I'd like to match three-character sequences of letters (only letters 'a', 'b', 'c' are allowed) separated by comma (last group is not ended with comma).
Examples:
abc,bca,cbb
ccc,abc,aab,baa
bcb
I have written following regular expression:
re.match('([abc][abc][abc],)+', "abc,defx,df")
However it doesn't work correctly, because for above example:
>>> print bool(re.match('([abc][abc][abc],)+', "abc,defx,df")) # defx in second group
True
>>> print bool(re.match('([abc][abc][abc],)+', "axc,defx,df")) # 'x' in first group
False
It seems only to check first group of three letters but it ignores the rest. How to write this regular expression correctly?
Try following regex:
^[abc]{3}(,[abc]{3})*$
^...$ from the start till the end of the string
[...] one of the given character
...{3} three time of the phrase before
(...)* 0 till n times of the characters in the brackets
What you're asking it to find with your regex is "at least one triple of letters a, b, c" - that's what "+" gives you. Whatever follows after that doesn't really matter to the regex. You might want to include "$", which means "end of the line", to be sure that the line must all consist of allowed triples. However in the current form your regex would also demand that the last triple ends in a comma, so you should explicitly code that it's not so.
Try this:
re.match('([abc][abc][abc],)*([abc][abc][abc])$'
This finds any number of allowed triples followed by a comma (maybe zero), then a triple without a comma, then the end of the line.
Edit: including the "^" (start of string) symbol is not necessary, because the match method already checks for a match only at the beginning of the string.
The obligatory "you don't need a regex" solution:
all(letter in 'abc,' for letter in data) and all(len(item) == 3 for item in data.split(','))
You need to iterate over sequence of found values.
data_string = "abc,bca,df"
imatch = re.finditer(r'(?P<value>[abc]{3})(,|$)', data_string)
for match in imatch:
print match.group('value')
So the regex to check if the string matches pattern will be
data_string = "abc,bca,df"
match = re.match(r'^([abc]{3}(,|$))+', data_string)
if match:
print "data string is correct"
Your result is not surprising since the regular expression
([abc][abc][abc],)+
tries to match a string containing three characters of [abc] followed by a comma one ore more times anywhere in the string. So the most important part is to make sure that there is nothing more in the string - as scessor suggests with adding ^ (start of string) and $ (end of string) to the regular expression.
An alternative without using regex (albeit a brute force way):
>>> def matcher(x):
total = ["".join(p) for p in itertools.product(('a','b','c'),repeat=3)]
for i in x.split(','):
if i not in total:
return False
return True
>>> matcher("abc,bca,aaa")
True
>>> matcher("abc,bca,xyz")
False
>>> matcher("abc,aaa,bb")
False
If your aim is to validate a string as being composed of triplet of letters a,b,and c:
for ss in ("abc,bbc,abb,baa,bbb",
"acc",
"abc,bbc,abb,bXa,bbb",
"abc,bbc,ab,baa,bbb"):
print ss,' ',bool(re.match('([abc]{3},?)+\Z',ss))
result
abc,bbc,abb,baa,bbb True
acc True
abc,bbc,abb,bXa,bbb False
abc,bbc,ab,baa,bbb False
\Z means: the end of the string. Its presence obliges the match to be until the very end of the string
By the way, I like the form of Sonya too, in a way it is clearer:
bool(re.match('([abc]{3},)*[abc]{3}\Z',ss))
To just repeat a sequence of patterns, you need to use a non-capturing group, a (?:...) like contruct, and apply a quantifier right after the closing parenthesis. The question mark and the colon after the opening parenthesis are the syntax that creates a non-capturing group (SO post).
For example:
(?:abc)+ matches strings like abc, abcabc, abcabcabc, etc.
(?:\d+\.){3} matches strings like 1.12.2., 000.00000.0., etc.
