I have a string like hello /world today/
I need to replace /world today/ with /MY NEW STRING/
Reading the manual I have found
newString = string.match("hello /world today/","%b//")
which I can use with gsub to replace, but I wondered is there also an elegant way to return just the text between the /, I know I could just trim it, but I wondered if there was a pattern.
Try something like one of the following:
slashed_text = string.match("hello /world today/", "/([^/]*)/")
slashed_text = string.match("hello /world today/", "/(.-)/")
slashed_text = string.match("hello /world today/", "/(.*)/")
This works because string.match returns any captures from the pattern, or the entire matched text if there are no captures. The key then is to make sure that the pattern has the right amount of greediness, remembering that Lua patterns are not a complete regular expression language.
The first two should match the same texts. In the first, I've expressly required that the pattern match as many non-slashes as possible. The second (thanks lhf) matches the shortest span of any characters at all followed by a slash. The third is greedier, it matches the longest span of characters that can still be followed by a slash.
The %b// in the original question doesn't have any advantages over /.-/ since the the two delimiters are the same character.
Edit: Added a pattern suggested by lhf, and more explanations.
Related
I have an html text. With my regex:
r'(http[\S]?://[\S]+/favicon\.ico[\S^,]+)"'
and with re.findall(), I get this result from it:
['https://cdn.sstatic.net/Sites/stackoverflow/Img/favicon.ico?v=ec617d715196', 'https://stackoverflow.com/favicon.ico,https://cdn.sstatic.net/Sites/stackoverflow/Img/favicon.ico?v=ec617d715196']
But i dont want this second result in list, i understand that it has coma inside, but i have no idea how to exclude coma from my regex. I use re.findall() in order to find necessery link in any place in html text because i dont know where it could be.
Note that [\S]+ contains redundant character class, it is the same as \S+. In http[\S]?://, [\S]? is most likely a human error, as [\S]? matches any optional non-whitespace char. I doubt you implied to match http§:// protocol. Just use s to match s, or S to match S.
You can use
https?://[^\s",]*/favicon\.ico[^",]+
See the regex demo.
Details:
https?:// - http:// or https://
[^\s",]* - zero or more chars other than whitespace, " and , chars
/favicon\.ico - a fixed /favicon.ico string
[^",]+ - one or more chars other than a " and , chars.
I am working on a wordle bot and I am trying to match words using regex. I am stuck at a problem where I need to look for specific permutations of a given word.
For example, if the word is "steal" these are all the permutations:
'tesla', 'stale', 'steal', 'taels', 'leats', 'setal', 'tales', 'slate', 'teals', 'stela', 'least', 'salet'.
I had some trouble creating a regex for this, but eventually stumbled on positive lookaheads which solved the issue. regex -
'(?=.*[s])(?=.*[l])(?=.*[a])(?=.*[t])(?=.*[e])'
But, if we are looking for specific permutations, how do we go about it?
For example words that look like 's[lt]a[lt]e'. The matching words are 'steal', 'stale', 'state'. But I want to limit the count of l and t in the matched word, which means the output should be 'steal' & 'stale'. 1 obvious solution is this regex r'slate|stale', but this is not a general solution. I am trying to arrive at a general solution for any scenario and the use of positive lookahead above seemed like a starting point. But I am unable to arrive at a solution.
Do we combine positive lookaheads with normal regex?
s(?=.*[lt])a(?=.*[lt])e (Did not work)
Or do we write nested lookaheads or something?
A few more regex that did not work -
s(?=.*[lt]a[tl]e)
s(?=.*[lt])(?=.*[a])(?=.*[lt])(?=.*[e])
I tried to look through the available posts on SO, but could not find anything that would help me understand this. Any help is appreciated.
You could append the regex which matches the permutations of interest to your existing regex. In your sample case, you would use:
(?=.*s)(?=.*l)(?=.*a)(?=.*t)(?=.*e)s[lt]a[lt]e
This will match only stale and slate; it won't match state because it fails the lookahead that requires an l in the word.
Note that you don't need the (?=.*s)(?=.*a)(?=.*e) in the above regex as they are required by the part that matches the permutations of interest. I've left them in to keep that part of the regex generic and not dependent on what follows it.
Demo on regex101
Note that to allow for duplicated characters you might want to change your lookaheads to something in this form:
(?=(?:[^s]*s){1}[^s]*)
You would change the quantifier on the group to match the number of occurrences of that character which are required.
