RegEx to match any character except strings with surrounding prefixes - regex

I'm trying to create a regex which will validate input to match any character but it should exclude input which is surrounded with prefixes (i.e. {#..#}, {#..#} or {$..$}).
Given is the example:
Free text which is fine // should return true
{#some other text#} // should return false
Text with numbers 671 // should return true
{#Hello world#} // should return false
{$Hello Mars$} // should return false
{$some text which i do not close // should return true
This should be possible with the use of negative look-around, something along the lines of:
^(\?!=(\{#[^.]+#\}))(\?!=(\{$[^.]+$\}))(\?!=(\{#[^.]+#\})).*
Any help would be greatly appreciated : )

You syntax is a bit weird if you ask me... I would suggest:
^(?!\{(?:[$##]).*(?:[$##])\}).*
Demo
First 'group' is \{(?:[$##]) which looks for the opening prefix, then .* to match everything in the middle and (?:[$##])\} to match the closing suffix.
Note that it will not allow things like:
{$Hello Mars$} how are you?
If you want it to accept this as well, add an end of line anchor:
^(?!\{(?:[$##]).*(?:[$##])\}$).*
^
Demo
You can use the character class to have the different symbols [#$#] and is shorter than having multiple negative lookarounds or | operators in side :)
EDIT: To prevent things like {#Free text which is fine$}' you could use:
^(?!\{([$##]).*\1\}).*
Or
^(?!\{([$##]).*\1\}$).*
For the second version.
\1 is a backreference and refers to the first captured group (any of $, #, or #).

Related

Shorten Regular Expression (\n) [duplicate]

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.

Python Regex - How to extract the third portion?

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

Regex to grab formulas

I am trying to parse a file that contains parameter attributes. The attributes are setup like this:
w=(nf*40e-9)*ng
but also like this:
par_nf=(1) * (ng)
The issue is, all of these parameter definitions are on a single line in the source file, and they are separated by spaces. So you might have a situation like this:
pd=2.0*(84e-9+(1.0*nf)*40e-9) nf=ng m=1 par=(1) par_nf=(1) * (ng) plorient=0
The current algorithm just splits the line on spaces and then for each token, the name is extracted from the LHS of the = and the value from the RHS. My thought is if I can create a Regex match based on spaces within parameter declarations, I can then remove just those spaces before feeding the line to the splitter/parser. I am having a tough time coming up with the appropriate Regex, however. Is it possible to create a regex that matches only spaces within parameter declarations, but ignores the spaces between parameter declarations?
Try this RegEx:
(?<=^|\s) # Start of each formula (start of line OR [space])
(?:.*?) # Attribute Name
= # =
(?: # Formula
(?!\s\w+=) # DO NOT Match [space] Word Characters = (Attr. Name)
[^=] # Any Character except =
)* # Formula Characters repeated any number of times
When checking formula characters, it uses a negative lookahead to check for a Space, followed by Word Characters (Attribute Name) and an =. If this is found, it will stop the match. The fact that the negative lookahead checks for a space means that it will stop without a trailing space at the end of the formula.
Live Demo on Regex101
Thanks to #Andy for the tip:
In this case I'll probably just match on the parameter name and equals, but replace the preceding whitespace with some other "parse-able" character to split on, like so:
(\s*)\w+[a-zA-Z_]=
Now my first capturing group can be used to insert something like a colon, semicolon, or line-break.
You need to add Perl tag. :-( Maybe this will help:
I ended up using this in C#. The idea was to break it into name value pairs, using a negative lookahead specified as the key to stop a match and start a new one. If this helps
var data = #"pd=2.0*(84e-9+(1.0*nf)*40e-9) nf=ng m=1 par=(1) par_nf=(1) * (ng) plorient=0";
var pattern = #"
(?<Key>[a-zA-Z_\s\d]+) # Key is any alpha, digit and _
= # = is a hard anchor
(?<Value>[.*+\-\\\/()\w\s]+) # Value is any combinations of text with space(s)
(\s|$) # Soft anchor of either a \s or EOB
((?!\s[a-zA-Z_\d\s]+\=)|$) # Negative lookahead to stop matching if a space then key then equal found or EOB
";
Regex.Matches(data, pattern, RegexOptions.IgnorePatternWhitespace | RegexOptions.ExplicitCapture)
.OfType<Match>()
.Select(mt => new
{
LHS = mt.Groups["Key"].Value,
RHS = mt.Groups["Value"].Value
});
Results:

