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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 string formulas like this:
?{a,b,c,d}
It can be can be embedded like this:
?{a,b,c,?{x,y,z}}
or this is the same:
?{a,b,c,
?{x,y,z}
}
So I have to find those commas, what are in the second and greather "level" brackets.
In the example below I marked the "levels" where I have to find all commas:
?{a,b,c,
?{x,y, <--Those
?{1,2,3} <--Those
}
}
I've tried with lookahead and lookbehind, but I'm totally confused now :/
Here is my latest working try, but it is not good at all:
OnlineRegex
Update:
To avoid misunderstanding, I don't want to count the commas.
I'd like to get groups of commas to replace them.
The condition is find the commas where more than one "open tags" before it like this: ?{
.. without closing tag like this: }
Examlpe.:
In this case I have not replace any commas:
?{1,2,3} ?{a,b,c}
But in this case I have to replace commas between a b c
?{1,2,3,?{a,b,c}}
For the examples which you have provided, the following regex works(gives the desired output as mentioned by you):
(?<!^\?{[^{}]*),(?=[\s\S]*(?:\s*}){2,})
For String ?{a,b,c,d}, see Demo1 No Match
For String, ?{a,b,c,?{x,y,z}}, see Demo2 Match successful
For String,
?{a,b,c,
?{x,y,z}
}
see Demo3 Match Successful
For String,
?{a,b,c,
?{x,y,
?{1,2,3}
}
}
see Demo4 Match Successful
For String ?{1,2,3} ?{a,b,c} ?{1,2,3} ?{a,b,c}, see Demo5 No Match
Explanation:
(?<!^\?{[^{}]*), - negative lookbehind to discard the 1st level commas. The logic applied here is it should not match the comma which is preceded by start of the string followed by ?{ followed by 0+ occurrences of any character except { or }
(?=[\s\S]*(?:\s*}){2,}) - The comma matched above must be followed by atleast 2 occurrences of }(consecutive or having only whitespaces between them)
Your question is rather unclear #norbre, but I presume you'd like to extract (i.e. "count") the number of commas.
You can't do this with a regex. Regexps can't count number of occurences. However, you can use this to extract the "internal part" and then use a spreadsheet formula to count number of commas:
^(?:\?{[a-zA-Z0-9,]+?,\n??\s*?\?{)([a-zA-Z0-9,?{}\n\s]+?(?:\n*?\s*?|})+)(?:[a-zA-Z0-9,\n\s]*})$
Try: https://regex101.com/r/Rr0eFo/5
Examples
1.
Input:
?{a,b,c,?{e,f},1,2,3}
Output:
e,f}
2.
Input:
?{a,b,c,
?{x,y,z,e,
?{1,2,3,?{f,g,3},4,5,6}
}
,d,e,f}
Output:
x,y,z,e,
?{1,2,3,?{f,g,3},4,5,6}
}
3.
Input:
?{a,b,c,?{e},1,2,3}
Output:
e}
(note that there are no commas here!)
One caveat however. As I have said, regexps can't count number of occurences.
Hence, the following sample (don't know if it's valid or not for your case) would return wrong match:
?{a,b,c,?{e,f}
,1,2,3,?{a,b}
}
Output:
e,f}
,1,2,3,?{a,b}
OK replacing commas is another story so I'll add another answer.
Your regexp engine would need to support recursion.
Still I don't see a way to do it with one regex - one match would either contain the first comma or contain everything between the braces!
What I suggest is to use one regexp to get "what is inside the inner braces", run a replace (, => "") and assemble the whole line again using submatches from the regexp.
Here it is: (\?{[^?{}]*)((?>[^?{}]|(?R))+?)([^?{}]*?\})
Try: https://regex101.com/r/IzTeY0/3
Example 1:
Input:
?{a,b,c,
?{x,y,z,e,
?{1,2,3,?{f,g,3},4,5,6}
}
,d,e,f}
Submatches:
1. ?{a,b,c,
2. ?{x,y,z,e,
?{1,2,3,?{f,g,3},4,5,6}
}
3.
,d,e,f}
Replace all commas in submatch 2 with anything you want, then reassamble the whole string using submatches 1 and 3.
Again, this would break the regexp:
?{a,b,c,?{e,f}
,1,2,3,?{a,b}
}
Submatch 2 would look like this:
?{e,f}
,1,2,3,?{a,b}
I don't write many regular expressions so I'm going to need some help on the one.
I need a regular expression that can validate that a string is an alphanumeric comma delimited string.
Examples:
123, 4A67, GGG, 767 would be valid.
12333, 78787&*, GH778 would be invalid
fghkjhfdg8797< would be invalid
This is what I have so far, but isn't quite right: ^(?=.*[a-zA-Z0-9][,]).*$
Any suggestions?
