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Finding correct regex for a bolded/underlined strings (Python) [duplicate]

This question already has answers here:
How to extract the substring between two markers?
(22 answers)
Closed 2 years ago.
So I have 2 sets of criterias that I would like to find in a string. For example:
import re
bold_pattern = re.compile() #pattern for finding all words in between ** **
underline_pattern = re.compile() # pattern for finding all words in between __ __
a = "__Hello__ **This** __is__ **Lego**"
How would I go abouts doing that on regex?

Use capture patterns to capture words between two patterns:
bold_pattern = re.compile(r'\*\*(.*?)\*\*') # pattern for finding all words in between ** **
underline_pattern = re.compile(r'__(.*?)__') # pattern for finding all words in between __ __
Then use them in a re.findall:
bolds = re.findall(bold_pattern, a)
# or: bold_pattern.findall(a)
underlines = re.findall(underline_pattern, a)
# or: underline_pattern.findall(a)

Using re.findall we can try:
a = "__Hello__ **This** __is__ **Lego**"
terms = re.findall(r'\*\*(.*?)\*\*', a)
print(terms)
This prints:
['This', 'Lego']

Hope this helps :) You need to first define the pattern in compile and further use the find all function to extract the string. You can also do it in one line by defining the pattern in findall function as #Tim Biegeleisen suggested.
import re
bold_pattern = re.compile(r'\*\*(.*?)\*\*')
underline_pattern = re.compile(r'\_\_(.*?)\_\_')
a = "__Hello__ **This** __is__ **Lego**"
print(bold_pattern.findall(a))
print(underline_pattern.findall(a))

Suggestion:
If you're dealing with multiline text (i.e. \n), then you'll need to pass the argument: flags=re.DOTALL to your re.findall() method.
Case: Multiline text
# string to be searched
a = """
__Hello__ **This
is a multiline test** __it is__ **Lego
**
"""
# pattern variations
bold_pattern = r'\*\*(.*?)\*\*'
# call re functions
match = re.findall(pattern=bold_pattern, string=a)
flag_match = re.findall(pattern=bold_pattern, string=a, flags=re.DOTALL)
# print results for observation
print(match)
print(flag_match) # using the flag
Returns:
[' __it is__ ']
['This \nis a multiline test', 'Lego\n']
From the Python 3.8.2 documentation:
"The expression’s behaviour can be modified by specifying a flags value."
Dealing with (\n)
Depending on your needs, there are a few different ways you can deal with \n. If I need to, I'll use re.sub() on the entire text body prior to doing anything else to remove them all.
To Compile or Not to Compile?
From the Python 3.8.2 documentation:
"Some of the functions are simplified versions of the full featured methods for compiled regular expressions. Most non-trivial applications always use the compiled form...
...but using re.compile() and saving the resulting regular expression object for reuse is more efficient when the expression will be used several times in a single program."
and
"The compiled versions of the most recent patterns passed to re.compile() and the module-level matching functions are cached, so programs that use only a few regular expressions at a time needn’t worry about compiling regular expressions."
So unless you're using a whole bunch of patterns, you shouldn't see a noticable improvement from compiling.
You can also use the %%time magic command to test both options and see if you notice an advantage locally!
Good luck!

find substrings between two string [duplicate]

This question already has answers here:
Parsing HTML using Python
(7 answers)
Closed 3 years ago.
I have a string like this:
string = r'''<img height="233" src="monline/" title="email example" width="500" ..
title="second example title" width="600"...
title="one more title"...> '''
I am trying to get anything that appears as title (title="Anything here")
I have already tried this but it does not work correctly.
re.findall(r'title=\"(.*)\"',string)

I think your Regex is too Greedy. You can try something like this
re.findall(r'title=\"(?P<title>[\w\s]+)\"', string)
As #Austin and #Plato77 said in the comments, there is a better way to parse HTML in python. See other SO Answers for more context. There are a few common tools for this like:
https://docs.python.org/3/library/html.parser.html
https://www.simplifiedpython.net/parsing-html-in-python/
https://github.com/psf/requests-html / Get html using Python requests?
If you would like to read more on performance testing of different python HTML parsers you can learn more here

As #Austin and #Plato77 said in the comments, there is a better way to parse HTML in python. I stand by this too, but if you want to get it done through regex this may help
c = re.finditer(r'title=[\"]([a-zA-Z0-9\s]+)[\" ]', string)
for i in c:
print(i.group(1))

The problem here is that the next " symbol is parsed as a character and is considered part of the (.*) of your RE. For your usecase, you can use only letters and numbers.