How can I remove all characters inside angle brackets including the brackets in a string? How can I also remove all the text between ("\r\n") and ("."+"any 3 characters") Is this possible? I am currently using the solution by #xkcdjerry
e.g
body = """Dear Students roads etc. you place a tree take a snapshot, then when you place a\r\nbuilding, take a snapshot. Place at least 5-6 objects and then have 5-6\r\nsnapshots. Please keep these snapshots with you as everyone will be asked\r\nto share them during the class.\r\n\r\nI am attaching one PowerPoint containing instructions and one video of\r\nexplanation for your reference.\r\n\r\nKind regards,\r\nTeacher Name\r\n zoom_0.mp4\r\n<https://drive.google.com/file/d/1UX-klOfVhbefvbhZvIWijaBdQuLgh_-Uru4_1QTkth/view?usp=drive_web>"""
d = re.compile("\r\n.+?\\....")
body = d.sub('', body)
a = re.compile("<.*?>")
body = a.sub('', body)
print(body)```
For some reason the output is fine except that it has:
```gle.com/file/d/1UX-klOfVhbefvbhZvIWijaBdQuLgh_-Uru4_1QTkth/view?usp=drive_web>
randomly attached to the end How can I fix it.
Answer
Your problem can be solved by a regex:
Put this into the shell:
import re
a=re.compile("<.*?>")
a.sub('',"Keep this part of the string< Remove this part>Keep This part as well")
Output:
'Keep this part of the stringKeep This part as well'
Second question:
import re
re.compile("\r\n.*?\\..{3}")
a.sub('',"Hello\r\nFilename.png")
Output:
'Hello'
Breakdown
Regex is a robust way of finding, replacing, and mutating small strings inside bigger ones, for further reading,consult https://docs.python.org/3/library/re.html. Meanwhile, here are the breakdowns of the regex information used in this answer:
. means any char.
*? means as many of the before as needed but as little as possible(non-greedy match)
So .*? means any number of characters but as little as possible.
Note: The reason there is a \\. in the second regex is that a . in the match needs to be escaped by a \, which in its turn needs to be escaped as \\
The methods:
re.compile(patten:str) compiles a regex for farther use.
regex.sub(repl:str,string:str) replaces every match of regex in string with repl.
Hope it helps.
I have this for example:
<#445288012218368010>
And I want to get from between <# > symbols the value.
I tried so:
string.replace(/^(?:\<\#)(?:.*)(?:\>)$/gim, '');
But then I don't get any result. It will delete/remove the whole string.
I want only this part: 445288012218368010 (it will be dynamic, so yeah it will be not the same numbers).
Anyway it is for the discord chat bot and I know that there is other methods for check the mentioned names but I want to do that in regex because which I am trying to do can't go the common method.
So yeah how can I get the value from between those symbols?
I need this in node.js regex.
You can use String#match which will return regular expression matches for the string (in this case the RegExp would be <#(\d+)> (the parenthesis around the \d+ make \d+ become its own group). This way you can use <string>.match(/<#(\d+)>/) to get the regular expression results and <string>.match(/<#(\d+)>/)[1] to get the first group of the regex (in this case the number).
You regex matches but you use a non capturing group (?:.*) so you get the full match and replace that with an empty string. Note that you could omit the first and the third non capturing group and use <# and > instead.
You could match what is between the brackets using a capturing group ([^>]+) or (\d+) and use replace and refer the first capturing group $1 in the replacement.
console.log("<#445288012218368010>".replace(/^<#([^>]+)>$/gim, '$1'));
Given a regular expression re and an input string str, I want to find the maximal substring of str, which starts at the minimal position, which matches re.
Special case:
re = Regex("a+|[ax](bc)*"); str = "yyabcbcb"
matching re with str should return the matching string "abcbc" (and not "a", as PCRE does). I also have in mind, that the result is as I want, if the order of the alternations is changed.
The options I found were:
POSIX extended RE - probably outdated, used by egrep ...
RE2 by Google - open source RE2 - C++ - also C-wrapper available
From my point of view, there are two problems with your question.
First is that changing the order of alternations the results are supposed to change.
For each single 'a' in the string, it can either match 'a+' or "ax*".
So it is ambiguous for matching 'a' to alternations in your regular expression.
Second, for finding the maximal substring, it requires the matching pattern of the longest match. As far as I know, only RE2 has provided such a feature, as mentioned by #Cosinus.
So my recommendation is that separating "a+|ax*" into two regexes, finding the maximal substring in each of them, and then comparing the positions of both substrings.
As to find the longest match, you can also refer to a previous regex post description here. The main idea is to search for substrings starting from string position 0 to len(str) and to keep track of the length and position when matched substrings are found.
P.S. Some languages provide regex functions similar to "findall()". Be careful of using them since the returns may be non-overlapping matches. And non-overlapping matches do not necessarily contain the longest matching substring.