Consider the below
Case 1: [Success]
Input : X(P)~AK,X(MV)~AK
Replace with: AP
Output: X(P)~AP,X(MV)~AP
Case 2: [Failure]
Input: X(P)~$B,X(MV)~$B
Replace with: C$
Output: X(P)~C$,X(MV)~C$
Actual Output: X(P)~C$B,X(MV)~C$B
I am using the below REGEXP
#"~(\w*[A-Z$%])"
This works fine for case 1 but falied for the second.
Need help
I am using C#3.0
Thanks
It's unclear what exactly your matching requirements are, but changing the regex to #"~(\w*[A-Z$%]+)" should do the trick. (For the examples given, just plain #"~([A-Z$%]+)" should work too.)
It looks like you want something like this:
public static String replaceWith(String input, String repl) {
return Regex.Replace(
input,
#"(?<=~)[A-Z$%]+",
repl
);
}
The (?<=…) is what is called a lookbehind. It's used to assert that to the left there's a tilde, but that tilde is not part of the match.
Now we can test it as follows (as seen on ideone.com):
Console.WriteLine(replaceWith(
"X(P)~AK,X(MV)~AK", "AP"
));
// X(P)~AP,X(MV)~AP
Console.WriteLine(replaceWith(
"X(P)~$B,X(MV)~$B", "C$"
));
// X(P)~C$,X(MV)~C$
Console.WriteLine(replaceWith(
"X(P)~THIS,X(MV)~THAT", "$$$$"
));
// X(P)~$$,X(MV)~$$
Note the last example: $ is a special symbol in substitutions and can have special meanings. $$ actually gets you one dollar sign.
Related questions
How does the regular expression (?<=#)[^#]+(?=#) work?
Your expression (being greedy) replaces the first string that starts with zero or more work characters that ends in [A-Z$%] after an '~' with your substitution.
In the first case you have ~AK, so \w*[A-Z$%] evaluates to the 'AK', matching \w* -> A, and [A-Z$%] -> K
In the second case you cae ~$C so \w*[A-Z$%] evaluates to '$', matching \w* -> nothing, and [A-Z$%] -> $
I think the important thing is that \w is optional (zero or more), but the [A-Z$%] is mandatory. This is why the second case gives '$' not '$C' as the matched part.
Since I don't know what you're trying to achieve I cannot tell you how to fix your expression.
Related
I'd like to match three-character sequences of letters (only letters 'a', 'b', 'c' are allowed) separated by comma (last group is not ended with comma).
Examples:
abc,bca,cbb
ccc,abc,aab,baa
bcb
I have written following regular expression:
re.match('([abc][abc][abc],)+', "abc,defx,df")
However it doesn't work correctly, because for above example:
>>> print bool(re.match('([abc][abc][abc],)+', "abc,defx,df")) # defx in second group
True
>>> print bool(re.match('([abc][abc][abc],)+', "axc,defx,df")) # 'x' in first group
False
It seems only to check first group of three letters but it ignores the rest. How to write this regular expression correctly?
Try following regex:
^[abc]{3}(,[abc]{3})*$
^...$ 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.
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
I am not able to understand the practical difference between ? and * in regular expressions. I know that ? means to check if previous character/group is present 0 or 1 times and * means to check if the previous character/group is present 0 or more times.
But this code
while(<>) {
chomp($_);
if(/hello?/) {
print "metch $_ \n";
}
else {
print "naot metch $_ \n";
}
}
gives the same out put for both hello? and hello*. The external file that is given to this Perl program contains
hello
helloooo
hell
And the output is
metch hello
metch helloooo
metch hell
for both hello? and hello*. I am not able to understand the exact difference between ? and *
In Perl (and unlike Java), the m//-match operator is not anchored by default.
As such all of the input it trivially matched by both /hello?/ and /hello*/. That is, these will match any string that contains "hell" (as both quantifiers make the "o" optional) anywhere.
Compare with /^hello?$/ and /^hello*$/, respectively. Since these employ anchors the former will not match "helloo" (as at most one "o" is allowed) while the latter will.
Under Regexp Quote-like Operators:
m/PATTERN/ searches [anywhere in] a string for a pattern match, and in scalar context returns true if it succeeds, false if it fails.
What is confusing you is that, without anchors like ^ and $ a regex pattern match checks only whether the pattern appears anywhere in the target string.
