Contextual Regular Expression - regex

I have a list of comma separated words that I want to remove the comma from and replace with a space:
elements-(a,b,c,d)
becomes:
elements-(a b c d)
The question is how can I do this using a regular expression if and only if that list is within a specific context, e.g. only prefixed by element-():
The following:
There are a number of elements-(a,b,c,d) and a number of other elements-(e,f,g,h)
should become:
There are a number of elements-(a b c d) and a number of other elements-(e f g h)
What would be the correct way to do this with regex?

For contextual regular expressions, you can use zero-width look-around assertions. Look-around assertions are used to assert that something must be true in order for the match to succeed, but they do not consume any characters (hence "zero-width").
In your case, you want to use positive look-behind and look-ahead assertions. In C#, you can do the following:
static string Replace(string text)
{
return Regex.Replace(
text,
#"(?<=elements\-\((\w+,)*)(\w+),(?=(\w+,)*\w+\))",
"$2 "
);
}
There are three basic parts to the pattern here (in order):
(?<=elements\-\((\w+,)*) - this is the positive look-behind assertion. It says that the pattern will only match if it is preceded by the text elements-( and zero-or-more comma-separated strings.
(\w+), - this is the actual match. It's the text that's being replaced.
(?=(\w+,)*\w+\)) - this is the positive look-ahead assertion. It says that the pattern will only match if it is followed by one-or-more comma-separated strings.
In C#, for matching the inner comma-separated contents, you can alternatively do the following:
static string Replace(string text)
{
return Regex.Replace(
text,
#"(?<=elements\-)\(((\w+,)+\w+)\)",
m => string.Format("({0})", m.Groups[1].Value.Replace(',', ' '))
);
}
The basic approach with the positive look-ahead assertion is still the same.
Example output:
"(x,y,z) elements-(a,b) (m,m,m) elements-(c,d,e,f,g,h)"
...becomes...
"(x,y,z) elements-(a b) (m,m,m) elements-(c d e f g h)"

Related

Why does the regex [a-zA-Z]{5} return true for non-matching string?

I defined a regular expression to check if the string only contains alphabetic characters and with length 5:
use regex::Regex;
fn main() {
let re = Regex::new("[a-zA-Z]{5}").unwrap();
println!("{}", re.is_match("this-shouldn't-return-true#"));
}
The text I use contains many illegal characters and is longer than 5 characters, so why does this return true?
You have to put it inside ^...$ to match the whole string and not just parts:
use regex::Regex;
fn main() {
let re = Regex::new("^[a-zA-Z]{5}$").unwrap();
println!("{}", re.is_match("this-shouldn't-return-true#"));
}
Playground.
As explained in the docs:
Notice the use of the ^ and $ anchors. In this crate, every expression is executed with an implicit .*? at the beginning and end, which allows it to match anywhere in the text. Anchors can be used to ensure that the full text matches an expression.
Your pattern returns true because it matches any consecutive 5 alpha chars, in your case it matches both 'shouldn't' and 'return'.
Change your regex to: ^[a-zA-Z]{5}$
^ start of string
[a-zA-Z]{5} matches 5 alpha chars
$ end of string
This will match a string only if the string has a length of 5 chars and all of the chars from start to end fall in range a-z and A-Z.

regex match longest substring with equal first and last char

/(\w)(\w*)\1/
For this string:"mgntdygtxrvxjnwksqhxuxtrv" I match "txrvxjnwksqhxuxt" (using Ruby), but not the even longer valid substring "tdygtxrvxjnwksqhxuxt".
For a given string, here are two ways to find the longest substring that begins and ends with the same character.
Suppose
str = "mgntdygtxrvxjnwksqhxuxtrv"
Use a regular expression
r = /(.)(?=(.*\1))/
str.gsub(r).map { $1 + $2 }.max_by(&:length)
#=> "tdygtxrvxjnwksqhxuxt".
When, as here, the regular expression contains capture groups, it may be more convenient to use String#gsub without a second argument or block (in which case it returns an enumerator, which can be chained) than String#scan (" If the pattern contains groups, each individual result is itself an array containing one entry per group.") Here gsub performs no substitutions; it merely generates matches of the regular expression.
The regular expression can be made self-documenting by writing it in free-spacing mode.
r = /
(.) # match any char and save to capture group 1
(?= # begin a positive lookahead
(.*\1) # match >= 0 characters followed by the contents of capture group 1
) # end the postive lookahead
/x # free-spacing regex definition mode
The following intermediate calculation is performed:
str.gsub(r).map { $1 + $2 }
#=> ["gntdyg", "ntdygtxrvxjn", "tdygtxrvxjnwksqhxuxt", "txrvxjnwksqhxuxt",
# "xrvxjnwksqhxux", "rvxjnwksqhxuxtr", "vxjnwksqhxuxtrv", "xjnwksqhxux",
# "xux"]
Notice that this does not enumerate all substrings beginning and ending with the same character (because .* is greedy). It does not generate, for example, the substring "xrvx".
Do not use a regular expression
v = str.each_char.with_index.with_object({}) do |(c,i),h|
if h.key?(c)
h[c][:size] = i - h[c][:start] + 1
else
h[c] = { start: i, size: 1 }
end
end.max_by { |_,h| h[:size] }.last
str[v[:start], v[:size]]
#=> "tdygtxrvxjnwksqhxuxt"

