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 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 have written a small program to whir through a textfile and find and replace regex where 9 digits \d{9}. It works fine, except what I need is a little more complicated.
I am finding the right data correctly. theFile is just a string with the text file streamread into it. I do this and then create and write it to another file.
But I need to find each string match individually, and replace that match with only the last 5 digits of that individual number (currently this is just replacing with FOUND). Keeping the file otherwise identical.
I am not sure how/what is the best way of doing this? would i have to split into an array of strings rather than one mass string? (it's quite a big file)
Any questions let me know, thanks in advance.
Dim regexString As String = "(\d{9})"
Dim replacement1 As String = "FOUND"
Dim rgx As New Regex(regexString)
Try
theFile = rgx.Replace(theFile, replacement1)
Catch
End try
Instead of using just one replacement pattern \d{9} split and group with two patterns, the first is 4 numbers long, the second 5 numbers. Then in the replace use only the last 5 numbers from the last group
Dim k = "abcd 123456789 abcf"
Dim ptn = "(\d{4})(\d{5})"
Dim result = Regex.Replace(k, ptn, "$2")
This approach leaves unchanged the sequences with less than 9 consecutive numbers, but if you have sequences with more than 9 numbers and don't want to change them, then you need a pattern with
Dim ptn = "(\b\d{4})(\d{5}\b)"
to fix the two groups inside a sequence of exactly nine numbers.
The question appears to ask for matches on exactly nine digits and wants the first four to be removed. Ie to replace the nine digits with the last five.
Splitting the regular expression in the question into two parts, for the unwanted and the wanted parts gives
regexString = "\d{4}(\d{5})"
which captures the wanted five digits, so then the replacement is
replacement1 ="$1"
Or in some other regular expression implementations it would be replacement1 ="\1". Additionally the replace method in some regular expression system may have additional options (parameters) for replace first versus replace n-th versus replace all occurrences.
Suppose there are more than nine digits and only the final five are wanted. In this case the regular expression can be written as one of the following (as different regular expression languages support different features). The replacement expression is the same as above.
regexString = "\d{4,}(\d{5})"
regexString = "\d\d\d\d+(\d{5})"
regexString = "\d\d\d\d\d*(\d{5})"
Because regular expressions are normally "greedy" the \d{5} should always match the final 5 digits but it may be worth finishing the regular expression with ...(\d{5})([^\d]|$) and changing the replace to be $1$2. That way it looks for a trailing non-digit or end-of-string.
I would like to construct regular expression which will match password if there is no character repeating 4 or more times.
I have come up with regex which will match if there is character or group of characters repeating 4 times:
(?:([a-zA-Z\d]{1,})\1\1\1)
Is there any way how to match only if the string doesn't contain the repetitions? I tried the approach suggested in Regular expression to match a line that doesn't contain a word? as I thought some combination of positive/negative lookaheads will make it. But I haven't found working example yet.
By repetition I mean any number of characters anywhere in the string
Example - should not match
aaaaxbc
abababab
x14aaaabc
Example - should match
abcaxaxaz
(a is here 4 times but it is not problem, I want to filter out repeating patterns)
That link was very helpful, and I was able to use it to create the regular expression from your original expression.
^(?:(?!(?<char>[a-zA-Z\d]+)\k<char>{3,}).)+$
or
^(?:(?!([a-zA-Z\d]+)\1{3,}).)+$
Nota Bene: this solution doesn't answer exaactly to the question, it does too much relatively to the expressed need.
-----
In Python language:
import re
pat = '(?:(.)(?!.*?\\1.*?\\1.*?\\1.*\Z))+\Z'
regx = re.compile(pat)
for s in (':1*2-3=4#',
':1*1-3=4#5',
':1*1-1=4#5!6',
':1*1-1=1#',
':1*2-a=14#a~7&1{g}1'):
m = regx.match(s)
if m:
print m.group()
else:
print '--No match--'
result
:1*2-3=4#
:1*1-3=4#5
:1*1-1=4#5!6
--No match--
--No match--
It will give a lot of work to the regex motor because the principle of the pattern is that for each character of the string it runs through, it must verify that the current character isn't found three other times in the remaining sequence of characters that follow the current character.
But it works, apparently.
I try to extract the name1 (first-row), name2 (second-row), name3 (third-row) and the street-name (last-row) with regex:
Company Inc.
JohnDoe
Foobar
Industrieterrein 13
The very last row is the street name and this part is already working (the text is stored in the variable "S2").
REGEXREPLACE(S2, "(.*\n)+(?!(.*\n))", "")
This expression will return me the very last line. I am also able the extract the first row:
REGEXREPLACE(S2, "(\n.*)", "")
My problem is, that I do not know how to extract the second and third row....
Also how do I test if the text contains one, two, three or more rows?
Update:
The regex is used in the context of Scribe (a ETL tool). The problem is I can not execute sourcecode, I only have the following functions:
REGEXMATCH(input, pattern)
REGEXREPLACE(input, pattern, replacement)
If the regex language provides support for lookaheads you may count rows backwards and thus get (assuming . does not match newline)
(.*)$ # matching the last line
(.*)(?=(\n.*){1}$) # matching the second last line (excl. newline)
(.*)(?=(\n.*){2}$) # matching the third last line (excl. newline)
just use this regex:
(.+)+
explain:
.
Wildcard: Matches any single character except \n.
+
Matches the previous element one or more times.
As for a regular expression that will match each of four rows, how about this:
(.*?)\n(.*?)\n(.*?)\n(.*)
The parentheses will match, and the \n will match a new line. Note: you may have to use \r\n instead of just \n depending; try both.
You can try the following:
((.*?)\n){3}