This the text file abc.txt
abc.txt
aa:s0:education.gov.in
bb:s1:defence.gov.in
cc:s2:finance.gov.in
I'm trying to parse this file by tokenizing (correct me if this is the incorrect term :) ) at every ":" using the following regular expression.
parser.py
import re,sys,os,subprocess
path = "C:\abc.txt"
site_list = open(path,'r')
for line in site_list:
site_line = re.search(r'(\w)*:(\w)*:([\w\W]*\.[\W\w]*\.[\W\w]*)',line)
print('Regex found that site_line.group(2) = '+str(site_line.group(2))
Why is the output
Regex found that site_line.group(2) = 0
Regex found that site_line.group(2) = 1
Regex found that site_line.group(2) = 2
Can someone please help me understand why it matches the last character of the second group ? I think its matching 0 from s0 , 1 from s1 & 2 from s2
But Why ?
Let's show a simplified example:
>>> re.search(r'(.)*', 'asdf').group(1)
'f'
>>> re.search(r'(.*)', 'asdf').group(1)
'asdf'
If you have a repetition operator around a capturing group, the group stores the last repetition. Putting the group around the repetition operator does what you want.
If you were expecting to see data from the third group, that would be group(3). group(0) is the whole match, and group(1), group(2), etc. count through the actual parenthesized capturing groups.
That said, as the comments suggest, regexes are overkill for this.
>>> 'aa:s0:education.gov.in'.split(':')
['aa', 's0', 'education.gov.in']
And first group is entire match by default.
If a groupN argument is zero, the corresponding return value is the
entire matching string.
So you should skip it. And check group(3), if you want last one.
Also, you should compile regexp before for-loop. It increase performance of your parser.
And you can replace (\w)* to (\w*), if you want match all symbols between :.
Related
I need to capture multiple groups of the same pattern. Suppose, I have the following string:
HELLO,THERE,WORLD
And I've written the following pattern
^(?:([A-Z]+),?)+$
What I want it to do is to capture every single word, so that Group 1 is : "HELLO", Group 2 is "THERE" and Group 3 is "WORLD". What my regex is actually capturing is only the last one, which is "WORLD".
I'm testing my regular expression here and I want to use it with Swift (maybe there's a way in Swift to get intermediate results somehow, so that I can use them?)
UPDATE: I don't want to use split. I just need to now how to capture all the groups that match the pattern, not only the last one.
With one group in the pattern, you can only get one exact result in that group. If your capture group gets repeated by the pattern (you used the + quantifier on the surrounding non-capturing group), only the last value that matches it gets stored.
You have to use your language's regex implementation functions to find all matches of a pattern, then you would have to remove the anchors and the quantifier of the non-capturing group (and you could omit the non-capturing group itself as well).
Alternatively, expand your regex and let the pattern contain one capturing group per group you want to get in the result:
^([A-Z]+),([A-Z]+),([A-Z]+)$
The key distinction is repeating a captured group instead of capturing a repeated group.
As you have already found out, the difference is that repeating a captured group captures only the last iteration. Capturing a repeated group captures all iterations.
In PCRE (PHP):
((?:\w+)+),?
Match 1, Group 1. 0-5 HELLO
Match 2, Group 1. 6-11 THERE
Match 3, Group 1. 12-20 BRUTALLY
Match 4, Group 1. 21-26 CRUEL
Match 5, Group 1. 27-32 WORLD
Since all captures are in Group 1, you only need $1 for substitution.
I used the following general form of this regular expression:
((?:{{RE}})+)
Example at regex101
I think you need something like this....
b="HELLO,THERE,WORLD"
re.findall('[\w]+',b)
Which in Python3 will return
['HELLO', 'THERE', 'WORLD']
After reading Byte Commander's answer, I want to introduce a tiny possible improvement:
You can generate a regexp that will match either n words, as long as your n is predetermined. For instance, if I want to match between 1 and 3 words, the regexp:
^([A-Z]+)(?:,([A-Z]+))?(?:,([A-Z]+))?$
will match the next sentences, with one, two or three capturing groups.
HELLO,LITTLE,WORLD
HELLO,WORLD
HELLO
You can see a fully detailed explanation about this regular expression on Regex101.
As I said, it is pretty easy to generate this regexp for any groups you want using your favorite language. Since I'm not much of a swift guy, here's a ruby example:
def make_regexp(group_regexp, count: 3, delimiter: ",")
regexp_str = "^(#{group_regexp})"
(count - 1).times.each do
regexp_str += "(?:#{delimiter}(#{group_regexp}))?"
end
regexp_str += "$"
return regexp_str
end
puts make_regexp("[A-Z]+")
That being said, I'd suggest not using regular expression in that case, there are many other great tools from a simple split to some tokenization patterns depending on your needs. IMHO, a regular expression is not one of them. For instance in ruby I'd use something like str.split(",") or str.scan(/[A-Z]+/)
Just to provide additional example of paragraph 2 in the answer. I'm not sure how critical it is for you to get three groups in one match rather than three matches using one group. E.g., in groovy:
def subject = "HELLO,THERE,WORLD"
def pat = "([A-Z]+)"
def m = (subject =~ pat)
m.eachWithIndex{ g,i ->
println "Match #$i: ${g[1]}"
}
Match #0: HELLO
Match #1: THERE
Match #2: WORLD
The problem with the attempted code, as discussed, is that there is one capture group matching repeatedly so in the end only the last match can be kept.