Ignore String containing special words (Months)

I am trying to find alphanumeric strings by using the following regular expression:
^(?=.*\d)(?=.*[a-zA-Z]).{3,90}$
Alphanumeric string: an alphanumeric string is any string that contains at least a number and a letter plus any other special characters it can be # - _ [] () {} ç _ \ ù %
I want to add an extra constraint to ignore all alphanumerical strings containing the following month formats :
JANVIER|FEVRIER|MARS|AVRIL|MAI|JUIN|JUILLET|AOUT|SEPTEMBRE|OCTOBRE|NOVEMBRE|DECEMBRE|Jan|Feb|Mar|Apr|May|Jun|JUN|Jul|Aug|Sep|Oct|Nov|Dec|[jJ]anvier|[fF][ée]vrier|[mM]ars|[aA]vril|[mM]ai|[jJ]uin|[jJ]uillet|[aA]o[éû]t|aout|[sS]eptembre|[oO]ctobre|[nN]ovembre|[dD][eé]cembre
One solution is to actually match an alphanumerical string. Then check if this string contains one of these names by using the following function:
vector<string> findString(string s)
{
vector<string> vec;
boost::regex rgx("JANVIER|FEVRIER|MARS|AVRIL|MAI|JUIN|JUILLET|AOUT|SEPTEMBRE|OCTOBRE|NOVEMBRE|DECEMBRE|Jan|Feb|Mar|Apr|May|Jun|JUN|Jul|Aug|Sep|Oct|Nov|Dec|[jJ]anvier|[fF][ée]vrier|[mM]ars|[aA]vril|[mM]ai|[jJ]uin|[jJ]uillet|[aA]o[éû]t|aout|[sS]eptembre|[oO]ctobre|[nN]ovembre|[dD][eé]cembre
");
boost::smatch match;
boost::sregex_iterator begin {s.begin(), s.end(), rgx},
end {};
for (boost::sregex_iterator& i = begin; i != end; ++i)
{
boost::smatch m = *i;
vec.push_back(m.str());
}
return vec;
}
Question: How can I add this constraint directly into the regular expression instead of using this function.
One solution is to use negative lookahead as mentioned in How to ignore words in string using Regular Expressions.
I used it as follows:
String : 2-hello-001
Regular expression : ^(?=.*\d)(?=.*[a-zA-Z]^(?!Jan|Feb|Mar)).{3,90}$
Result: no match
Test website: http://regexlib.com/
The edit provided by #Robin and #RyanCarlson : ^[][\w#_(){}ç\\ù%-]{3,90}$ works perfectly in detecting alphanumeric strings with special characters. It's just the negative lookahead part that isn't working.
You can use negative look ahead, the same way you're using positive lookahead:
(?=.*\d)(?=.*[a-zA-Z])
(?!.*(?:JANVIER|FEVRIER|MARS|AVRIL|MAI|JUIN|JUILLET|AOUT|SEPTEMBRE|OCTOBRE|NOVEMBRE|DECEMBRE|Jan|Feb|Mar|Apr|May|Jun|JUN|Jul|Aug|Sep|Oct|Nov|Dec|[jJ]anvier|[fF][ée]vrier|[mM]ars|[aA]vril|[mM]ai|[jJ]uin|[jJ]uillet|[aA]o[éû]t|aout|[sS]eptembre|[oO]ctobre|[nN]ovembre|[dD][eé]cembre)).{3,90}$
Also you regex is pretty unclear. If you want alphanumerical strings with a length between 3 and 90, you can just do:
/^(?!.*(?:JANVIER|F[Eé]VRIER|MARS|AVRIL|MAI|JUIN|JUILLET|AO[Uù]T|SEPTEMBRE|OCTOBRE|NOVEMBRE|D[Eé]CEMBRE|Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec))
[][\w#_(){}ç\\ù%-]{3,90}$/i
the i flag means it will match upper and lower case (so you can reduce your forbidden list), \w is a shortcut for [0-9a-zA-Z_] (careful if you copy-paste, there's a linebreak here for readability between (?! ) and [ ]). Just add in the final [...] whatever special characters you wanna match.

