I need to extract specific characters like brackets (not the elements within it), *, # etc and replace it with ' '. So I compiled my pattern like below
p = re.compile(r'\s([\[]).*|\s([\(]).*|\s([#]).*|\s([\{]).*|\s([\*]).*|\s([\<]).*|\s.*(\>)\s|\s.*
(\])\s|\s.*(\))\s|\s.*(#)\s|\s.*(\*)\s|\s.*(\})\s')
string = "hello (you) "
for match in re.finditer(p, string):
print(match.group())
This gives the output:
(you)
But what I am expecting is match to give the output list with the captured group like below
["(",")"]
so that I can replace it with ' ' and have the desired output as
hello you
Input: Abnormal heart rate (with fever) should be monitored. Insert your <Name> here.
Output:Abnormal heart rate with fever should be monitored. Insert your Name here.
This answer assumes that you want to replace terms in parentheses or angle brackets with only the content inside them. That is:
(with fever) -> with fever
<Name> -> Name
We can try using re.sub here with a callback function:
inp = "Abnormal heart rate (with fever) should be monitored. Insert your <Name> here."
print(re.sub(r'\(.*?\)|<.*?>', lambda x: re.sub(r'[()<>]', '', x.group(0)), inp))
This prints:
Abnormal heart rate with fever should be monitored. Insert your Name here.
The logic here is that we selectively target the (...) and <...> terms using an alternation. Then, we pass the entire match to a lambda callback which then replaces the surrounding symbols with just the content.
Just list all the characters you want to remove in a single character set, and use re.sub() to remove them.
print(re.sub(r'[[\](){}<>#*]', '', string))
I think you can proceed with replace all with space expect A-Z a-z if you also want digits 0-9 you can specify.
public class MyClass {
public static void main(String args[]) {
String string = "hello (you) hai";
String result =string.replaceAll("[^A-Z a-z]","");
System.out.println(result);
}
}
This will work but here we are using replaceAll();
Related
I'm trying to allow users to filter strings of text using a glob pattern whose only control character is *. Under the hood, I figured the easiest thing to filter the list strings would be to use Js.Re.test[https://rescript-lang.org/docs/manual/latest/api/js/re#test_], and it is (easy).
Ignoring the * on the user filter string for now, what I'm having difficulty with is escaping all the RegEx control characters. Specifically, I don't know how to replace the capture groups within the input text to create a new string.
So far, I've got this, but it's not quite right:
let input = "test^ing?123[foo";
let escapeRegExCtrl = searchStr => {
let re = [%re("/([\\^\\[\\]\\.\\|\\\\\\?\\{\\}\\+][^\\^\\[\\]\\.\\|\\\\\\?\\{\\}\\+]*)/g")];
let break = ref(false);
while (!break.contents) {
switch (Js.Re.exec_ (re, searchStr)) {
| Some(result) => {
let match = Js.Re.captures(result)[0];
Js.log2("Matching: ", match)
}
| None => {
break := true;
}
}
}
};
search -> escapeRegExCtrl
If I disregard the "test" portion of the string being skipped, the above output will produce:
Matching: ^ing
Matching: ?123
Matching: [foo
With the above example, at the end of the day, what I'm trying to produce is this (with leading and following .*:
.*test\^ing\?123\[foo.*
But I'm unsure how to achieve creating a contiguous string from the matched capture groups.
(echo "test^ing?123[foo" | sed -r 's_([\^\?\[])_\\\1_g' would get the work done on the command line)
EDIT
Based on Chris Maurer's answer, there is a method in the JS library that does what I was looking for. A little digging exposed the ReasonML proxy for that method:
https://rescript-lang.org/docs/manual/latest/api/js/string#replacebyre
Let me see if I have this right; you want to implement a character matcher where everything is literal except *. Presumably the * is supposed to work like that in Windows dir commands, matching zero or more characters.
Furthermore, you want to implement it by passing a user-entered character string directly to a Regexp match function after suitably sanitizing it to only deal with the *.
