I have something like this
AD ABCDEFG HIJKLMN
AB HIJKLMN
AC DJKEJKW SJKLAJL JSHELSJ
Rule: Always 2 Chars Code (AB|AC|AD) at line beginning then any number of 7 Chars codes following.
With this regex:
^(AB|AC|AD)|((\S{7})?
in this groovy code sample:
def m= Pattern.compile(/^(AB|AC|AD)|((\S{7})?)/).matcher("AC DJKEJKW SJKLAJL JSHELSJ")
println m.getCount()
I always get 8 as count, means it counts the spaces.
How do I get 4 groups (as expected) without spaces ?
Thanks from a not-yet-regex-expert
Sven
Using this code:
def input = [ 'AD ABCDEFG HIJKLMN', 'AB HIJKLMN', 'AC DJKEJKW SJKLAJL JSHELSJ' ]
def regexp = /^(AB|AC|AD)|((\S{7})+)/
def result = input.collect {
matcher = ( it =~ regexp )
println "got $matcher.count for $it"
matcher.collect { it[0] }
}
println result
I get the output
got 3 for AD ABCDEFG HIJKLMN
got 2 for AB HIJKLMN
got 4 for AC DJKEJKW SJKLAJL JSHELSJ
[[AD, ABCDEFG, HIJKLMN], [AB, HIJKLMN], [AC, DJKEJKW, SJKLAJL, JSHELSJ]]
Is this more what you wanted?
This pattern will match your requirements
^A[BCD](?:\s\S{7})+
See it here online on Regexr
Meaning start with A then either a B or a C or a D. This is followed by at least one group consisting of a whitespace followed by 7 non whitespaces.
Related
I have the following pattern that I could parse using pandas in Python, but struggle with translating the code into Scala.
grade string_column
85 (str:ann smith,14)(str:frank chase,15)
86 (str:john foo,15)(str:al more,14)
In python I used:
df.set_index('grade')['string_column']\
.str.extractall(r'\((str:[^,]+),(\d+)\)')\
.droplevel(1)
with the output:
grade 0 1
85 str:ann smith 14
85 str:frank chase 15
86 str:john foo 15
86 str:al more 14
In Scala I tried to duplicate the approach, but it's failing:
import scala.util.matching.Regex
val pattern = new Regex("((str:[^,]+),(\d+)\)")
val str = "(str:ann smith,14)(str:frank chase,15)"
println(pattern findAllIn(str)).mkString(","))
There are a few notes about the code:
There is an unmatched parenthesis for a group, but that one should be escaped
The backslashes should be double escaped
In the println you don't have to use all the parenthesis and the dot
findAllIn returns a MatchIterator, and looping those will expose a matched string. Joining those matched strings with a comma, will in this case give back the same string again.
For example
import scala.util.matching.Regex
val pattern = new Regex("\\((str:[^,]+),(\\d+)\\)")
val str = "(str:ann smith,14)(str:frank chase,15)"
println(pattern findAllIn str mkString ",")
Output
(str:ann smith,14),(str:frank chase,15)
But if you want to print out the group 1 and group 2 values, you can use findAllMatchIn that returns a collection of Regex Matches:
import scala.util.matching.Regex
val pattern = new Regex("\\((str:[^,]+),(\\d+)\\)")
val str = "(str:ann smith,14)(str:frank chase,15)"
pattern findAllMatchIn str foreach(m => {
println(m.group(1))
println(m.group(2))
}
)
Output
str:ann smith
14
str:frank chase
15
In Python, Series.str.extractall only returns captured substrings. In Scala, findAllIn returns the matched values if you do not query its matchData property that in its turn contains a subgroups property.
So, to get the captures only in Scala, you need to use
val pattern = """\((str:[^,()]+),(\d+)\)""".r
val str = "(str:ann smith,14)(str:frank chase,15)"
(pattern findAllIn str).matchData foreach {
m => println(m.subgroups.mkString(","))
}
Output:
str:ann smith,14
str:frank chase,15
See the Scala online demo.
Here, m.subgroups accesses all subgroups (captures) of each match (m).
Also, note you do not need to double backslashes in triple-quoted string literals. \((str:[^,()]+),(\d+)\) matches
\( - a ( char
(str:[^,()]+) - Group 1: str: and one or more chars other than ,, ( and )
, - a comma
(\d+) - Group 2: one or more digits
\) - a ) char.
