Beautiful soup if class not like "string" or regex - python

I know that beautiful soup has a function to match classes based on regex that contains certain strings, based on a post here. Below is a code example from that post:
regex = re.compile('.*listing-col-.*')
for EachPart in soup.find_all("div", {"class" : regex}):
print EachPart.get_text()
Now, is it possible to do the opposite? Basically, find classes that do not contain a certain regex. In SQL language, it's like:
where class not like '%test%'
Thanks in advance!

This actually can be done by using Negative Lookahead
Negative Lookahead has the following syntax (?!«pattern») and matches if pattern does not match what comes before the current location in the input string.
In your case, you could use the following regex to match all classes that don’t contain listing-col- in their name:
regex = re.compile('^((?!listing-col-).)*$')
Here’s the pretty simple and straightforward explanation of this regex ^((?!listing-col-).)*$:
^ asserts position at start of a line
Capturing Group ((?!listing-col-).)*
* matches the previous token between zero and unlimited times, as many times as possible, giving back as needed
Negative Lookahead (?!listing-col-).
Assert that the Regex below does not match.
listing-col- matches the characters listing-col- literally (case sensitive)
. matches any character
$ asserts position at the end of a line
Also, you may find the https://regex101.com site useful
It will help you test your patterns and show you a detailed explanation of each step. It's your best friend in writing regular expressions.

One possible solution is utilizing regex directly.
You can refer to Regular expression to match a line that doesn't contain a word.
Or you can introduce a function to implement the logic and pass it to find_all as a parameter.
You can refer to https://beautiful-soup-4.readthedocs.io/en/latest/index.html?highlight=find_all#find-all

You can use css selector syntax with :not() pseudo class and * contains operator
data = [i.text() for i in soup.select('div[class]:not([class*="listing-col-"])')]