Instead, instruct the regex to match (and capture) all pattern instances in the string, what can be done in any regex implementation (language). So come up with the regex pattern for this.
The defining property of the shown sample data is that the patterns of interest are separated by commas so we can match anything-but-a-comma, using a negated character class
[^,]+
and match (capture) globally, to get all matches in the string.
If your pattern need be more restrictive then adjust the exclusion list. For example, to capture words separated by any of the listed punctuation
[^,.!-]+
This extracts all words from hi,there-again!, without the punctuation. (The - itself should be given first or last in a character class, unless it's used in a range like a-z or 0-9.)
In Python
import re
string = "HELLO,THERE,WORLD"
pattern = r"([^,]+)"
matches = re.findall(pattern,string)
print(matches)
In Perl (and many other compatible systems)
use warnings;
use strict;
use feature 'say';
my $string = 'HELLO,THERE,WORLD';
my #matches = $string =~ /([^,]+)/g;
say "#matches";
(In this specific example the capturing () in fact aren't needed since we collect everything that is matched. But they don't hurt and in general they are needed.)
The approach above works as it stands for other patterns as well, including the one attempted in the question (as long as you remove the anchors which make it too specific). The most common one is to capture all words (usually meaning [a-zA-Z0-9_]), with the pattern \w+. Or, as in the question, get only the substrings of upper-case ascii letters[A-Z]+.
I know that my answer came late but it happens to me today and I solved it with the following approach:
^(([A-Z]+),)+([A-Z]+)$
So the first group (([A-Z]+),)+ will match all the repeated patterns except the final one ([A-Z]+) that will match the final one. and this will be dynamic no matter how many repeated groups in the string.
You actually have one capture group that will match multiple times. Not multiple capture groups.
javascript (js) solution:
let string = "HI,THERE,TOM";
let myRegexp = /([A-Z]+),?/g; // modify as you like
let match = myRegexp.exec(string); // js function, output described below
while (match != null) { // loops through matches
console.log(match[1]); // do whatever you want with each match
match = myRegexp.exec(string); // find next match
}
Syntax:
// matched text: match[0]
// match start: match.index
// capturing group n: match[n]
As you can see, this will work for any number of matches.
Sorry, not Swift, just a proof of concept in the closest language at hand.
// JavaScript POC. Output:
// Matches: ["GOODBYE","CRUEL","WORLD","IM","LEAVING","U","TODAY"]
let str = `GOODBYE,CRUEL,WORLD,IM,LEAVING,U,TODAY`
let matches = [];
function recurse(str, matches) {
let regex = /^((,?([A-Z]+))+)$/gm
let m
while ((m = regex.exec(str)) !== null) {
matches.unshift(m[3])
return str.replace(m[2], '')
}
return "bzzt!"
}
while ((str = recurse(str, matches)) != "bzzt!") ;
console.log("Matches: ", JSON.stringify(matches))
Note: If you were really going to use this, you would use the position of the match as given by the regex match function, not a string replace.
Design a regex that matches each particular element of the list rather then a list as a whole. Apply it with /g
Iterate throught the matches, cleaning them from any garbage such as list separators that got mixed in. You may require another regex, or you can get by with simple replace substring method.
The sample code is in JS, sorry :) The idea must be clear enough.
const string = 'HELLO,THERE,WORLD';
// First use following regex matches each of the list items separately:
const captureListElement = /^[^,]+|,\w+/g;
const matches = string.match(captureListElement);
// Some of the matches may include the separator, so we have to clean them:
const cleanMatches = matches.map(match => match.replace(',',''));
console.log(cleanMatches);
repeat the A-Z pattern in the group for the regular expression.
data="HELLO,THERE,WORLD"
pattern=r"([a-zA-Z]+)"
matches=re.findall(pattern,data)
print(matches)
output
['HELLO', 'THERE', 'WORLD']
I have a string as follows:
my_str = "808c000003a185c50cd9b00285e78220500ac56a1c5ca5a1004b2404aa412f058c0a1ba85820cc8208080813c7040a228e0133ca5aca03a2829012533208704411004010808c001003a1c5c50cd9b00285e7822"
I want to group together the strings if they meet a condition
Whenever there is a sequence of '0808', I want this and the following text UP TO the point of the next 0808 pattern etc..
result = re.findall(r'(0808)', my_str)
This just gives me a list of the pattern itself.
