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How to use regular expression to remove all math expression in latex file

Suppose I have a string which consists of a part of latex file. How can I use python re module to remove any math expression in it?
e.g:
text="This is an example $$a \text{$a$}$$. How to remove it? Another random math expression $\mathbb{R}$..."
I would like my function to return ans="This is an example . How to remove it? Another random math expression ...".
Thank you!

Try this Regex:
(\$+)(?:(?!\1)[\s\S])*\1
Click for Demo
Code
Explanation:
(\$+) - matches 1+ occurrences of $ and captures it in Group 1
(?:(?!\1)[\s\S])* - matches 0+ occurrences of any character that does not start with what was captured in Group 1
\1 - matches the contents of Group 1 again
Replace each match with a blank string.
As suggested by #torek, we should not match 3 or more consecutive $, hence changing the expression to (\${1,2})(?:(?!\1)[\s\S])*\1

It's commonly said that regular expressions cannot count, which is kind of a loose way of describing a problem more formally discussed in Count parentheses with regular expression. See that for what this means.
Now, with that in mind, note that LaTeX math expressions can include nested sub-equations, which can include further nested sub-equations, and so on. This is analogous to the problem of detecting whether a closing parenthesis closes an inner parenthesized expression (as in (for instance) this example, where the first one does not) or an outer parenthesis. Therefore, regular expressions are not going to be powerful enough to handle the full general case.
If you're willing to do a less-than-complete job, you can construct a regular expression that finds $...$ and $$...$$. You will need to pay attention to the particular regular expression language available. Python's is essentially the same as Perl's here.
Importantly, these $-matchers will completely miss \begin{equation} ... \end{equation}, \begin{eqnarray} ... \end{eqnarray}, and so on. We've already noted that handling LaTeX expression parsing with a mere regular expression recognizer is inadequate, so if you want to do a good job—while ignoring the complexity of lower-level TeX manipulation of token types, where one can change any individual character's category code —you will want a more general parser. You can then tokenize \begin, {, }, and words, and match up the begin/end pairs. You can also tokenize $ and $$ and match those up. Since parsers can count, in exactly the way that regular expressions can't, you can do a much better job this way.

regex- capturing text between matches

In the following text, I try to match a number followed by ")" and number followed by a period. I am trying to retrieve the text between the matches.
Example:
"1) there is a dsfsdfsd and 2) there is another one and 3) yet another
case"
so I am trying to output: ["there is a dsfsdfsd and", "there is another one and", yet another case"]
I've used this regex: (?:\d)|\d.)
Adding a .* at the end matches the entire string, I only want it to match the words between
also in this string:
"we will give 4. there needs to be another option and 6.99 USD is a
bit amount"
I want to only match the 4. and not the 6.99
Any pointers will be appreciated. Thank you. r