Related

Regex Statement to only match parts of a string for comparison - Python

What I am trying to do is match values from one file to another, but I only need to match the first portion of the string and the last portion.
I am reading each file into a list, and manipulating these based on different Regex patterns I have created. Everything works, except when it comes to these type of values:
V-1\ZDS\R\EMBO-20-1:24
V-1\ZDS\R\EMBO-20-6:24
In this example, I only want to match 'V-1\ZDS\R\EMBO-20' and then compare the '24' value at the end of the string. The number x in '20-x:', can vary and doesn't matter in terms of comparisons, as long as the first and last parts of this string match.
This is the Regex I am using:
re.compile(r"(?:.*V-1\\ZDS\\R\\EMBO-20-\d.*)(:\d*\w.*)")
Once I filter down the list, I use the following function to return the difference between the two sets:
funcDiff = lambda x, y: list((set(x)- set(y))) + list((set(y)- set(x)))
Is there a way to take the list of differences and filter out the ones that have matching values after the
:
as mentioned above?
I apologize is this is an obvious answer, I'm new to Python and Regex!
The output I get is the differences between the entire strings, so even if the first and last part of the string match, if the number following the 'EMBO-20-x' doesn't also match, it returns it as being different.
Before discussing your question, regex101 is an incredibly useful tool for this type of thing.
Your issue stems from two issues:
1.) The way you used .*
2.) Greedy vs. Nongreedy matches
.* kinda sucks
.* is a regex expression that is very rarely what you actually want.
As a quick aside, a useful regex expression is [^c]* or [^c]+. These expressions match any character except the letter c, with the first expression matching 0 or more, and the second matched 1 or more.
.* will match all characters as many times as it can. Instead, try to start your regex patterns with more concrete starting points. Two good ways to do this are lookbehind expressions and anchors.
Another quick aside, it's likely that you are misusing regex.match and regex.find. match will only return a match that begins at the start of the string, while find will return matches anywhere in the input string. This could be the reason you included the .* in the first place, to allow a .match call to return a match deeper in the string.
Lookbehind Expressions
There are more complete explanations online, but in short, regex patterns like:
(?<=test)foo
will match the text foo, but only if test is right in front of it. To be more clear, the following strings will not match that regex:
foo
test-foo
test foo
but the following string will match:
testfoo
This will only match the text foo, though.
Anchors
Another option is anchors. ^ and $ are special characters, matching the start and end of a line of text. If you know your regex pattern will match exactly one line of text, start it with ^ and end it with $.
Leading patterns with .* and ending with .* are likely the source of your issue. Although you did not include full examples of your input or your code, you likely used match as opposed to find.
In regex, . matches any character, and * means 0 or more times. This means that for any input, your pattern will match the entire string.
Greedy vs. Non-Greedy qualifiers
The second issue is related to greediness. When your regex patterns have a * in them, they can match 0 or more characters. This can hide problems, as entire * expressions can be skipped. Your regex is likely matched several lines of text as one match, and hiding multiple records in a single .*.
The Actual Answer
Taking all of this in to consideration, let's assume that your input data looks like this:
V-1\ZDS\R\EMBO-20-1:24
V-1\ZDS\R\EMBO-20-6:24
V-1\ZDS\R\EMBO-20-3:93
V-1\ZDS\R\EMBO-20-6:22309
V-1\ZDS\R\EMBO-20-8:2238
V-1\ZDS\R\EMBO-20-3:28
A better regular expression would be:
^V-1\\ZDS\\R\\EMBO-20-\d:(\d+)$
To visualize this regex in action, follow this link.
There are several differences I would like to highlight:
Starting the expression with ^ and ending with $. This forces the regex to match exactly one line. Even though the pattern works without these characters, it's good practice when working with regex to be as explicit as possible.
No useless non-capturing group. Your example had a (?:) group at the start. This denotes a group that does not capture it's match. It's useful if you want to match a subpattern multiple times ((?:ab){5} matches ababababab without capturing anything). However, in your example, it did nothing :)
Only capturing the number. This makes it easier to extract the value of the capture groups.
No use of *, one use of +. + works like *, but it matches 1 or more. This is often more correct, as it prevents 'skipping' entire characters.

Python regex to exclude several words

I try to search for URLS and want to exclude some. In the variable download_artist I stored the base URL and wanto to find additional links, but not upload, favorites, followers or listens.
So I tried different versions with the mentioned words and a |. Like:
urls = re.findall(rf'^{download_artist}uploads/|{download_artist}^favorites/|^{download_artist}followers/|^{download_artist}listens/|{download_artist}\S+"', response.text, re.IGNORECASE)
or:
urls = re.findall(rf'{download_artist}^uploads/|^favorites/|^followers/|^listens/|\S+"', response.text, re.IGNORECASE)
But it ignores my ^ for excluding the words. Where is my mistake?
You need use "lookaround" in this case, can see more details in https://www.regular-expressions.info/lookaround.html.
So, i think wich this regex solve your problem:
{download_artist}(?!uploads/|favorites/|followers/|listens/)\S+\"
You can test if regex working in https://regex101.com/. This site is very useful when you work with regex.
^ only works as a negation in character classes inside [], outside it represents the beginning of the input.
I suggest you do two matches: One to match all urls and another one to match the ones to exclude. Then remove the second set of urls from the first one.
That will keep the regexes simple and readable.
If you have to do it in one regex for whatever reason you can try to solve it with (negative) lookaround pattern (see https://www.rexegg.com/regex-lookarounds.html).

