online regex tool is a helpful way to test your regex as you build it. You can confirm this solution there! :)

Python Regex to Remove Special Characters from Middle of String and Disregard Anything Else

Using the python re.sub, is there a way I can extract the first alpha numeric characters and disregard the rest form a string that starts with a special character and might have special characters in the middle of the string? For example:
re.sub('[^A-Za-z0-9]','', '#my,name')
How do I just get "my"?
re.sub('[^A-Za-z0-9]','', '#my')
Here I would also want it to just return 'my'.

re.sub(".*?([A-Za-z0-9]+).*", r"\1", str)
The \1 in the replacement is equivalent to matchobj.group(1). In other words it replaces the whole string with just what was matched by the part of the regexp inside the brackets. $ could be added at the end of the regexp for clarity, but it is not necessary because the final .* will be greedy (match as many characters as possible).
This solution does suffer from the problem that if the string doesn't match (which would happen if it contains no alphanumeric characters), then it will simply return the original string. It might be better to attempt a match, then test whether it actually matches, and handle separately the case that it doesn't. Such a solution might look like:
matchobj = re.match(".*?([A-Za-z0-9]+).*", str)
if matchobj:
print(matchobj.group(1))
else:
print("did not match")
But the question called for the use of re.sub.

Instead of re.sub it is easier to do matching using re.search or re.findall.
Using re.search:
>>> s = '#my,name'
>>> res = re.search(r'[a-zA-Z\d]+', s)
>>> if res:
... print (res.group())
...
my
Code Demo

This is not a complete answer. [A-Za-z]+ will give give you ['my','name']
Use this to further explore: https://regex101.com/

Python - Parsing JSON formatted text file with regex

I have a text file formatted like a JSON file however everything is on a single line (could be a MongoDB File). Could someone please point me in the direction of how I could extract values using a Python regex method please?
The text shows up like this:
{"d":{"__type":"WikiFileNodeContent:http:\/\/samplesite.com.‌​au\/ns\/business\/wi‌​ki","author":null,"d‌​escription":null,"fi‌​leAssetId":"034b9317‌​-60d9-45c2-b6d6-0f24‌​b59e1991","filename"‌​:"Reports.pdf"},"cre‌​atedBy":1531,"create‌​dByUsername":"John Cash","icon":"\/Assets10.37.5.0\/pix\/16x16\/page_white_acro‌​bat.png","id":3041,"‌​inheritedPermissions‌​":false,"name":"map"‌​,"permissions":[23,8‌​7,35,49,65],"type":3‌​,"viewLevel":2},{"__‌​type":"WikiNode:http‌​:\/\/samplesite.com.‌​au\/ns\/business\/wi‌​ki","children":[],"c‌​ontent":
I am wanting to get the "fileAssetId" and filename". Ive tried to load the like with Pythons JSON module but I get an error
For the FileAssetid I tried this regex:
regex = re.compile(r"([0-9a-f]{8})\S*-\S*([0-9a-f]{4})\S*-\S*([0-9a-f]{4})\S*-\S*([0-9a-f]{4})\S*-\S*([0-9a-f]{12})")
But i get the following 034b9317‌​, 60d9, 45c2, b6d6, 0f24‌​b59e1991
Im not to sure how to get the data as its displayed.

How about using positive lookahead and lookbehind:
(?<=\"fileAssetId\":\")[a-fA-F0-9-]+?(?=\")
captures the fileAssetId and
(?<=\"filename\":\").+?(?=\")
matches the filename.
For a detailed explanation of the regex have a look at the Regex101-Example. (Note: I combined both in the example with an OR-Operator | to show both matches at once)
To get a list of all matches use re.findall or re.finditer instead of re.match.
re.findall(pattern, string) returns a list of matching strings.
re.finditer(pattern, string) returns an iterator with the objects.

You can use python's walk method and check each entry with re.match.
In case that the string you got is not convertable to a python dict, you can use just regex:
print re.match(r'.*fileAssetId\":\"([^\"]+)\".*', your_pattern).group(1)
Solution for your example:
import re
example_string = '{"d":{"__type":"WikiFileNodeContent:http:\/\/samplesite.com.u\/ns\/business\/wiki","author":null,"description":null,"fileAssetId":"034b9317-60d9-45c2-b6d6-0f24b59e1991","filename":"Reports.pdf"},"createdBy":1531,"createdByUsername":"John Cash","icon":"\/Assets10.37.5.0\/pix\/16x16\/page_white_acrobat.png","id":3041,"inheritedPermissions":false,"name":"map","permissions":[23,87,35,49,65],"type":3,"viewLevel":2},{"__type":"WikiNode:http:\/\/samplesite.com.au\/ns\/business\/wiki","children":[],"content"'
regex_pattern = r'.*fileAssetId\":\"([^\"]+)\".*'
match = re.match(regex_pattern, example_string)
fileAssetId = match.group(1)
print('fileAssetId: {}'.format(fileAssetId))
executing this yields:
34b9317‌​-60d9-45c2-b6d6-0f24‌​b59e1991