tldr
Regular expressions are tricky beasts and you should avoid them if at all possible.
If you can't avoid them, then make sure you have lots of test cases for all the edge cases that can occur.
Build up your regular expression slowly and systematically, testing your assumptions at every step.
If this code will go intro production, then please write unit tests that explain the thinking process to the poor soul who has to maintain it one day
The long version
Regular expressions are finicky. Your best approach may be to solve the problem a different way.
For example, your language might have a library function that allows you to split up strings using a regular expression to define what comes between the numbers. That will let you get away with writing a simpler regex to match the numbers and brackets/dots.
If you still decide to use regular expressions, then you need to be very structured about how you build up your regular expressions. It's extremely easy to miss edge cases.
So let's break this down piece by piece...
Set up a test environment for quickly experimenting with your regex.
There are lots of options here, depending on your programming language and OS. Ones I sometimes use are:
a Powershell window for testing .Net regexes (NB: the cli gives you a history of past attempts, so you can go back a few steps if you mess things up too badly)
a Python console for testing Python regexes (which are slightly different to .Net regexes in their syntax for named capture groups).
an html page with JavaScript to test the regex
an online or desktop regex tool (I still use the ancient Regular Expression Workbench from Eric Gunnerson, but I'm sure there are better alternatives these days)
Since you didn't specify a language or regex version, I'll assume .Net regular expressions
Create a single test string for testing a wider variety of options.
Your goal is to include as many edge cases as you can think of. Here's what I would use: "ab 1. there is a dsfsdfsd costing $6.99 and 2) there is another one and 3. yet another case 4)5) 6)10."
Note that I've added a few extra cases you didn't mention:
empty strings between two round bracket numbers: "4)" and "5)"
white space string between two round bracket numbers: "5)" and "6)"
empty strings between a round bracket number and a dotted number: "6)" and "10."
empty string after the dotted number "10." at the end of the string
random text and empty space, which should be ignored, before the first number
I'm going to make a few assumptions here, which you will need to vary based on your actual requirements:
You DO want to capture the white space after the dot or round bracket.
You DO want to capture the white space before the next dotted number or round bracket number.
You might have numbers that go beyond 9, so I've included "10" in the test cases.
You want to capture empty strings at the end e.g. after the "10."
NOTES:
Thinking through this test case forces you to be more rigorous about your requirements.
It will also help you be more efficient while you are manually testing your regular expression.
HOWEVER, this is assuming you aren't following a TDD approach. If you are, then you should probably do things a little differently... create unit tests for each scenario separately and get the regex working incrementally.
This test string doesn't cover all cases. For example, there are no newline or tab characters in the test string. Also it can't test for an empty string following a round bracket number at the very end.
First get a regex working that just captures the round brackets and dotted brackets.
Don't worry about the $6.99 edge case yet.
Drop the "(?:" non-capturing group syntax from your regex for now: "\d)|\d."
This doesn't even parse, because you have an unescaped round bracket.
The revised string is "\d\)|\d.", which parses, but which also matches "99" which you probably weren't expecting. That's because you forgot to escape the "."
The revised string is "\d\)|\d\.". This no longer matches "99", but it now matches "0." at the end instead of "10.". That's because it assumes that numbers will be single digit only.
The following string seems to work: "\d+\)|\d+\."
Time to deal with that pesky "$6.99" now...
Modify the regex so that it doesn't capture a floating point number.
You need to use a negative look ahead pattern to prevent a digit being after the decimal point.
Result: "\d+\)|\d+\.(?!\d)"
Count how many matches this produces. You're going to use this number for checking later results.
Hint: Save the regex pattern somewhere. You want to be able to go back to it any time you mess up your regex pattern beyond repair.
If you found a string splitting function, then you should use it now and avoid the complexity that follows. [I've included an example of this at the end.]
Simple is better, but I'm going to continue with the longer solution in the interests of showing an approach to staying in control of regex'es that start getting horribly complicated
Decide how to exclude that pattern
You used the non-capture group pattern in your question i.e. "(?:"
That approach can work. But it's a bit cumbersome, because you need to have a capturing group after it that you will look for instead.
It would be much nicer if your entire pattern matched what you are looking for.
So wrap the number pattern inside a zero-width positive look behind pattern (if your language supports it) i.e. "(?<=".
This checks for the pattern, but doesn't include it in what gets captured.
So now your regex looks like this: "(?<=\d+\)|\d+\.(?!\d))"
Test it!
It might seem silly to test this on its own - all the matches are empty strings.
Do it anyway. You want to sanity check every step of the way.
Make sure that it still produces the same number of matches as in step 4.
Decide how to match the text in between the numbers.
You rightly mention that ".*" will match the entire string, not just the parts in between.
There's a neat trick that allows you to reuse the pattern from step 5 to get the text in between.
Start by just matching the next character
The trick is that you want to match any character unless it's the start of the next number
That sounds like a negative look ahead pattern again: "(?!"
Let X be the pattern you saved in step 4. Matching a single character will look like this: "(?!X)."
You want to match lots of those characters. So put that pattern into a non-capturing group and repeat it: "(?:(?!X).)*"
This assumes you want to capture empty text.
If you're not, then change the "*" to a "+".
Hint: This is such a common pattern that you will want to reuse it in future pasting in different patterns in place of X
I used a non-capturing group instead of a normal group so that you can also embed this pattern in regexes where you do care about the capturing groups
Resulting pattern: "(?:(?!\d+\)|\d+\.(?!\d)).)*"
I suggest testing this pattern on its own to see what it does
Now put parts 5 and 7 together: "(?<=\d+\)|\d+\.(?!\d))(?:(?!\d+\)|\d+\.(?!\d)).)*"
Test it!
Unit tests!
If this is going into production, then please write lots of unit tests that will explain each step of this thought process
Have pity on the poor soul who has to maintain your regex in future!
By rights that person should be you
I suggest putting a note in your calendar to return to this code in 6 months' time and make sure you can still understand it from the unit tests alone!
Refactor
In six months' time, if you can't understand the code any more, use your newfound insight (and incentive) to solve the problem without using regular expressions (or only very simple ones)
Addendum
As an example of using a string splitting function to get away with a simpler regex, here's a solution in Powershell:
$string = 'ab 1. there is a dsfsdfsd costing $6.99 and 2) there is another one and 3. yet another case 4)5) 6)10.'
$pattern = [regex] '\d+\)|\d+\.(?!\d)'
$string -split $pattern | select-object -skip 1