I want it to include the pattern and the following text. Something fundamentally wrong with the regex i'm inputting.
Help appreciated.
0808.*?(?=0808)
This looks for the 0808 string then lazily matches 0 or more chars afterwards up to the point of the lookahead to 0808
This will return a match for each selection of 0808foobar matches
I am trying to write a regex for matching a text file that has multiple lines such as :
* 964 0050.56aa.3480 dynamic 200 F F Veth1379
* 930 0025.b52a.dd7e static 0 F F Veth1469
My intention is to match the "0050.56aa.3480 " and "Veth1379" and put them in group(1) & group(2) for using later on.
The regex I wrote is :
\*\s*\d{1,}\s*(\d{1,}\.(?:[a-z][a-z]*[0-9]+[a-z0-9]*)\.\d{1,})\s*(?:[a-z][a-z]+)\s*\d{1,}\s*.\s*.\s*((?:[a-z][a-z]*[0-9]+[a-z0-9]*))
But it does not seem to be working when I test at:
http://www.pythonregex.com/
Could someone point to any obvious error I am doing here.
Thanks,
~Newbie
Try this:
^\* [0-9]{3} +([0-9]{4}.[0-9a-z]{4}.[0-9a-z]{4}).*(Veth[0-9]{4})$
Debuggex Demo
The first part is in capture group one, the "Veth" code in capture group two.
Please consider bookmarking the Stack Overflow Regular Expressions FAQ for future reference. There's a list of online testers in the bottom section.
I don't think you need a regex for this:
for line in open('myfile','r').readlines():
fields = line.split( )
print "\n" + fields[1] + "\n" +fields[6]
A very strict version would look something like this:
^\*\s+\d{3}\s+(\d{4}(?:\.[0-9a-f]{4}){2})\s+\w+\s+\d+\s+\w\s+\w\s+([0-9A-Za-z]+)$
Debuggex Demo
Here I assume that:
the columns will be pretty much the same,
your first match group contains a group of decimal digits and two groups of lower-case hex digits,
and the last word can be anything.
A few notes:
\d+ is equivalent to \d{1,} or [0-9]{1,}, but reads better (imo)
use \. to match a literal ., as . would simply match anything
[a-z]{2} is equivalent to [a-z][a-z], but reads better (my opinion, again)
however, you might want to use \w instead to match a word character
This will do it:
reobj = re.compile(r"^.*?([\w]{4}\.[\w]{4}\.[\w]{4}).*?([\w]+)$", re.IGNORECASE | re.MULTILINE)
match = reobj.search(subject)
if match:
group1 = match.group(1)
group2 = match.group(2)
else:
result = ""
I have the following 2 strings of train station IDs (showing the direction of travel) separated by "-".
String A (strA):
NS1-NS2-NS3-NS4-NS5-NS7-NS8-NS9-NS10-NS11-NS13-NS14-NS15-NS16-NS17-NS18-NS19-NS20-NS21-NS22-NS23-NS24-NS25-NS26-NS27
String B (strB):
NS27-NS26-NS25-NS24-NS23-NS22-NS21-NS20-NS19-NS18-NS17-NS16-NS15-NS14-NS13-NS11-NS10-NS9-NS8-NS7-NS5-NS4-NS3-NS2-NS1
I want to find out which of String A or B contains stations "NS4" followed by "NS1" (answer should be String B).
My current code as follows:
searchStr = ".*NS4-.*NS1(-.*|)"
re.search(searchStr, strA)
re.search(searchStr, strB)
But the result keep returning a match in String A.
May I know how to specify 'searchStr' in order to match only String B?
Two ways to do it: tokenizing and improving the regex.
Tokenizing
tokA = strA.split('-')
tokB = strB.split('-')
print('NS4' in tokA and tokA.index('NS1') > tokA.index('NS4'))
print('NS4' in tokB and tokB.index('NS1') > tokB.index('NS4'))
# False
# True
Regex
import re
pattern = '(^|-)NS4.+NS1(-|$)'
print(re.search(pattern, strA) is not None)
print(re.search(pattern, strB) is not None)
# False
# True
Performance
Tokenization: 2.3072939129997394
Regex: 11.138173280000046
But if you really need performance, I'm sure there are faster ways. Even the tokenization method does multiple passes.
As an alternative to tokenizing, you could use the following expression.
NS4(?=.*?NS1(?!\d))
It literally means:
The characters "NS4" literally.
Followed by any characters, until it finds NS1.
NS1 cannot be followed by a digit.
To educate readers as to what I've used:
(?=) is a Positive Lookahead.
Whatever you place inside this token must be found for the match to be True.
I placed .*? to match anything, as few times as possible using the ? quantifier, followed by NS1 since that is what we want to find.
(?!) is a Negative Lookahead
Whatever you place inside this token, as you might guess, must NOT be found for the match to be True.
I placed a digit in here, so that things like NS10 or NS11 or NS19 are never matched.
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.