REGEX: Negative lookbehind with multiple whitespaces [duplicate]

I am trying to use lookbehinds in a regular expression and it doesn't seem to work as I expected. So, this is not my real usage, but to simplify I will put an example. Imagine I want to match "example" on a string that says "this is an example". So, according to my understanding of lookbehinds this should work:
(?<=this\sis\san\s*?)example
What this should do is find "this is an", then space characters and finally match the word "example". Now, it doesn't work and I don't understand why, is it impossible to use '+' or '*' inside lookbehinds?
I also tried those two and they work correctly, but don't fulfill my needs:
(?<=this\sis\san\s)example
this\sis\san\s*?example
I am using this site to test my regular expressions: http://gskinner.com/RegExr/
Many regular expression libraries do only allow strict expressions to be used in look behind assertions like:
only match strings of the same fixed length: (?<=foo|bar|\s,\s) (three characters each)
only match strings of fixed lengths: (?<=foobar|\r\n) (each branch with fixed length)
only match strings with a upper bound length: (?<=\s{,4}) (up to four repetitions)
The reason for these limitations are mainly because those libraries can’t process regular expressions backwards at all or only a limited subset.
Another reason could be to avoid authors to build too complex regular expressions that are heavy to process as they have a so called pathological behavior (see also ReDoS).
See also section about limitations of look-behind assertions on Regular-Expressions.info.
Hey if your not using python variable look behind assertion you can trick the regex engine by escaping the match and starting over by using \K.
This site explains it well .. http://www.phpfreaks.com/blog/pcre-regex-spotlight-k ..
But pretty much when you have an expression that you match and you want to get everything behind it using \K will force it to start over again...
Example:
string = '<a this is a tag> with some information <div this is another tag > LOOK FOR ME </div>'
matching /(\<a).+?(\<div).+?(\>)\K.+?(?=\<div)/ will cause the regex to restart after you match the ending div tag so the regex won't include that in the result. The (?=\div) will make the engine get everything in front of ending div tag
What Amber said is true, but you can work around it with another approach: A non-capturing parentheses group
(?<=this\sis\san)(?:\s*)example
That make it a fixed length look behind, so it should work.
You can use sub-expressions.
(this\sis\san\s*?)(example)
So to retrieve group 2, "example", $2 for regex, or \2 if you're using a format string (like for python's re.sub)
Most regex engines don't support variable-length expressions for lookbehind assertions.