Judging by the task you have, it might be easier to match the delimiters and use re.split (as also pointed out by bobblebubble in the comments).
I dsuggest a mere
\d+[.)]\B\s*
See it in action (demo)
It matches 1 or more digits, then a . or a ), then it makes sure there is no word letter (digit, letter or underscore) after it and then matches zero or more whitespace.
Python demo:
import re
rx = r'\d+[.)]\B\s*'
test_str = "1) there is a dsfsdfsd and 2) there is another one and 3) yet another case\n\"we will give 4. there needs to be another option and 6.99 USD is a bit amount"
print([x for x in re.split(rx,test_str) if x])

Try the following regex with the g modifier:
([A-Za-z\s\-_]+|\d(?!(\)|\.)\D)|\.\d)
Example: https://regex101.com/r/kB1xI0/3
[A-Za-z\s\-_]+ automatically matches all alphabetical characters + whitespace
\d(?!(\)|\.)\D) match any numeric sequence of digits not followed by a closing parenthesis ) or decimal value (.99)
\.\d match any period followed by numeric digit.

I used this pattern:
(?<=\d.\s)(.*?)(?=\d.\s)
demo
This looks for the contents between any digit, any character, then a space.
Edit: Updated pattern to handle the currency issue and line ends better:
This is with flag 'g'
(?<=[0-9].\s)(.*?)(?=\s[0-9].\s|\n|\r)
Demo 2

import re
s = "1) there is a dsfsdfsd and 2) there is another one and 3) yet another case"
s1 = "we will give 4. there needs to be another option and 6.99 USD is a bit amount"
regex = re.compile("\d\)\s.*?|\s\d\.\D.*?")
print ([x for x in regex.split(s) if x])
print regex.split(s1)
Output:
['there is a dsfsdfsd and ', 'there is another one and ', 'yet another case']
['we will give', 'there needs to be another option and 6.99 USD is a bit amount']

parsing string with specific name in python

i have string like this
<name:john student male age=23 subject=\computer\sience_{20092973}>
i am confused ":","="
i want to parsing this string!
so i want to split to list like this
name:john
job:student
sex:male
age:23
subject:{20092973}
parsing string with specific name(name, job, sex.. etc) in python
i already searching... but i can't find.. sorry..
how can i this?
thank you.

It's generally a good idea to give more than one example of the strings you're trying to parse. But I'll take a guess. It looks like your format is pretty simple, and primarily whitespace-separated. It's simple enough that using regular expressions should work, like this, where line_to_parse is the string you want to parse:
import re
matchval = re.match("<name:(\S+)\s+(\S+)\s+(\S+)\s+age=(\S+)\s+subject=[^\{]*(\{\S+\})", line_to_parse)
matchgroups = matchval.groups()
Now matchgroups will be a tuple of the values you want. It should be trivial for you to take those and get them into the desired format.
If you want to do many of these, it may be worth compiling the regular expression; take a look at the re documentation for more on this.
As for the way the expression works: I won't go into regular expressions in general (that's what the re docs are for) but in this case, we want to get a bunch of strings that don't have any whitespace in them, and have whitespace between them, and we want to do something odd with the subject, ignoring all the text except the part between { and }.
Each "(...)" in the expression saves whatever is inside it as a group. Each "\S+" stands for one or more ("+") characters that aren't whitespace ("\S"), so "(\S+)" will match and save a string of length at least one that has no whitespace in it. Each "\s+" does the opposite: it has not parentheses around it, so it doesn't save what it matches, and it matches at one or more ("+") whitespace characters ("\s"). This suffices for most of what we want. At the end, though, we need to deal with the subject. "[...]" allows us to list multiple types of characters. "[^...]" is special, and matches anything that isn't in there. {, like [, (, and so on, needs to be escaped to be normal in the string, so we escape it with \, and in the end, that means "[^{]*" matches zero or more ("*") characters that aren't "{" ("[^{]"). Since "*" and "+" are "greedy", and will try to match as much as they can and still have the expression match, we now only need to deal with the last part. From what I've talked about before, it should be pretty clear what "({\S+})" does.

