regex

问题 I have a need to replace this: fixed variable 123 with this: fixed variable 234 In VSCode this matches fine: fixed(.*)123 I can't find any way to make it put the capture in the output if a number follows: fixed$1234 fixed${1}234 But the find replace window just looks like this: I read that VSCode uses rust flavoured rexes.. Here indicates ${1}234 should work, but VSCode just puts it in the output.. Tried named capture in a style according to here fixed(?P<n>.*)123 //"invalid regular

问题 I have a string with which I would like to remove all punctuation. I currently use: import string translator = str.maketrans('','', string.punctuation) name = name.translate(translator) However, for strings which are names this removed the hyphen also, which I would like to keep in the string. For Instance '\Fred-Daniels!" Should become "Fred-Daniels". How can I modify the above code to achieve this? 回答1: If you'd like to exclude some punctuation characters from string.puncation , you can

问题 I want to match expression of the pattern a space followed by a (addition operator or subtraction operator) For example: " +" should return True I have tried the using std::regex_match on the following regular exp: " [+-]" , "\\s[+-]" , "\\s[+\\-]" , "\\s[\\+-]" but they all return false. What should be the correct expression ? EDIT Here's the test code: #include<iostream> #include<string> #include<regex> using std::cout; int main() { std::string input; std::cin>>input; const std::regex ex("

问题 While using regex to help solve a problem in the Python Challenge, I came across some behaviour that confused me. from here: (...) Matches whatever regular expression is inside the parentheses. and '+' Causes the resulting RE to match 1 or more repetitions of the preceding RE. So this makes sense: >>>import re >>>re.findall(r"(\d+)", "1111112") ['1111112'] But this doesn't: >>> re.findall(r"(\d)+", "1111112") ['2'] I realise that findall returns only groups when groups are present in the

问题 I wish to extract numbers with any decimals (at least one number both sides of the decimal), but not patterns followed by percentages. Therefore, I believe I need a negative lookahead (so it can see if the number is followed by a percentage sign). For clarity, I would want to extract "123.123" , but would not like to extract "123.123%" I have tried a dozen syntax arrangements but cannot find the one that works. This successfully extracts the decimal pattern. c("123.123%", "123.123") %>% str

问题 I'm using the following regular expression to validate CSS sizes: ([0-9]*\.?[0-9]+)(em|px|%) The following sizes are therefore valid: 10px 10.2px 1.5em 100% How can I change the regex to make the unit (em|px|%) optional to allow a number only? 回答1: You can simply add ? at the end to make the unit optional. ([0-9]*\.?[0-9]+)(em|px|%)? As the unit is now optional, this pattern will allow numbers only entities as well. Be aware however, that doing some will also partial match the content making

问题 I would like to know if there is a way to associate regular expression for the value of an attribute. like in XML Schema: <xs:pattern value="([0-9]|[A-Z]){3}" /> to make the DTD match with this XML line: <airport iata="LAE" name="Nadzab Airport" city="Nadzab" country="Papua New Guinea"> 回答1: No, DTDs do not support regex. Use XSD for vastly superior datatyping. DTD attribute values can be enumerations, however: <!ELEMENT airport EMPTY> <!ATTLIST airport iata (LAE|LAX|LGA) #IMPLIED> 来源： https:

问题 I am really bad with regex but here is what i am trying to acheive StringOne = [5, -, e, 4, e, e, 0, 5, 3, 5, e, b, e, e, 5, 0, a, 4, 3, 3, 1, 9, 0, 8, 1, b, 3, 6, 1, b, 3, 6, 4, d, 3, 3, -, 2, 0, c, c, 1, c, 1, -, ., 8, 3, -, 4, 8, 4, 3]; And I want to remove everything but numbers, characters and '-' I found an answer to save characters and number by doing this StringOne = StringOne.replaceAll("[^a-zA-Z0-9]", ""); But I also want to save '-' Is there some way I can add that to the regex or

问题 The pattern I'm looking for is this: TXT.*\.txt That pattern can occur multiple times in any given line. I would like to either extract each instance of the pattern out or alternatively delete the text that surrounds each instance using sed (or anything, really). Thanks! 回答1: You can use grep as: grep -o 'TXT[^.]*\.txt' file 回答2: You can use Perl as: $ cat file foo TXT1.txt bar TXT2.txt baz foo TXT3.txt bar TXT4.txt baz $ perl -ne 'print "$1\n" while(/(TXT.*?\.txt)/g)' file TXT1.txt TXT2.txt

问题 As the title suggests, I want to run code like this (top_url_list is just a list of urls I'm looping through to find instances of these filename conventions that I'm looking for with regex: name_files = [] for i in top_url_list: result = re.findall("\/([a-z]+[0-9][0-9]\W[a-z]+)", str(urlopen(i).read())) Where the objective is to grab all of the instances where the regex checks out, hence the 'findall()" function. The problem is, it's important that I only get distinct/uniques of each instance