If I have this right, then it sounds like you need to do two things to get the string ready for js.re.test:
Quote all the special regex characters, and
Turn all instances of * into .* or maybe .*?
Let's keep this simple and process the string in two steps, each one using Js.re.replace. So the list of special characters in regex are [^$.|?*+(). Suitably quoting these for replace:
str.replace(/[\[\\\^\$\.\|\?\+\(\)]/g, '\$&')
This is just all those special characters quoted. The $& in the replacement specifications says to insert whatever matched.
Then pass that result to a second replace for the * to .*? transformation.
str.replace(/*+/g, '.*?')
I am trying to match a string in the format of domain\username using Lua and then mask the pattern with #.
So if the input is sample.com\admin; the output should be ######.###\#####;. The string can end with either a ;, ,, . or whitespace.
More examples:
sample.net\user1,hello -> ######.###\#####,hello
test.org\testuser. Next -> ####.###\########. Next
I tried ([a-zA-Z][a-zA-Z0-9.-]+)\.?([a-zA-Z0-9]+)\\([a-zA-Z0-9 ]+)\b which works perfectly with http://regexr.com/. But with Lua demo it doesn't. What is wrong with the pattern?
Below is the code I used to check in Lua:
test_text="I have the 123 name as domain.com\admin as 172.19.202.52 the credentials"
pattern="([a-zA-Z][a-zA-Z0-9.-]+).?([a-zA-Z0-9]+)\\([a-zA-Z0-9 ]+)\b"
res=string.match(test_text,pattern)
print (res)
It is printing nil.
Lua pattern isn't regular expression, that's why your regex doesn't work.
\b isn't supported, you can use the more powerful %f frontier pattern if needed.
In the string test_text, \ isn't escaped, so it's interpreted as \a.
. is a magic character in patterns, it needs to be escaped.
This code isn't exactly equivalent to your pattern, you can tweek it if needed:
test_text = "I have the 123 name as domain.com\\admin as 172.19.202.52 the credentials"
pattern = "(%a%w+)%.?(%w+)\\([%w]+)"
print(string.match(test_text,pattern))
Output: domain com admin
After fixing the pattern, the task of replacing them with # is easy, you might need string.sub or string.gsub.
Like already mentioned pure Lua does not have regex, only patterns.
Your regex however can be matched with the following code and pattern:
--[[
sample.net\user1,hello -> ######.###\#####,hello
test.org\testuser. Next -> ####.###\########. Next
]]
s1 = [[sample.net\user1,hello]]
s2 = [[test.org\testuser. Next]]
s3 = [[abc.domain.org\user1]]
function mask_domain(s)
s = s:gsub('(%a[%a%d%.%-]-)%.?([%a%d]+)\\([%a%d]+)([%;%,%.%s]?)',
function(a,b,c,d)
return ('#'):rep(#a)..'.'..('#'):rep(#b)..'\\'..('#'):rep(#c)..d
end)
return s
end
print(s1,'=>',mask_domain(s1))
print(s2,'=>',mask_domain(s2))
print(s3,'=>',mask_domain(s3))
The last example does not end with ; , . or whitespace. If it must follow this, then simply remove the final ? from pattern.
UPDATE: If in the domain (e.g. abc.domain.org) you need to also reveal any dots before that last one you can replace the above function with this one:
function mask_domain(s)
s = s:gsub('(%a[%a%d%.%-]-)%.?([%a%d]+)\\([%a%d]+)([%;%,%.%s]?)',
function(a,b,c,d)
a = a:gsub('[^%.]','#')
return a..'.'..('#'):rep(#b)..'\\'..('#'):rep(#c)..d
end)
return s
end
I need to select a small portion of a string.
Here's an example string: http://itunes.apple.com/app/eyelashes/id564783832?uo=5
I need: 564783832
A couple of things to keep in mind:
The number will always be preceded by id (ie. id564783832)
There may or may not be a ?uo=5 following the number (and it could be other parameters besides uo)
The string I need can be different lengths (won't always be 9 digits)
The text preceding id will have similar formatting (same # of slashes, but text will be different)
This will ultimately be implemented with Ruby.
without knowing your language/tool, just assume look behind was supported.