Try adding \n to the string that you are entering in to the file (\n means new line)

Based on the idea given here https://stackoverflow.com/a/3845829 and by following the JSON standard https://www.json.org/json-en.html, we can use Python + regex https://pypi.org/project/regex/ and do the following:
json_pattern = (
r'(?(DEFINE)'
r'(?P<whitespace>( |\n|\r|\t)*)'
r'(?P<boolean>true|false)'
r'(?P<number>-?(0|([1-9]\d*))(\.\d*[1-9])?([eE][+-]?\d+)?)'
r'(?P<string>"([^"\\]|\\("|\\|/|b|f|n|r|t|u[0-9a-fA-F]{4}))*")'
r'(?P<array>\[((?&whitespace)|(?&value)(,(?&value))*)\])'
r'(?P<key>(?&whitespace)(?&string)(?&whitespace))'
r'(?P<value>(?&whitespace)((?&boolean)|(?&number)|(?&string)|(?&array)|(? &object)|null)(?&whitespace))'
r'(?P<object>\{((?&whitespace)|(?&key):(?&value)(,(?&key):(?&value))*)\})'
r'(?P<document>(?&object)|(?&array))'
r')'
r'(?&document)'
)
json_regex = regex.compile(json_pattern)
match = json_regex.match(json_document_text)
You can change last line in json_pattern to match not document but individual objects replacing (?&document) by (?&object). I think the regex is easier than I expected, but I did not run extensive tests on this. It works fine for me and I have tested hundreds of files. I wil try to improve my answer in case I find any issue when running it.

Regex in python, repeated fragment finding

I try find in text using regex the elements like this: abs=abs , 1=1 etc.
i wrote this i this way:
opis="Some text abs=abs sfsdvc"
wyn=re.search('([\w]*)=\1',opis)
print(wyn.group(0))
And this find nothing, when i tried this code in the websites like www.regexr.com it was working correctly.
Am I doing something wrong in python re ?

You must specify the regex as raw string r'..'
>>> opis="Some text abs=abs sfsdvc"
>>> wyn=re.search(r'([\w]*)=\1',opis)
>>> print wyn.group(0)
abs=abs
From re documentation
Raw string notation (r"text") keeps regular expressions sane. Without it, every backslash ('\') in a regular expression would have to be prefixed with another one to escape it. For example, the two following lines of code are functionally identical:
Meaning, if you are not planing to use raw string, then all the \ in the string must be escaped as
>>> opis="Some text abs=abs sfsdvc"
>>> wyn=re.search('([\\w]*)=\\1',opis)
>>> print wyn.group(0)
abs=abs

Change your regex to:
re.search(r'(\w+)=\1', opis).group()
↑
Note that you don't really need character class here, the [ and ] are redundant, also it's better to have \w+ if you don't want to match the string "=" (lonely equal sign).

python regex and replace

I am trying to learn python and regex at the same time and I am having some trouble in finding how to match till end of string and make a replacement on the fly.
So, I have a string like so:
ss="this_is_my_awesome_string/mysuperid=687y98jhAlsji"
What I'd want is to first find 687y98jhAlsji (I do not know this content before hand) and then replace it to myreplacedstuff like so:
ss="this_is_my_awesome_string/mysuperid=myreplacedstuff"
Ideally, I'd want to do a regex and replace by first finding the contents after mysuperid= (till the end of string) and then perform a .replace or .sub if this makes sense.
I would appreciate any guidance on this.

You can try this:
re.sub(r'[^=]+$', 'myreplacedstuff', ss)
The idea is to use a character class that exclude the delimiter (here =) and to anchor the pattern with $
explanation:
[^=] is a character class and means all characters that are not =
[^=]+ one or more characters from this class
$ end of the string
Since the regex engine works from the left to the right, only characters that are not an = at the end of the string are matched.

You can use regular expressions:
>>> import re
>>> mymatch = re.search(r'mysuperid=(.*)', ss)
>>> ss.replace(mymatch.group(1), 'replacing_stuff')
'this_is_my_awesome_string/mysuperid=replacing_stuff'
You should probably use #Casimir's answer though. It looks cleaner, and I'm not that good at regex :p.