For a string such as:
abzyxcabkmqfcmkcde
Notice that there are string patterns between ab and c in bold. To capture the first string pattern:
ab([a-z]{3,5})c
Is it possible to match both of the groups from the sample string? Actually, there should be 1 to 5 groups.
Note: python style regex.
You can verify that a given string conforms to the 1-5 repetitions of ab([a-z]{3,5})c using this regex
(?:ab([a-z]{3,5})c){1,5}
or this one if there are characters expected between the groups
(?:ab([a-z]{3,5})c.*?){1,5}
You will only be able to extract the last matching group from that string however, not any of the previous ones. to get a previous one you need to use hsz's approach
Just match all results - i.e. with g flag:
/ab([a-z]{3,5})c/g
or some method like in Python:
re.findall(pattern, string, flags=0)
Related
My input is of this format: (xxx)yyyy(zz)(eee)fff where {x,y,z,e,f} are all numbers. But fff is optional though.
Input: x = (123)4567(89)(660)
Expected output: Only the eeepart i.e. the number inside 3rd "()" i.e. 660 in my example.
I am able to achieve this so far:
re.search("\((\d*)\)", x).group()
Output: (123)
Expected: (660)
I am surely missing something fundamental. Please advise.
Edit 1: Just added fff to the input data format.
You could find all those matches that have round braces (), and print the third match with findall
import re
n = "(123)4567(89)(660)999"
r = re.findall("\(\d*\)", n)
print(r[2])
Output:
(660)
The (eee) part is identical to the (xxx) part in your regex. If you don't provide an anchor, or some sequencing requirement, then an unanchored search will match the first thing it finds, which is (xxx) in your case.
If you know the (eee) always appears at the end of the string, you could append an "at-end" anchor ($) to force the match at the end. Or perhaps you could append a following character, like a space or comma or something.
Otherwise, you might do well to match the other parts of the pattern and not capture them:
pattern = r'[0-9()]{13}\((\d{3})\)'
If you want to get the third group of numbers in brackets, you need to skip the first two groups which you can do with a repeating non-capturing group which looks for a set of digits enclosed in () followed by some number of non ( characters:
x = '(123)4567(89)(660)'
print(re.search("(?:\(\d+\)[^(]*){2}(\(\d+\))", x).group(1))
Output:
(660)
Demo on rextester
EDIT: This question pertains to Oracle implementation of regex (POSIX ERE) which does not support 'lookaheads'
I need to separate a string of characters with a comma, however, the pattern is not consistent and I am not sure if this can be accomplished with Regex.
Corpus: 1710ABCD.131711ABCD.431711ABCD.41711ABCD.4041711ABCD.25
The pattern is basically 4 digits, followed by 4 characters, followed by a dot, followed by 1,2, or 3 digits! To make the string above clear, this is how it looks like separated by a space 1710ABCD.13 1711ABCD.43 1711ABCD.4 1711ABCD.404 1711ABCD.25
So the output of a replace operation should look like this:
1710ABCD.13,1711ABCD.43,1711ABCD.4,1711ABCD.404,1711ABCD.25
I was able to match the pattern using this regex:
(\d{4}\w{4}\.\d{1,3})
It does insert a comma but after the third digit beyond the dot (wrong, should have been after the second digit), but I cannot get it to do it in the right position and globally.
Here is a link to a fiddle
https://regex101.com/r/qQ2dE4/329
All you need is a lookahead at the end of the regular expression, so that the greedy \d{1,3} backtracks until it's followed by 4 digits (indicating the start of the next substring):
(\d{4}\w{4}\.\d{1,3})(?=\d{4})
^^^^^^^^^
https://regex101.com/r/qQ2dE4/330
To expand on #CertainPerformance's answer, if you want to be able to match the last token, you can use an alternative match of $:
(\d{4}\w{4}\.\d{1,3})(?=\d{4}|$)
Demo: https://regex101.com/r/qQ2dE4/331
EDIT: Since you now mentioned in the comment that you're using Oracle's implementation, you can simply do:
regexp_replace(corpus, '(\d{1,3})(\d{4})', '\1,\2')
to get your desired output:
1710ABCD.13,1711ABCD.43,1711ABCD.4,1711ABCD.404,1711ABCD.25
Demo: https://regex101.com/r/qQ2dE4/333
In order to continue finding matches after the first one you must use the global flag /g. The pattern is very tricky but it's feasible if you reverse the string.
Demo
var str = `1710ABCD.131711ABCD.431711ABCD.41711ABCD.4041711ABCD.25`;
// Reverse String
var rts = str.split("").reverse().join("");
// Do a reverse version of RegEx
/*In order to continue searching after the first match,
use the `g`lobal flag*/
var rgx = /(\d{1,3}\.\w{4}\d{4})/g;
// Replace on reversed String with a reversed substitution
var res = rts.replace(rgx, ` ,$1`);
// Revert the result back to normal direction
var ser = res.split("").reverse().join("");
console.log(ser);
Using Alteryx, I have a field called Address which consists of fields like A32C, GH2X, ABC19E. So basically where digits are pinned between sets of letters. I am trying to use the RegEx tool to extract the digits out into a new column called ADDRESS1.