'(?<=id)\d+'
With awk
awk '{print $2}' FS='(id|?)'
You can match some sequence of digits preceded by "id" - this assumes that those are the only sequence of digits preceded by "id":
(?<=id)\d++
A test case in Java:
public static void main(String[] args) {
String input = "http://itunes.apple.com/app/eyelashes/id564783832?uo=5";
Pattern pattern = Pattern.compile("(?<=id)\\d++");
Matcher matcher = pattern.matcher(input);
while (matcher.find()) {
System.out.println(matcher.group());
}
}
Output
564783832
Here's mine:
[\w\/]+id(\d+)(\?|$)
I know I can exclude outside characters in a string using look-ahead and look-behind, but I'm not sure about characters in the center.
What I want is to get a match of ABCDEF from the string ABC 123 DEF.
Is this possible with a Regex string? If not, can it be accomplished another way?
EDIT
For more clarification, in the example above I can use the regex string /ABC.*?DEF/ to sort of get what I want, but this includes everything matched by .*?. What I want is to match with something like ABC(match whatever, but then throw it out)DEF resulting in one single match of ABCDEF.
As another example, I can do the following (in sudo-code and regex):
string myStr = "ABC 123 DEF";
string tempMatch = RegexMatch(myStr, "(?<=ABC).*?(?=DEF)"); //Returns " 123 "
string FinalString = myStr.Replace(tempMatch, ""); //Returns "ABCDEF". This is what I want
Again, is there a way to do this with a single regex string?
Since the regex replace feature in most languages does not change the string it operates on (but produces a new one), you can do it as a one-liner in most languages. Firstly, you match everything, capturing the desired parts:
^.*(ABC).*(DEF).*$
(Make sure to use the single-line/"dotall" option if your input contains line breaks!)
And then you replace this with:
$1$2
That will give you ABCDEF in one assignment.
Still, as outlined in the comments and in Mark's answer, the engine does match the stuff in between ABC and DEF. It's only the replacement convenience function that throws it out. But that is supported in pretty much every language, I would say.
Important: this approach will of course only work if your input string contains the desired pattern only once (assuming ABC and DEF are actually variable).
Example implementation in PHP:
$output = preg_replace('/^.*(ABC).*(DEF).*$/s', '$1$2', $input);
Or JavaScript (which does not have single-line mode):
var output = input.replace(/^[\s\S]*(ABC)[\s\S]*(DEF)[\s\S]*$/, '$1$2');
Or C#:
string output = Regex.Replace(input, #"^.*(ABC).*(DEF).*$", "$1$2", RegexOptions.Singleline);
A regular expression can contain multiple capturing groups. Each group must consist of consecutive characters so it's not possible to have a single group that captures what you want, but the groups themselves do not have to be contiguous so you can combine multiple groups to get your desired result.
Regular expression
(ABC).*(DEF)
Captures
ABC
DEF
See it online: rubular
Example C# code
string myStr = "ABC 123 DEF";
Match m = Regex.Match(myStr, "(ABC).*(DEF)");
if (m.Success)
{
string result = m.Groups[1].Value + m.Groups[2].Value; // Gives "ABCDEF"
// ...
}
I am attempting to parse a string like the following using a .NET regular expression:
H3Y5NC8E-TGA5B6SB-2NVAQ4E0
and return the following using Split:
H3Y5NC8E
TGA5B6SB
2NVAQ4E0
I validate each character against a specific character set (note that the letters 'I', 'O', 'U' & 'W' are absent), so using string.Split is not an option. The number of characters in each group can vary and the number of groups can also vary. I am using the following expression:
([ABCDEFGHJKLMNPQRSTVXYZ0123456789]{8}-?){3}
This will match exactly 3 groups of 8 characters each. Any more or less will fail the match.