How to validate a string to have only certain letters by perl and regex

I am looking for a perl regex which will validate a string containing only the letters ACGT. For example "AACGGGTTA" should be valid while "AAYYGGTTA" should be invalid, since the second string has "YY" which is not one of A,C,G,T letters. I have the following code, but it validates both the above strings
if($userinput =~/[A|C|G|T]/i)
{
$validEntry = 1;
print "Valid\n";
}
Thanks
Use a character class, and make sure you check the whole string by using the start of string token, \A, and end of string token, \z.
You should also use * or + to indicate how many characters you want to match -- * means "zero or more" and + means "one or more."
Thus, the regex below is saying "between the start and the end of the (case insensitive) string, there should be one or more of the following characters only: a, c, g, t"
if($userinput =~ /\A[acgt]+\z/i)
{
$validEntry = 1;
print "Valid\n";
}
Using the character-counting tr operator:
if( $userinput !~ tr/ACGT//c )
{
$validEntry = 1;
print "Valid\n";
}
tr/characterset// counts how many characters in the string are in characterset; with the /c flag, it counts how many are not in the characterset. Using !~ instead of =~ negates the result, so it will be true if there are no characters not in characterset or false if there are characters not in characterset.
Your character class [A|C|G|T] contains |. | does not stand for alternation in a character class, it only stands for itself. Therefore, the character class would include the | character, which is not what you want.
Your pattern is not anchored. The pattern /[ACGT]+/ would match any string that contains one or more of any of those characters. Instead, you need to anchor your pattern, so that only strings that contain just those characters from beginning to end are matched.
$ can match a newline. To avoid that, use \z to anchor at the end. \A anchors at the beginning (although it doesn't make a difference whether you use that or ^ in this case, using \A provides a nice symmetry.
So, you check should be written:
if ($userinput =~ /\A [ACGT]+ \z/ix)
{
$validEntry = 1;
print "Valid\n";
}

Regular expression match decimal with letters

I have following string 3.14, 123.56f, .123e5f, 123D, 1234, 343E12, 32.
What I want to do is match any combination of above inputs. So far I started with the following:
^[0-9]\d*(\.\d+)
I realize I have to escape the . since its a regular expression itself.
Thanks.
This should also work, if not already proposed.
try {
Pattern regex = Pattern.compile("\\.?\\b[0-9]*\\.?[0-9]+(?:[eE][-+]?[0-9]+)?[fD]?\\b", Pattern.CASE_INSENSITIVE | Pattern.UNICODE_CASE);
Matcher regexMatcher = regex.matcher(subjectString);
while (regexMatcher.find()) {
// matched text: regexMatcher.group()
// match start: regexMatcher.start()
// match end: regexMatcher.end()
}
} catch (PatternSyntaxException ex) {
// Syntax error in the regular expression
}
Probably
^(\d+(\.\d+)?|\.\d+)([eE]\d+)?[fD]?$
http://regexr.com?2ut9t
^ start of the string
(\d+(\.\d+)?|\.\d+) one or more digits with an optional ( . and one or more digits)
or
. and one or more digits
([eE]\d+)? an optional ( e or E and one or more digits)
[fD]? an optional f or D
$ end of the string
As a sidenote, I've made the D compatible with everything but the f.
If you need positive and negative sign, add [+-]? after the ^
This will match all of those:
[0-9.]+(?:[Ee][0-9.]*)?[DdFf]?
Note that within a character class (square brackets), dot . is not a special character and should not be escaped.
Maybe that one ?
^\d*(?:\.\d+)?(?:[eE]\d+)?(?:[fD])?$
with
^\d* #possibly a digit or sequence of digits at the start
(?:\.\d+)? #possibly followed by a dot and at least one digit
(?:[eE]\d+)? #possibly a 'e' or 'E' followed by at least one digit
(?:[fD])?$ #optionnaly followed by 'f' or 'D' letters until the end
You can use regexpal to test it out, but this seems to work on all of those examples:
^\d*\.?(\d*[eE]?\d*)[fD]?$