Ignore special characters when creating a regular expression in python [duplicate]

This question already has answers here:
Escaping regex string
(4 answers)
Closed 6 years ago.
Is there a way to ignore special character meaning when creating a regular expression in python? In other words, take the string "as is".
I am writing code that uses internally the expect method from a Telnet object, which only accepts regular expressions. Therefore, the answer cannot be the obvious "use == instead of regular expression".
I tried this
import re
SPECIAL_CHARACTERS = "\\.^$*+?{}[]|():" # backslash must be placed first
def str_to_re(s):
result = s
for c in SPECIAL_CHARACTERS:
result = result.replace(c,'\\'+c)
return re.compile(result)
TEST = "Bob (laughing). Do you know 1/2 equals 2/4 [reference]?"
re_bad = re.compile(TEST)
re_good = str_to_re(TEST)
print re_bad.match(TEST)
print re_good.match(TEST)
It works, since the first one does not recognize the string, and the second one does. I looked at the options in the python documentation, and was not able to find a simpler way. Or are there any cases my solution does not cover (I used python docs to build SPECIAL_CHARACTERS)?
P.S. The problem can apply to other libraries. It does not apply to the pexpect library, because it provides the expect_exact method which solves this problem. However, someone could want to specify a mix of strings (as is) and regular expressions.

If 'reg' is the regex, you gotta use a raw string as follows
pat = re.compile(r'reg')
If reg is a name bound to a regex str, use
reg = re.escape(reg)
pat = re.compile(reg)

Python generate string based on regex format [duplicate]

This question already has answers here:
Reversing a regular expression in Python
(8 answers)
Closed 1 year ago.
I have some difficulties learning regex in python. I want to parse my tornado web route configuration along with arguments into a request path string without handlers request.path method.
For example, I have route with patterns like:
/entities/([0-9]+)
/product/([0-9]+/actions
The expected result combine with integer parameter (123) will be a string like:
/entities/123
/product/123/actions
How do I generate string based on that pattern?
Thank you very much in advance!

This might be a possible duplicate to:
Reversing a regular expression in Python
Generate a String that matches a RegEx in Python
Using the answer provided by #bjmc a solution works like this:
>>> import rstr
>>> intermediate = rstr.xeger(\d+)
>>> path = '/product/' + intermediate + '/actions'
Depending on how long you want your intermediate integer, you could replace the regex: \d{1,3}

python regex on variable

Please help with my regex problem
Here is my string
source="http://www.amazon.com/ref=s9_hps_bw_g200_t2?pf_rd_m=ATVPDKIKX0DER&pf_rd_i=3421"
source_resource="pf_rd_m=ATVPDKIKX0DER"
The source_resource is in the source may end with & or with .[for example].
So far,
regex = re.compile("pf_rd_m=ATVPDKIKX0DER+[&.]")
regex.findall(source)
[u'pf_rd_m=ATVPDKIKX0DER&']
I have used the text here. Rather using text, how can i use source_resource variable with & or . to find this out.

If the goal is to extract the pf_rd_m value (which it apparently is as you are using regex.findall), than I'm not sure regex are the easiest solution here:
>>> import urlparse
>>> qs = urlparse.urlparse(source).query
>>> urlparse.parse_qs(qs)
{'pf_rd_m': ['ATVPDKIKX0DER'], 'pf_rd_i': ['3421']}
>>> urlparse.parse_qs(qs)['pf_rd_m']
['ATVPDKIKX0DER']

You also have to escape the .
pattern=re.compile(source_resource + '[&\.]')

You can just build the string for the regular expression like a normal string, utilizing all string-formatting options available in Python:
import re
source_and="http://rads.stackoverflow.com/amzn/click/B0030DI8NA/pf_rd_m=ATVPDKIKX0DER&"
source_dot="http://rads.stackoverflow.com/amzn/click/B0030DI8NA/pf_rd_m=ATVPDKIKX0DER."
source_resource="pf_rd_m=ATVPDKIKX0DER"
regex_string = source_resource + "[&\.]"
regex = re.compile(regex_string)
print regex.findall(source_and)
print regex.findall(source_dot)
>>> ['pf_rd_m=ATVPDKIKX0DER&']
['pf_rd_m=ATVPDKIKX0DER.']
I hope this is what you mean.
Just take note that I modified your regular expression: the . is a special symbol and needs to be escaped, as is the + (I just assumed the string will only occur once, which makes the use of + unnecessary).