heavy regex - really time consuming

I have the following regex to detect start and end script tags in the html file:
<script(?:[^<]+|<(?:[^/]|/(?:[^s])))*>(?:[^<]+|<(?:[^/]|/(?:[^s]))*)</script>
meaning in short it will catch: <script "NOT THIS</s" > "NOT THIS</s" </script>
it works but needs really long time to detect <script>,
even minutes or hours for long strings
The lite version works perfectly even for long string:
<script[^<]*>[^<]*</script>
however, the extended pattern I use as well for other tags like <a> where < and > are possible to appears also as values of attributes.
python test:
import re
pattern = re.compile('<script(?:[^<]+|<(?:[^/]|/(?:[^s])))*>(?:[^<]+|<(?:[^/]|/(?:^s]))*)</script>', re.I + re.DOTALL)
re.search(pattern, '11<script type="text/javascript"> easy>example</script>22').group()
re.search(pattern, '<script type="text/javascript">' + ('hard example' * 50) + '</script>').group()
how can I fix it?
The inner part of regex (after <script>) should be changed and simplified.
PS :) Anticipate your answers about the wrong approach like using regex in html parsing,
I know very well many html/xml parsers, and what I can expect in often broken html code, and regex is really useful here.
comment:
well, I need to handle:
each <a < document like this.border="5px;">
and approach is to use parsers and regex together
BeautifulSoup is only 2k lines, which not handling every html and just extends regex from sgmllib.
and the main reason is that I must know exact the position where every tag starts and stop. and every broken html must be handled.
BS is not perfect, sometimes happens:
BeautifulSoup('< scriPt\n\n>a<aa>s< /script>').findAll('script') == []
#Cylian:
atomic grouping as you know is not available in python's re.
so non-geedy everything .*? until <\s/\stag\s*>** is a winner at this time.
I know that is not perfect in that case:
re.search('<\sscript.?<\s*/\sscript\s>','< script </script> shit </script>').group()
but I can handle refused tail in the next parsing.
It's pretty obvious that html parsing with regex is not one battle figthing.
Use an HTML parser like beautifulsoup.
See the great answers for "Can I remove script tags with beautifulsoup?".
If your only tool is a hammer, every problem starts looking like a nail. Regular expressions are a powerful hammer but not always the best solution for some problems.
I guess you want to remove scripts from HTML posted by users for security reasons. If security is the main concern, regular expressions are hard to implement because there are so many things a hacker can modify to fool your regex, yet most browsers will happily evaluate... An specialized parser is easier to use, performs better and is safer.
If you are still thinking "why can't I use regex", read this answer pointed by mayhewr's comment. I could not put it better, the guy nailed it, and his 4433 upvotes are well deserved.
I don't know python, but I know regular expressions:
if you use the greedy/non-greedy operators you get a much simpler regex:
<script.*?>.*?</script>
This is assuming there are no nested scripts.
The problem in pattern is that it is backtracking. Using atomic groups this issue could be solved. Change your pattern to this**
<script(?>[^<]+?|<(?:[^/]|/(?:[^s])))*>(?>[^<]+|<(?:[^/]|/(?:[^s]))*)</script>
^^^^^ ^^^^^
Explanation
<!--
<script(?>[^<]+?|<(?:[^/]|/(?:[^s])))*>(?>[^<]+|<(?:[^/]|/(?:[^s]))*)</script>
Match the characters “<script” literally «<script»
Python does not support atomic grouping «(?>[^<]+?|<(?:[^/]|/(?:[^s])))*»
Match either the regular expression below (attempting the next alternative only if this one fails) «[^<]+?»
Match any character that is NOT a “<” «[^<]+?»
Between one and unlimited times, as few times as possible, expanding as needed (lazy) «+?»
Or match regular expression number 2 below (the entire group fails if this one fails to match) «<(?:[^/]|/(?:[^s]))»
Match the character “<” literally «<»
Match the regular expression below «(?:[^/]|/(?:[^s]))»
Match either the regular expression below (attempting the next alternative only if this one fails) «[^/]»
Match any character that is NOT a “/” «[^/]»
Or match regular expression number 2 below (the entire group fails if this one fails to match) «/(?:[^s])»
Match the character “/” literally «/»
Match the regular expression below «(?:[^s])»
Match any character that is NOT a “s” «[^s]»
Match the character “>” literally «>»
Python does not support atomic grouping «(?>[^<]+|<(?:[^/]|/(?:[^s]))*)»
Match either the regular expression below (attempting the next alternative only if this one fails) «[^<]+»
Match any character that is NOT a “<” «[^<]+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
Or match regular expression number 2 below (the entire group fails if this one fails to match) «<(?:[^/]|/(?:[^s]))*»
Match the character “<” literally «<»
Match the regular expression below «(?:[^/]|/(?:[^s]))*»
Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
Match either the regular expression below (attempting the next alternative only if this one fails) «[^/]»
Match any character that is NOT a “/” «[^/]»
Or match regular expression number 2 below (the entire group fails if this one fails to match) «/(?:[^s])»
Match the character “/” literally «/»
Match the regular expression below «(?:[^s])»
Match any character that is NOT a “s” «[^s]»
Match the characters “</script>” literally «</script>»
-->

How can I match any substring except a particular one in python

I want to write a regular expression that will match the following string
a (any substring except 'ABC') ABC
An example for this would be a pqrs h js ABC
The tricky part is to match any substring except 'ABC'. Since the document in which I am searching for, can contain multiple lines that contain such pattern and I want to find all the lines separately I can't use the following expression
a.*ABC
because this would just give me the line where the first a is found extending uptill where the last 'ABC' is found in the document.
There is this answer which says I can use look ahead negation but that is not working in python, or maybe in my case because there is substring before and I have not tested simply using that expression because it will not serve my purpose
Use the non greedy quantifier i.e ?
^a.*?ABC

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