Sounds like you need an expression like this:
^[0-9a-zA-Z]+(,[0-9a-zA-Z]+)*$
Posix allows for the more self-descriptive version:
^[[:alnum:]]+(,[[:alnum:]]+)*$
^[[:alnum:]]+([[:space:]]*,[[:space:]]*[[:alnum:]]+)*$ // allow whitespace
If you're willing to admit underscores, too, search for entire words (\w+):
^\w+(,\w+)*$
^\w+(\s*,\s*\w+)*$ // allow whitespaces around the comma
Try this pattern: ^([a-zA-Z0-9]+,?\s*)+$
I tested it with your cases, as well as just a single number "123". I don't know if you will always have a comma or not.
The [a-zA-Z0-9]+ means match 1 or more of these symbols
The ,? means match 0 or 1 commas (basically, the comma is optional)
The \s* handles 1 or more spaces after the comma
and finally the outer + says match 1 or more of the pattern.
This will also match
123 123 abc (no commas) which might be a problem
This will also match 123, (ends with a comma) which might be a problem.
Try the following expression:
/^([a-z0-9\s]+,)*([a-z0-9\s]+){1}$/i
This will work for:
test
test, test
test123,Test 123,test
I would strongly suggest trimming the whitespaces at the beginning and end of each item in the comma-separated list.
You seem to be lacking repetition. How about:
^(?:[a-zA-Z0-9 ]+,)*[a-zA-Z0-9 ]+$
I'm not sure how you'd express that in VB.Net, but in Python:
>>> import re
>>> x [ "123, $a67, GGG, 767", "12333, 78787&*, GH778" ]
>>> r = '^(?:[a-zA-Z0-9 ]+,)*[a-zA-Z0-9 ]+$'
>>> for s in x:
... print re.match( r, s )
...
<_sre.SRE_Match object at 0xb75c8218>
None
>>>>
You can use shortcuts instead of listing the [a-zA-Z0-9 ] part, but this is probably easier to understand.
Analyzing the highlights:
[a-zA-Z0-9 ]+ : capture one or more (but not zero) of the listed ranges, and space.
(?:[...]+,)* : In non-capturing parenthesis, match one or more of the characters, plus a comma at the end. Match such sequences zero or more times. Capturing zero times allows for no comma.
[...]+ : capture at least one of these. This does not include a comma. This is to ensure that it does not accept a trailing comma. If a trailing comma is acceptable, then the expression is easier: ^[a-zA-Z0-9 ,]+
Yes, when you want to catch comma separated things where a comma at the end is not legal, and the things match to $LONGSTUFF, you have to repeat $LONGSTUFF:
$LONGSTUFF(,$LONGSTUFF)*
If $LONGSTUFF is really long and contains comma repeated items itself etc., it might be a good idea to not build the regexp by hand and instead rely on a computer for doing that for you, even if it's just through string concatenation. For example, I just wanted to build a regular expression to validate the CPUID parameter of a XEN configuration file, of the ['1:a=b,c=d','2:e=f,g=h'] type. I... believe this mostly fits the bill: (whitespace notwithstanding!)
xend_fudge_item_re = r"""
e[a-d]x= #register of the call return value to fudge
(
0x[0-9A-F]+ | #either hardcode the reply
[10xks]{32} #or edit the bitfield directly
)
"""
xend_string_item_re = r"""
(0x)?[0-9A-F]+: #leafnum (the contents of EAX before the call)
%s #one fudge
(,%s)* #repeated multiple times
""" % (xend_fudge_item_re, xend_fudge_item_re)
xend_syntax = re.compile(r"""
\[ #a list of
'%s' #string elements
(,'%s')* #repeated multiple times
\]
$ #and nothing else
""" % (xend_string_item_re, xend_string_item_re), re.VERBOSE | re.MULTILINE)
Try ^(?!,)((, *)?([a-zA-Z0-9])\b)*$
Step by step description:
Don't match a beginning comma (good for the upcoming "loop").
Match optional comma and spaces.
Match characters you like.
The match of a word boundary make sure that a comma is necessary if more arguments are stacked in string.