If you add something to the pattern after the hello, like
if (/hello?, Ashwin/) { ... }
Then the strings
hello, Ashwin
and
hell, Ashwin
will match, but
helloooo, Ashwin
will not, because there are too many o characters between hell and the comma ,.
However, if you use a star * instead, like
if (/hello*, Ashwin/) { ... }
then all three strings will match.
? Means the last item is optional. * Means it is both optional and you can have multiple items.
ie.
hello? matches hell, hello
hello* matches hell, hello, helloo, hellooo, ....
But not using either ^ or $ means these matches can occur anywhere in the string
Here's an example I came up with that makes it quite clear:
What if you wanted to only match up to tens of people and your data was like below:
2 people. 20 people. 200 people. 2000 people.
Only ? would be useful in that case, whereas * would incorrectly capture larger numbers.
Sorry in advance that this might be a little challenging to read...
I'm trying to parse a line (actually a subject line from an IMAP server) that looks like this:
=?utf-8?Q?Here is som?= =?utf-8?Q?e text.?=
It's a little hard to see, but there are two =?/?= pairs in the above line. (There will always be one pair; there can theoretically be many.) In each of those =?/?= pairs, I want the third argument (as defined by a ? delimiter) extracted. (In the first pair, it's "Here is som", and in the second it's "e text.")
Here's the regex I'm using:
=\?(.+)\?.\?(.*?)\?=
I want it to return two matches, one for each =?/?= pair. Instead, it's returning the entire line as a single match. I would have thought that the ? in the (.*?), to make the * operator lazy, would have kept this from happening, but obviously it doesn't.
Any suggestions?
EDIT: Per suggestions below to replace ".?" with "[^(\?=)]?" I'm now trying to do:
=\?(.+)\?.\?([^(\?=)]*?)\?=
...but it's not working, either. (I'm unsure whether [^(\?=)]*? is the proper way to test for exclusion of a two-character sequence like "?=". Is it correct?)
Try this:
\=\?([^?]+)\?.\?(.*?)\?\=
I changed the .+ to [^?]+, which means "everything except ?"
A good practice in my experience is not to use .*? but instead do use the * without the ?, but refine the character class. In this case [^?]* to match a sequence of non-question mark characters.
You can also match more complex endmarkers this way, for instance, in this case your end-limiter is ?=, so you want to match nonquestionmarks, and questionmarks followed by non-equals:
([^?]*\?[^=])*[^?]*
At this point it becomes harder to choose though. I like that this solution is stricter, but readability decreases in this case.
One solution:
=\?(.*?)\?=\s*=\?(.*?)\?=
Explanation:
=\? # Literal characters '=?'
(.*?) # Match each character until find next one in the regular expression. A '?' in this case.
\?= # Literal characters '?='
\s* # Match spaces.
=\? # Literal characters '=?'
(.*?) # Match each character until find next one in the regular expression. A '?' in this case.
\?= # Literal characters '?='
Test in a 'perl' program:
use warnings;
use strict;
while ( <DATA> ) {
printf qq[Group 1 -> %s\nGroup 2 -> %s\n], $1, $2 if m/=\?(.*?)\?=\s*=\?(.*?)\?=/;
}
__DATA__
=?utf-8?Q?Here is som?= =?utf-8?Q?e text.?=
Running:
perl script.pl
Results:
Group 1 -> utf-8?Q?Here is som
Group 2 -> utf-8?Q?e text.
EDIT to comment:
I would use the global modifier /.../g. Regular expression would be:
/=\?(?:[^?]*\?){2}([^?]*)/g
Explanation:
=\? # Literal characters '=?'
(?:[^?]*\?){2} # Any number of characters except '?' with a '?' after them. This process twice to omit the string 'utf-8?Q?'
([^?]*) # Save in a group next characters until found a '?'
/g # Repeat this process multiple times until end of string.
Tested in a Perl script:
use warnings;
use strict;
while ( <DATA> ) {
printf qq[Group -> %s\n], $1 while m/=\?(?:[^?]*\?){2}([^?]*)/g;
}
__DATA__
=?utf-8?Q?Here is som?= =?utf-8?Q?e text.?= =?utf-8?Q?more text?=
Running and results:
Group -> Here is som
Group -> e text.
Group -> more text
Thanks for everyone's answers! The simplest expression that solved my issue was this:
=\?(.*?)\?.\?(.*?)\?=
The only difference between this and my originally-posted expression was the addition of a ? (non-greedy) operator on the first ".*". Critical, and I'd forgotten it.
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.