If you just want to get all matches without captures, you can use
val pattern = """\((str:[^,]+),(\d+)\)""".r
println((pattern findAllIn str).matchData.mkString(","))
Output:
(str:ann smith,14),(str:frank chase,15)
See the online demo.
I would like to select all long words from a string: re.findall("[a-z]{3,}")
However, for a reason I can use substitute only. Hence I need to substitute everything but words of 3 and more letters by space. (e.g. abc de1 fgh ij -> abc fgh)
How would such a regex look like?
The result should be all "[a-z]{3,}" concatenated by spaces. However, you can use substitution only.
Or in Python: Find a regex such that
re.sub(regex, " ", text) == " ".join(re.findall("[a-z]{3,}", text))
Here is some test cases
import re
solution_regex="..."
for test_str in ["aaa aa aaa aa",
"aaa aa11",
"11aaa11 11aa11",
"aa aa1aa aaaa"
]:
expected_str = " ".join(re.findall("[a-z]{3,}", test_str))
print(test_str, "->", expected_str)
if re.sub(solution_regex, " ", test_str)!=expected_str:
print("ERROR")
->
aaa aa aaa aa -> aaa aaa
aaa aa11 -> aaa
11aaa11 11aa11 -> aaa
aa aa1aa aaaa -> aaaa
Note that space is no different than any other symbol.
\b(?:[a-z,A-Z,_]{1,2}|\w*\d+\w*)\b
Explanation:
\b means that the substring you are looking for start and end by border of word
(?: ) - non captured group
\w*\d+\w* Any word that contains at least one digit and consists of digits, '_' and letters
Here you can see the test.
You can use the regex
(\s\b(\d*[a-z]\d*){1,2}\b)|(\s\b\d+\b)
and replace with an empty string, here is a python code for the same
import re
regex = r"(\s\b(\d*[a-z]\d*){1,2}\b)|(\s\b\d+\b)"
test_str = "abcd abc ad1r ab a11b a1 11a 1111 1111abcd a1b2c3d"
subst = ""
# You can manually specify the number of replacements by changing the 4th argument
result = re.sub(regex, subst, test_str, 0)
if result:
print (result)
here is a demo
In Autoit this works for me
#include <Array.au3>
$a = StringRegExp('abc de1 fgh ij 234234324 sdfsdfsdf wfwfwe', '(?i)[a-z]{3,}', 3)
ConsoleWrite(_ArrayToString($a, ' ') & #CRLF)
Result ==> abc fgh sdfsdfsdf wfwfwe
import re
regex = r"(?:^|\s)[^a-z\s]*[a-z]{0,2}[^a-z\s]*(?:\s|$)"
str = "abc de1 fgh ij"
subst = " "
result = re.sub(regex, subst, str)
print (result)
Output:
abc fgh
Explanation:
(?:^|\s) : non capture group, start of string or space
[^a-z\s]* : 0 or more any character that is not letter or space
[a-z]{0,2} : 0, 1 or 2 letters
[^a-z\s]* : 0 or more any character that is not letter or space
(?:\s|$) : non capture group, space or end of string
With the other ideas posted here, I came up with an answer. I can't believe I missed that:
([^a-z]+|(?<![a-z])[a-z]{1,2}(?![a-z]))+
https://regex101.com/r/IIxkki/2
Match either non-letters, or up to two letters bounded by non-letters.
Consider the string "AB 1 BA 2 AB 3 BA". How can I match the content between "AB" and "BA" in a non-greedy fashion (in awk)?
I have tried the following:
awk '
BEGIN {
str="AB 1 BA 2 AB 3 BA"
regex="AB([^B][^A]|B[^A]|[^B]A)*BA"
if (match(str,regex))
print substr(str,RSTART,RLENGTH)
}'
with no output. I believe the reason for no match is that there is an odd number of characters between "AB" and "BA". If I replace str with "AB 11 BA 22 AB 33 BA" the regex seems to work..
Merge your two negated character classes and remove the [^A] from the second alternation:
regex = "AB([^AB]|B|[^B]A)*BA"
This regex fails on the string ABABA, though - not sure if that is a problem.
Explanation:
AB # Match AB
( # Group 1 (could also be non-capturing)
[^AB] # Match any character except A or B
| # or
B # Match B
| # or
[^B]A # Match any character except B, then A
)* # Repeat as needed
BA # Match BA
Since the only way to match an A in the alternation is by matching a character except B before it, we can safely use the simple B as one of the alternatives.