Please use - ^((([a-zA-Z0-9\s]){1,45},)+([a-zA-Z0-9\s]){1,45})$
Here, I have set max word size to 45, as longest word in english is 45 characters, can be changed as per requirement
I need a regex to get numeric values that can be
111.111,11
111,111.11
111,111
And separate the integer and decimal portions so I can store in a DB with the correct syntax
I tried ([0-9]{1,3}[,.]?)+([,.][0-9]{2})? With no success since it doesn't detect the second part :(
The result should look like:
111.111,11 -> $1 = 111111; $2 = 11
First Answer:
This matches #,###,##0.00:
^[+-]?[0-9]{1,3}(?:\,?[0-9]{3})*(?:\.[0-9]{2})?$
And this matches #.###.##0,00:
^[+-]?[0-9]{1,3}(?:\.?[0-9]{3})*(?:\,[0-9]{2})?$
Joining the two (there are smarter/shorter ways to write it, but it works):
(?:^[+-]?[0-9]{1,3}(?:\,?[0-9]{3})*(?:\.[0-9]{2})?$)
|(?:^[+-]?[0-9]{1,3}(?:\.?[0-9]{3})*(?:\,[0-9]{2})?$)
You can also, add a capturing group to the last comma (or dot) to check which one was used.
Second Answer:
As pointed by Alan M, my previous solution could fail to reject a value like 11,111111.00 where a comma is missing, but the other isn't. After some tests I reached the following regex that avoids this problem:
^[+-]?[0-9]{1,3}
(?:(?<comma>\,?)[0-9]{3})?
(?:\k<comma>[0-9]{3})*
(?:\.[0-9]{2})?$
This deserves some explanation:
^[+-]?[0-9]{1,3} matches the first (1 to 3) digits;
(?:(?<comma>\,?)[0-9]{3})? matches on optional comma followed by more 3 digits, and captures the comma (or the inexistence of one) in a group called 'comma';
(?:\k<comma>[0-9]{3})* matches zero-to-any repetitions of the comma used before (if any) followed by 3 digits;
(?:\.[0-9]{2})?$ matches optional "cents" at the end of the string.
Of course, that will only cover #,###,##0.00 (not #.###.##0,00), but you can always join the regexes like I did above.
Final Answer:
Now, a complete solution. Indentations and line breaks are there for readability only.
^[+-]?[0-9]{1,3}
(?:
(?:\,[0-9]{3})*
(?:.[0-9]{2})?
|
(?:\.[0-9]{3})*
(?:\,[0-9]{2})?
|
[0-9]*
(?:[\.\,][0-9]{2})?
)$
And this variation captures the separators used:
^[+-]?[0-9]{1,3}
(?:
(?:(?<thousand>\,)[0-9]{3})*
(?:(?<decimal>\.)[0-9]{2})?
|
(?:(?<thousand>\.)[0-9]{3})*
(?:(?<decimal>\,)[0-9]{2})?
|
[0-9]*
(?:(?<decimal>[\.\,])[0-9]{2})?
)$
edit 1: "cents" are now optional;
edit 2: text added;
edit 3: second solution added;
edit 4: complete solution added;
edit 5: headings added;
edit 6: capturing added;
edit 7: last answer broke in two versions;
I would at first use this regex to determine wether a comma or a dot is used as a comma delimiter (It fetches the last of the two):
[0-9,\.]*([,\.])[0-9]*
I would then strip all of the other sign (which the previous didn't match). If there were no matches, you already have an integer and can skip the next steps. The removal of the chosen sign can easily be done with a regex, but there are also many other functions which can do this faster/better.
You are then left with a number in the form of an integer possible followed by a comma or a dot and then the decimals, where the integer- and decimal-part easily can be separated from eachother with the following regex.
([0-9]+)[,\.]?([0-9]*)
Good luck!
Edit:
Here is an example made in python, I assume the code should be self-explaining, if it is not, just ask.
import re
input = str(raw_input())
delimiterRegex = re.compile('[0-9,\.]*([,\.])[0-9]*')
splitRegex = re.compile('([0-9]+)[,\.]?([0-9]*)')
delimiter = re.findall(delimiterRegex, input)
if (delimiter[0] == ','):
input = re.sub('[\.]*','', input)
elif (delimiter[0] == '.'):
input = re.sub('[,]*','', input)
print input
With this code, the following inputs gives this:
111.111,11
111111,11
111,111.11
111111.11
111,111
111,111
After this step, one can now easily modify the string to match your needs.
How about
/(\d{1,3}(?:,\d{3})*)(\.\d{2})?/
if you care about validating that the commas separate every 3 digits exactly,
or
/(\d[\d,]*)(\.\d{2})?/
if you don't.
If I'm interpreting your question correctly so that you are saying the result SHOULD look like what you say is "would" look like, then I think you just need to leave the comma out of the character class, since it is used as a separator and not a part of what is to be matched.
So get rid of the "." first, then match the two parts.
$value = "111,111.11";
$value =~ s/\.//g;
$value =~ m/(\d+)(?:,(\d+))?/;
$1 = leading integers with periods removed
$2 = either undef if it didn't exist, or the post-comma digits if they do exist.
See Perl's Regexp::Common::number.