Regular expression help - comma delimited string

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

Regex: match whole line except first string and #comment lines

I tried (\s|\t).*[\b\w*\s\b], this one is almost ok but I want also except lines with #.
#Name Type Allowable values
#========================== ========= ========================================
_absolute-path-base-uri String -
add-xml-decl Boolean y/n, yes/no, t/f, true/false, 1/0
As #anubhava said in his answer, it looks you just need to check for # at the beginning of the line. The regex for that is simple, but the mechanics of applying the regex varies wildly, so it would help if we knew which regex flavor/tool you're using (e.g. PHP, .NET, Notepad++, EditPad Pro, etc.). Here's a JavaScript version:
/^[^#].*$/mg
Notice the modifiers: m ("multiline") allows ^ and $ to match at line boundaries, and g ("global") allows you to find all the matches, not just the first one.
Now let's look at your regex. [\b\w*\s\b] is a character class that matches a word character (\w), a whitespace character (\s), an asterisk (*), or a backspace (\b). In other words, both * and \b lose their special meanings when the appear in a character class.
\s matches any whitespace character including \t, so (\s|\t) is needlessly redundant, and may not be needed at all. What it's actually doing in your case is matching the newline before each matched line. There's no need for that when you can use ^ in multiline mode. If you want to allow for horizontal whitespace (i.e., spaces and tabs) before the #, you can do this:
/^(?![ \t]*#).*$/mg
(?![ \t]*#) is a negative lookahead; it means "from this position, it is impossible to match zero or more tabs or spaces followed by #". Coming right after the ^ line anchor as it does, "this position" means the beginning of a line.
Try this:
^[A-z0-9_-]+\s+(.+)$
Assuming your first string will consist of only letters, numbers, underscores or hyphens, the first part will match that. Then we match whitespace, and then capture the rest. However, this is all dependent on the regular expression engine being used. Is this using language support for regexes, a specific editor, or a certain library? Which one? There isn't a standard: each regex engine works slightly differently.
Try this:
^[^#].*?(\s|\t)(?<Group>.*)$
After a match is found, the Group group will contain your string.
I would use this regex. In English, this says "First character is not a pound sign (#), then non-white space to match the first 'word', then white space, then match the whole line.
^[^#]\S*\s+(.+)$
Can I suggest another approach though? It looks like there are tabs between each field in the text, so why not just read the text line-by-line and split by tab into an array?
Here is an example in C# (untested):
using(StreamReader sr = new StreamReader("C:\\Path\\to\\file.txt"))
{
string line = sr.ReadLine();
while(!sr.EndOfStream)
{
//skip the comment lines
if(line.StartsWith("#"))
continue;
string[] fields = line.Split(new string[] {"\t"}, StringSplitOptions.RemoveEmptyEntries);
//now fields[0] contains the Name field
//fields[1] contains the Type field
//fields[2] contains the Allowable Values field
line = sr.ReadLine();
}
}
Try this code in php:
<?php
$s="#Name Type Allowable values
#========================== ========= ========================================
_absolute-path-base-uri String -
add-xml-decl Boolean y/n, yes/no, t/f, true/false, 1/0 ";
$a = explode("\n", $s);
foreach($a as $str) {
preg_match('~^[^#].*$~', $str, $m);
var_dump($m);
}
?>
OUTPUT
array(0) {
}
array(0) {
}
array(1) {
[0]=>
string(79) "_absolute-path-base-uri String - "
}
array(1) {
[0]=>
string(77) "add-xml-decl Boolean y/n, yes/no, t/f, true/false, 1/0 "
}
Code is pretty simple, it just ignores matching # at the start of a line thus ingoring those lines completely.