re2dot | 根据正则表达式生成其对应 DFA 的状态转移图 | Regex library
kandi X-RAY | re2dot Summary
kandi X-RAY | re2dot Summary
根据正则表达式生成其对应 DFA 的状态转移图, 简单来说就是输入一个正则表达式, 然后会构造其对应的 NFA, 然后再构造对应的 DFA, 然后再最小化 DFA, 最后使用 dot 语法绘制出最小 DFA. emmm.. 精力所限, 这里只支持闭包, 连接, 选择三种正则运算符. 据说剩余的正则运算符均可只用这三种运算符重写.
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Top functions reviewed by kandi - BETA
- Convert a lexer to RPN
- Execute RPN tokens
- Alternate
- Create a state
- Concatenate two ranges
- Build a FAFA
- Create a new move
- Return a new state
- Generate a dot source
- Normalize the state_id
- Add an edge to the digraph
- Add a node to the digraph
- Return a dfa representation of the state
- Computes the e - closure of the given sequence
- Create a new state from a list
- Return a list of states that match ch
- Return a list of all nodes in q
re2dot Key Features
re2dot Examples and Code Snippets
Community Discussions
Trending Discussions on Regex
QUESTION
I have the following regex: /\.([s]?[ac]ss)$/. The problem is, it matches .scss, .sass, .css, .ass. How would I make it not match .ass?
ANSWER
Answered 2022-Mar-02 at 08:35You can use
QUESTION
I have an array of 5 numbers, I'd like to match as long as there are three of the same number and two of the same different number in the array, placement does not matter. Number sequences can be any random string of 5 numbers between 1 - 5.
Examples of matches would be:
33322
24422
52225
44111
54545
*basically any grouping of 2 and 3 of the same numbers needs to match.
Best I've come up with so far:
^([0-9])\1{2}|([0-9])\1{1}$
I am not so good with regex, any help would be greatly appreciated.
...ANSWER
Answered 2022-Jan-16 at 23:38You can use
QUESTION
Currently, this is my code.
...ANSWER
Answered 2022-Jan-06 at 04:57A simple solution would be to convert all characters to lowercase, replace any character that isn't a-z, 0-9, or a space with a space character, and then replace multiple space characters with a single space character.
QUESTION
I am generating some meaningful name with the following rule in a JavaScript/Node JS program:
Input: "tenancy_account__accountPublicId__workspace__workspacePublicId__remove-user__userPublicId"
Expected output: "TenancyAccountAccountPublicIdWorkspaceWorkspacePublicIdRemove-userUserPublicId"
Rules:
- replace any character with zero or more underscore to the non-underscored uppercase
Example:
x | __*x => X - If exists remove last _
This is what is tried so far, looking for better alternatives, if any:
...ANSWER
Answered 2021-Dec-17 at 14:21Split and slice
QUESTION
Java 11 added some new methods to the Pattern class (a compiled version of a regular expression), including:
I am trying to understand the difference between the two and when I would want to use one over the other?
...ANSWER
Answered 2021-Nov-16 at 14:57Pattern.asPredicatewill return true if any part of the input string matches the Regular expression. You should use this method if you're testing some larger body of text for a certain pattern. For example, to test whether a comment from a user contains a hyperlink.Pattern.asMatchPredicatewill return true if the entire input string matches the Regular expression. You should use this method if you're testing the entire input for a certain pattern. For example, to validate the phone number of a user in their profile.
Pattern.asPredicate internally uses Matcher.find(), while Pattern.asMatchPrediate internally uses Matcher.matches(). So the difference between the two boils down to the difference between these two methods from the Matcher class.
Below are some examples to showcase the difference. You can copy & paste below code in an online Java sandbox like https://www.compilejava.net/ to play around with it yourself.
QUESTION
I am trying to replace a single occurrence of a character '1' in a String with a different character.
This same character can occur multiple times in the String which I am not interested in.
For example, in the below string I want to replace the single occurrence of 1 with 2.
...ANSWER
Answered 2021-Nov-13 at 09:22Use a negative lookahead in the regexp to match a 1 that isn't followed by another 1:
QUESTION
I'm attempting to solve a very simple problem - find strings in an array which only contain certain letters. However, I've run up against something in the behavior of regular expressions and/or grep that I don't get.
ANSWER
Answered 2021-Nov-02 at 13:15Both fails are fixed with the addition of anchors ^ and $ and quantifier +
These both work:
QUESTION
I'm looking for a regexp that matches any line that contains 'B', 'R', 'A' and 'S' (in any order) at the start. It would match all the following lines, except the last two.
...ANSWER
Answered 2021-Oct-30 at 21:05There are only 24 permutations :)
QUESTION
I have a Json file having multiple comments and I want to replace a value in it.
I tried the below and it gives me a json file without comments. But I don't understand how to change the value and save it back with comments. Is this even possible because we are replacing all the comments with empty lines?
...ANSWER
Answered 2021-Oct-22 at 22:32Use
QUESTION
The header question may not be easy to understand. Hope you can understand my detailed info below.
I have sentence data below, that has some tags, represented by [tn]tag[/tn]:
ANSWER
Answered 2021-Oct-04 at 08:47Here is a solution that assumes that there are no nested tags, that all tags open and close in the part. Also, this assumes that all characters from the sentence are in parts. For this last assumption, I had to add the . after it has to be in the second expected part. I also had to remove newline characters from the sentence but I think it was because of the copy/paste.
This solution will loop through all characters and store two parallel buffers : one with the tags, one without. We will use the second one to compare with the parts, and use the first one to generate the output.
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Install re2dot
You can use re2dot like any standard Python library. You will need to make sure that you have a development environment consisting of a Python distribution including header files, a compiler, pip, and git installed. Make sure that your pip, setuptools, and wheel are up to date. When using pip it is generally recommended to install packages in a virtual environment to avoid changes to the system.
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