I have Address set to Field to Parse. Output method Parse.
My regular expression is typed in as:
(?:[[alpha]]+)(/d+)(?:[[alpha]]+)
And then I have (/d+) outputting to ADDRESS1. However, when I run this it parses 0 records. What am I doing wrong?
To match a digit, use [0-9] or \d. To match a letter, use [[:alpha:]].
Use
[[:alpha:]]+(\d+)[[:alpha:]]+
See the regex demo.
You can try this :
let regex = /(?!([A-Z]+))(\d+)(?=[A-Z]+)/g;
let values = 'A32CZ, GH2X, ABC19E'
let result = values.match(regex);
console.log(result);
This question already has answers here:
How to capture an arbitrary number of groups in JavaScript Regexp?
(5 answers)
Closed 7 years ago.
I have a long strings taken from a VCF file such as (These are truncated for example purpose):
chr1 11189845 COSM462604;COSM893813 G C,T 158.16 PASS AF=0,0;AO=0,0;DP=1201;FAO=0,0;FDP=1201;FR=.;
chr1 11190804 COSM180789 C T 134.06 PASS AF=0;AO=0;DP=1016;FAO=0;FDP=1018;FR=.;FRO=1018;
I want to to write a single regex to return all values of FAO on a given line.
The valid format for FAO is: FAO=SomeNumber; or FAO=SomeNumber, SomeNumber, SomeNumber, etc...;
Is there a way to write a REGEX capture group that takes into account both a single value and an infinite number of values separated by a comma until you see a ';'?
I've tried
FAO=((([0-9]+);)|(([0-9]+),([0-9])+))
But it only takes into account up to 2 numbers and I need matcher group 1 to be the first value, matcher group 2 to be the second etc...
You can use a negated character class: [^;]+ This says to match any characters that are not a semicolon. Since it's a greedy match it will continue until it sees the first semicolon.
var strings = [
'chr1 11189845 COSM462604;COSM893813 G C,T 158.16 PASS AF=0,0;AO=0,0;DP=1201;FAO=0,0;FDP=1201;FR=.;',
'chr1 11190804 COSM180789 C T 134.06 PASS AF=0;AO=0;DP=1016;FAO=0;FDP=1018;FR=.;FRO=1018;'
];
strings.forEach(function(str) {
alert(str.match(/(FAO=[^;]+)/)[1]);
});
From there you can edit the group match to only grab the values /FAO=([^;]+)/ and then you can split that value on the comma delimiter.
var strings = [
'chr1 11189845 COSM462604;COSM893813 G C,T 158.16 PASS AF=0,0;AO=0,0;DP=1201;FAO=0,0;FDP=1201;FR=.;',
'chr1 11190804 COSM180789 C T 134.06 PASS AF=0;AO=0;DP=1016;FAO=0;FDP=1018;FR=.;FRO=1018;'
];
strings.forEach(function(str) {
alert(str.match(/FAO=([^;]+)/)[1].split(','));
});
As stated in this SO answer it's not possible in most languages to have an arbitrary number of group matches.
you could use a regex like this
FAO=([0-9]+(,[0-9]+)*);
the outer parentheses allow you to extract the value or values with the first matching group.
EDIT
considering that you want to capture the individual values with different matching groups this approach won't work (capturing groups inside * will only capture the last match). see the accepted answer to this question for a solution.
EDIT 2
see this demo based on that answer for an example of a pcre regex that will match each number with the same capturing group.
(?:FAO=|\G,)\K(\d+)
note that not all regex flavours support \G and \K. \G matches the end of the previous match (or the start of the string), and \K resets the start of current match.
I'm trying to make a Regular Expression that captures the following:
- XX or XX:XX, up to 6 repetitions (XX:XX:XX:XX:XX:XX), where X is a hexadecimal number.
In other words, I'm trying to capture MAC addresses than can range from 1 to 6 bytes.
regex = re.compile("^([0-9a-fA-F]{2})(?:(?:\:([0-9a-fA-F]{2})){0,5})$")
The problem is that if I enter for example "11:22:33", it only captures the first match and the last, which results in ["11", "22"].
The question: is there any method that {0,5} character will let me catch all repetitions, and not the last one?
Thanks!
Not in Python, no. But you can first check the correct format with your regex, and then simply split the string at ::
result = s.split(':')
Also note that you should always write regular expressions as raw strings (otherwise you get problems with escaping). And your outer non-capturing group does nothing.
Technically there is a way to do it with regex only, but the regex is quite horrible:
r"^([0-9a-fA-F]{2})(?:([0-9a-fA-F]{2}))?(?:([0-9a-fA-F]{2}))?(?:([0-9a-fA-F]{2}))?(?:([0-9a-fA-F]{2}))?(?:([0-9a-fA-F]{2}))?$"
But here you would always get six captures, just that some might be empty.