Here, you can use
^[abc]{3}(?:,[abc]{3})*$
^^
Note that using a capturing group is fraught with unwelcome effects in a lot of Python regex methods. See a classical issue described at re.findall behaves weird post, for example, where re.findall and all other regex methods using this function behind the scenes only return captured substrings if there is a capturing group in the pattern.
In Pandas, it is also important to use non-capturing groups when you just need to group a pattern sequence: Series.str.contains will complain that this pattern has match groups. To actually get the groups, use str.extract. and
the Series.str.extract, Series.str.extractall and Series.str.findall will behave as re.findall.
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);
I have a series of strings which are identifiable by finding a substring "p" tag followed by at least two CAPITAL letters.
Input:
<p>JIM <p>SALLY <p>ROBERT <p>Eric
I want to change the "p" tag to an "i" tag if it's followed by those two capital letters (so not the last one, 'Eric').
Desired output:
<i>JIM <i>SALLY <i>ROBERT <p>Eric
I've tried this using regular expressions in Python:
import re
Mytext = "<p>JIM <p>SALLY <p>ROBERT <p>Eric"
changeTags = re.sub('<p>[A-Z]{2}', '<i>' + re.search('<p>[A-Z]{2}', Mytext).group()[-2:], Mytext)
print changeTags
But the output uses "i" tag + JI in every instance, rather than interating through to use SA and then RO in entries 2 and 3.
<i>JIM <i>JILLY <i>JIBERT <p>Eric
I believe the problem is that I don't understand the .group() method properly. Can anyone advise what I've done wrong?
Thank you.
Another way using look-ahead assertion:
re.sub(r'<p>(?=[A-Z]{2,})','<i>',MyText)
Your inner re.search is only evaluted once, and the result is passed as one of the parameters to re.sub. This can't possible capture all the capital-letters-pairs, only the first one. This means your approach cannot work, not merely your understanding of groups.
Furthermore, using groups is unnecessary.
You need to capture the capital letters using parenthesis, and reference it as \1 in the substitution expression:
re.sub('<p>([A-Z]{2})', r'<i>\1', Mytext)
\1 here means: replace with the substring matched by the first (...) in the regular expression. (docs)
Note the leading r in front of the substitution string, to make it raw.
I would like to construct regular expression which will match password if there is no character repeating 4 or more times.
I have come up with regex which will match if there is character or group of characters repeating 4 times:
(?:([a-zA-Z\d]{1,})\1\1\1)
Is there any way how to match only if the string doesn't contain the repetitions? I tried the approach suggested in Regular expression to match a line that doesn't contain a word? as I thought some combination of positive/negative lookaheads will make it. But I haven't found working example yet.
By repetition I mean any number of characters anywhere in the string
Example - should not match
aaaaxbc
abababab
x14aaaabc
Example - should match
abcaxaxaz
(a is here 4 times but it is not problem, I want to filter out repeating patterns)
That link was very helpful, and I was able to use it to create the regular expression from your original expression.
^(?:(?!(?<char>[a-zA-Z\d]+)\k<char>{3,}).)+$
or
^(?:(?!([a-zA-Z\d]+)\1{3,}).)+$
Nota Bene: this solution doesn't answer exaactly to the question, it does too much relatively to the expressed need.
-----
In Python language:
import re
pat = '(?:(.)(?!.*?\\1.*?\\1.*?\\1.*\Z))+\Z'
regx = re.compile(pat)
for s in (':1*2-3=4#',
':1*1-3=4#5',
':1*1-1=4#5!6',
':1*1-1=1#',
':1*2-a=14#a~7&1{g}1'):
m = regx.match(s)
if m:
print m.group()
else:
print '--No match--'
result
:1*2-3=4#
:1*1-3=4#5
:1*1-1=4#5!6
--No match--
--No match--
It will give a lot of work to the regex motor because the principle of the pattern is that for each character of the string it runs through, it must verify that the current character isn't found three other times in the remaining sequence of characters that follow the current character.
But it works, apparently.
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.