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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.
I have a string, and I want to remove all special characters, including spaces. Except, I want to leave the colon if it exists in the string.
I was using this, and it was sort of working, but appears to not replace parens, or back slash or dashes.......
TRIM(REGEXP_REPLACE(REPLACE(REGEXP_REPLACE(c.category_name,'[^:^0-9A-Za-z ]',''),' : ','|'), '\s+', '_', 'g'))
Please advise
You can add the colon into the list of allowed characters
[^0-9A-Za-z:]: anything that is not a number, a letter or a colon.
'g': apply the replacement as many time as needed (else it would stop at the 1st one)
select REGEXP_REPLACE('0.1[2]?ab cd:ef g&*(h)/ij;','[^0-9A-Za-z:]','','g');
regexp_replace
----------------
012abcd:efghij
You are missing the 'g' ("global") flag, so only the first unwanted character was being removed (replaced by blank).
Change:
REGEXP_REPLACE(c.category_name, '[^:^0-9A-Za-z ]', '')
to:
REGEXP_REPLACE(c.category_name, '[^:0-9A-Za-z ]', '', 'g')
Note: I removed the extra ^ from the regex, but leave it in if you want to keep ^ characters too.
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 want a regix format that Must be alphabets and special characters (like space, ‘, -) but numeric value should not be taken.
I tried with this expression /^[a-zA-Z ]*$/ but it treats space as special character.
Please Help.
/^[a-zA-Z\s\-\'\"]*$/
use this.
This will contain any alphabet([upper/lower]case)
,space,
hiphen,
",
'
update
If you are using it inside NSPredicate
then make sure that you put the - in the end, as it throws error.
Move it to the end of the sequence to be the last character before the closing square bracket ].
like this [a-zA-Z '"-]
If you want only the alphabets and space, ' and - then:
/^[-a-zA-Z\s\']+$/
Notice the + from above instead of *. If you use * then it will match with empty string, where the + sign means to have at least one character in your input.
Now, if you want to match any alphabets with any special characters(not only those three which are mentioned), then I'll just you to use this one:
/^\D+$/
It means any characters other than digits!
Maybe try this:
\b[a-zA-Z \-\']+\b
http://regex101.com/r/oQ5nU9
You can use it defiantly work it
[a-zA-Z._^%$#!~#,-]+
this code work fine you can try it....
//Use this for allowing space as we all as other special character.
#"[a-zA-Z\\s\\-\\'\\"]"
//Following link will be help for further.
http://www.raywenderlich.com/30288/nsregularexpression-tutorial-and-cheat-sheet
Thanks for your response.. I finally resolved it with this
NSString characterRegex = #"^(\s[a-zA-Z]+(([\'\-\+\s]\s*[a-zA-Z])?[a-zA-Z])\s)+$";
NSPredicate *characterTest = [NSPredicate predicateWithFormat:#"SELF MATCHES %#",characterRegex];
return [characterTest evaluateWithObject:inputString];
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