The other answer didn't really answer: how to match non-greedily?
Looks like it can't be done in (G)AWK. The manual says this:
awk (and POSIX) regular expressions always match the leftmost, longest
sequence of input characters that can match.
https://www.gnu.org/software/gawk/manual/gawk.html#Leftmost-Longest
And the whole manual doesn't contain the words "greedy" nor "lazy". It mentions Extended Regular Expressions, but for greedy matching you'd need Perl-Compatible Regular Expressions. So… no, can't be done.
For general expressions, I'm using this as a non-greedy match:
function smatch(s, r) {
if (match(s, r)) {
m = RSTART
do {
n = RLENGTH
} while (match(substr(s, m, n - 1), r))
RSTART = m
RLENGTH = n
return RSTART
} else return 0
}
smatch behaves like match, returning:
the position in s where the regular expression r occurs, or 0 if it does not. The variables RSTART and RLENGTH are set to the position and length of the matched string.
I have a problem building a regex. this is a sample of the text:
text 123 12345 abc 12 def 67 i 89 o 0 t 2
The numbers are sometimes padded with blanks to the max length (3).
e.g.:
"1" can be "1" or "1 "
"13" can be "13" or "13 "
My regex is at the moment this:
\b([\d](\s*)){1,3}\b
The results of this regex are the following: (. = blank for better visibility)
123.
12....
67.
89.
0....
2
But I need this: (. = blank for better visibility)
123
12.
67.
89.
0..
2
How can I tell the regex engine to count the blanks into the {1,3} option?
Try this:
\b(?:\d[\d\s]{0,2})(?:(?<=\s)|\b)
This will also cover strings like text 123 1 23 12345 123abc 12 def 67 i 89 o 0 t 2 and results in:
123
1.
23.
12.
67.
89.
0..
2
Does this do what you want?
\b(\d){1,3}\s*\b
This will also include whitespace (if available) after the selection.
I think you want this
\b(?:\d[\d\s]{0,2})(?!\d)
See it here on Regexr
the word boundary will not work at the end, because if the end of the match is a whitespace, there is no word boundary. Therefor I use a negative lookahead (?!\d) to ensure that there is no digit following.
But if you have a string like this "1 23". It will match only the "2" and the "23", but not the whitespace after the first "2".
Assuming you want to use the padded numbers somewhere else, break the problem apart into two; (simple) parsing the numbers, and (simple) formatting the numbers (including padding).
while ( $text =~ /\b(\d{1,3})\b/g ) {
printf( "%-3d\n", $1 );
}
Alternatively:
#padded_numbers = map { sprintf( "%-3d", $_ ) } ( $text =~ /\b(\d{1,3})\b/g )
I want to split a string like this:
abc//def//ghi
into a part before and after the first occurrence of //:
a: abc
b: //def//ghi
I'm currently using this regex:
(?<a>.*?)(?<b>//.*)
Which works fine so far.
However, sometimes the // is missing in the source string and obviously the regex fails to match. How is it possible to make the second group optional?
An input like abc should be matched to:
a: abc
b: (empty)
I tried (?<a>.*?)(?<b>//.*)? but that left me with lots of NULL results in Expresso so I guess it's the wrong idea.
Try a ^ at the begining of your expression to match the begining of the string and a $ at the end to match the end of the string (this will make the ungreedy match work).
^(?<a>.*?)(?<b>//.*)?$
A proof of Stevo3000's answer (Python):
import re
test_strings = ['abc//def//ghi', 'abc//def', 'abc']
regex = re.compile("(?P<a>.*?)(?P<b>//.*)?$")
for ts in test_strings:
match = regex.match(ts)
print 'a:', match.group('a'), 'b:', match.group('b')
a: abc b: //def//ghi
a: abc b: //def
a: abc b: None
Why use group matching at all? Why not just split by "//", either as a regex or a plain string?
use strict;
my $str = 'abc//def//ghi';
my $short = 'abc';
print "The first:\n";
my #groups = split(/\/\//, $str, 2);
foreach my $val (#groups) {
print "$val\n";
}
print "The second:\n";
#groups = split(/\/\//, $short, 2);
foreach my $val (#groups) {
print "$val\n";
}
gives
The first:
abc
def//ghi
The second:
abc
[EDIT: Fixed to return max 2 groups]