Python regex: how to match anything up to a specific string and avoid backtraking when failin

I'm trying to craft a regex able to match anything up to a specific pattern. The regex then will continue looking for other patterns until the end of the string, but in some cases the pattern will not be present and the match will fail. Right now I'm stuck at:
.*?PATTERN
The problem is that, in cases where the string is not present, this takes too much time due to backtraking. In order to shorten this, I tried mimicking atomic grouping using positive lookahead as explained in this thread (btw, I'm using re module in python-2.7):
Do Python regular expressions have an equivalent to Ruby's atomic grouping?
So I wrote:
(?=(?P<aux1>.*?))(?P=aux1)PATTERN
Of course, this is faster than the previous version when STRING is not present but trouble is, it doesn't match STRING anymore as the . matches everyhing to the end of the string and the previous states are discarded after the lookahead.
So the question is, is there a way to do a match like .*?STRING and alse be able to fail faster when the match is not present?

You could try using split
If the results are of length 1 you got no match. If you get two or more you know that the first one is the first match. If you limit the split to size one you'll short-circuit the later matching:
"HI THERE THEO".split("TH", 1) # ['HI ', 'ERE THEO']
The first element of the results is up to the match.

One-Regex Solution
^(?=(?P<aux1>(?:[^P]|P(?!ATTERN))*))(?P=aux1)PATTERN
Explanation
You wanted to use the atomic grouping like this: (?>.*?)PATTERN, right? This won't work. Problem is, you can't use lazy quantifiers at the end of an atomic grouping: the definition of the AG is that once you're outside of it, the regex won't backtrack inside.
So the regex engine will match the .*?, because of the laziness it will step outside of the group to check if the next character is a P, and if it's not it won't be able to backtrack inside the group to match that next character inside the .*.
What's usually used in Perl are structures like this: (?>(?:[^P]|P(?!ATTERN))*)PATTERN. That way, the equivalent of .* (here (?:[^P]|P(?!ATTERN))) won't "eat up" the wanted pattern.
This pattern is easier to read in my opinion with possessive quantifiers, which are made just for these occasions: (?:[^P]|P(?!ATTERN))*+PATTERN.
Translated with your workaround, this would lead to the above regex (added ^ since you should anchor the regex, either to the start of the string or to another regex).

The Python documentation includes a brief outline of the differences between the re.search() and re.match() functions http://docs.python.org/2/library/re.html#search-vs-match. In particular, the following quote is relevant:
Sometimes you’ll be tempted to keep using re.match(), and just add .* to the front of your RE. Resist this temptation and use re.search() instead. The regular expression compiler does some analysis of REs in order to speed up the process of looking for a match. One such analysis figures out what the first character of a match must be; for example, a pattern starting with Crow must match starting with a 'C'. The analysis lets the engine quickly scan through the string looking for the starting character, only trying the full match if a 'C' is found.
Adding .* defeats this optimization, requiring scanning to the end of the string and then backtracking to find a match for the rest of the RE. Use re.search() instead.
In your case, it would be preferable to define your pattern simply as:
pattern = re.compile("PATTERN")
And then call pattern.search(...), which will not backtrack when the pattern is not found.

Small problem with reg exps in python

So I have one variable that has all the code from some file. I need to remove all comments from this file. One of my regexp lines is this
x=re.sub('\/\*.*\*\/','',x,re.M,re.S);
What I want this to be doing is to remove all multi line comments. For some odd reason though, its skipping two instances of */, and removing everything up to the third instance of */.
I'm pretty sure the reason is this third instance of */ has code after it, while the first two are by themselves on the line. I'm not sure why this matters, but I'm pretty sure thats why.
Any ideas?

.* will always match as many characters as possible. Try (.*?) - most implementations should try to match as few characters as possible then (should work without the brackets but not sure right now). So your whole pattern should look like this: \/\*.*?\*\/ or \/\*(.*?)\*\/

The expression .* is greedy, meaning that it will attempt to match as many characters as possible. Instead, use (.*?) which will stop matching characters as soon as possible.

The regular expression is "greedy" and when presented with several stopping points will take the farthest one. Regex has some patterns to help control this, in particular the
(?&gt!...)
which matches the following expression only if it is Not preceeded by a match of the pattern in parens. (put in a pointy brace for &gt in the above - I don't know the forum convention for getting on in my answer).
(?*...) was not in Python 2.4 but is a good choice if you are using a later version.