This works insofar as it correctly matches the input. However, when I use the Split method to extract each character group, I just get the final group. RegexBuddy complains that I have repeated the capturing group itself and that I should put a capture group around the repeated group. However, none of my attempts to do this achieve the desired result. I have been trying expressions like this:
(([ABCDEFGHJKLMNPQRSTVXYZ0123456789]{8})-?){4}
But this does not work.
Since I generate the regex in code, I could just expand it out by the number of groups, but I was hoping for a more elegant solution.
Please note that the character set does not include the entire alphabet. It is part of a product activation system. As such, any characters that can be accidentally interpreted as numbers or other characters are removed. e.g. The letters 'I', 'O', 'U' & 'W' are not in the character set.
The hyphens are optional since a user does not need top type them in, but they can be there if the user as done a copy & paste.
BTW, you can replace [ABCDEFGHJKLMNPQRSTVXYZ0123456789] character class with a more readable subtracted character class.
[[A-Z\d]-[IOUW]]
If you just want to match 3 groups like that, why don't you use this pattern 3 times in your regex and just use captured 1, 2, 3 subgroups to form the new string?
([[A-Z\d]-[IOUW]]){8}-([[A-Z\d]-[IOUW]]){8}-([[A-Z\d]-[IOUW]]){8}
In PHP I would return (I don't know .NET)
return "$1 $2 $3";
I have discovered the answer I was after. Here is my working code:
static void Main(string[] args)
{
string pattern = #"^\s*((?<group>[ABCDEFGHJKLMNPQRSTVXYZ0123456789]{8})-?){3}\s*$";
string input = "H3Y5NC8E-TGA5B6SB-2NVAQ4E0";
Regex re = new Regex(pattern);
Match m = re.Match(input);
if (m.Success)
foreach (Capture c in m.Groups["group"].Captures)
Console.WriteLine(c.Value);
}
After reviewing your question and the answers given, I came up with this:
RegexOptions options = RegexOptions.None;
Regex regex = new Regex(#"([ABCDEFGHJKLMNPQRSTVXYZ0123456789]{8})", options);
string input = #"H3Y5NC8E-TGA5B6SB-2NVAQ4E0";
MatchCollection matches = regex.Matches(input);
for (int i = 0; i != matches.Count; ++i)
{
string match = matches[i].Value;
}
Since the "-" is optional, you don't need to include it. I am not sure what you was using the {4} at the end for? This will find the matches based on what you want, then using the MatchCollection you can access each match to rebuild the string.
Why use Regex? If the groups are always split by a -, can't you use Split()?
Sorry if this isn't what you intended, but your string always has the hyphen separating the groups then instead of using regex couldn't you use the String.Split() method?
Dim stringArray As Array = someString.Split("-")
What are the defining characteristics of a valid block? We'd need to know that in order to really be helpful.
My generic suggestion, validate the charset in a first step, then split and parse in a seperate method based on what you expect. If this is in a web site/app then you can use the ASP Regex validation on the front end then break it up on the back end.
If you're just checking the value of the group, with group(i).value, then you will only get the last one. However, if you want to enumerate over all the times that group was captured, use group(2).captures(i).value, as shown below.
system.text.RegularExpressions.Regex.Match("H3Y5NC8E-TGA5B6SB-2NVAQ4E0","(([ABCDEFGHJKLMNPQRSTVXYZ0123456789]+)-?)*").Groups(2).Captures(i).Value
Mike,
You can use character set of your choice inside character group. All you need is to add "+" modifier to capture all groups. See my previous answer, just change [A-Z0-9] to whatever you need (i.e. [ABCDEFGHJKLMNPQRSTVXYZ0123456789])
You can use this pattern:
Regex.Split("H3Y5NC8E-TGA5B6SB-2NVAQ4E0", "([ABCDEFGHJKLMNPQRSTVXYZ0123456789]{8}+)-?")
But you will need to filter out empty strings from resulting array.
Citation from MSDN:
If multiple matches are adjacent to one another, an empty string is inserted into the array.