Please suggest a wildcard for below Firstjson list
Firstjson = { p10_7_8 , p10_7_2 , p10_7_3 p10_7_4}
I have tried p10.7.* wildcard for below Secondjson list, it worked. But when I tried p10_7_* for above Firstjson list it did not work
Secondjson = { p10.7.8 , p10.7.2 , p10.7.3 , p10.7.4 }
You are attempting to use wildcard syntax, but Groovy expects regular expression syntax for its pattern matching.
What went wrong with your attempt:
Attempt #1: p10.7.*
A regular expression of . matches any single character and .* matches 0 or more characters. This means:
p10{exactly one character of any kind here}7{zero or more characters of any
kind here}
You didn't realize it, but the . character in your first attempt was acting like a single-character wildcard too. This might match with p10x7abcdefg for example. It also does match p10.7.8 though. But be careful, it also matches p10.78, because the .* expression at the end of your pattern will happily match any sequence of characters, thus any and all characters following p10.7 are accepted.
Attempt #2: p10_7_*
_ matches only a literal underscore. But _* means to match zero or more underscores. It does not mean to match characters of any kind. So p10_7_* matches things like p10_7_______. Literally:
p10_7{zero or more underscores here}
What you can do instead:
You probably want a regular expression like p10_7_\d+
This will match things like p10_7_3 or p10_7_422. It works by matching the literal text p10_7_ followed by one or more digits where a digit is 0 through 9. \d matches any digit, and + means to match one or more of the preceding thing. Literally:
p10_7_{one or more digits here}
Related
I need to find mentions of compensations in emails. I am new to regex. Please see below the approach I am using.
sample_text = "Rate – $115k/yr. or $55/hr. - $60/hr"
My python code to find this,
impor re
PATTERN = r'((\$|\£) [0-9]*)|((\$|\£)[0-9]*)'
print(re.findall(PATTERN,sample_text))
The matches I am getting is
[('', '', '$115', '$'), ('', '', '$55', '$'), ('', '', '$60', '$')]
Expected match
["$115k/yr","$55/hr","$60/hr"]
Also the $ sign can be written as USD. How do I handle this in the same regex.
You can use
[$£]\d+[^.\s]*
[$£] Match either $ or £
\d+ Match 1+ digits
[^.\s]* Repeat 0+ times matching any char except . or a whitespace
Regex demo
import re
sample_text = "Rate – $115k/yr. or $55/hr. - $60/hr"
PATTERN = r'[$£]\d+[^.\s]*'
print(re.findall(PATTERN,sample_text))
Output
['$115k/yr', '$55/hr', '$60/hr']
If there must be a / present, you might also use
[$£]\d+[^\s/.]*/\w+
Regex demo
You can have something like:
[$£]\d+[^.]+
>>> PATTERN = '[$£]\d+[^.]+
>>> print(re.findall(PATTERN,sample_text))
['$115k/yr', '$55/hr', '$60/hr']
[$£] matches "$" or a "£"
\d+ matches one or more digits
[^.]+ matches everything that's not a "."
The parentheses in your regex cause the engine to report the contents of each parenthesized subexpression. You can use non-grouping parentheses (?:...) to avoid this, but of course, your expression can be rephrased to not have any parentheses at all:
PATTERN = r'[$£]\s*\d+'
Notice also how I changed the last quantifier to a + -- your attempt would also find isolated currency symbols with no numbers after them.
To point out the hopefully obvious, \s matches whitespace and \s* matches an arbitrary run of whitespace, including none at all; and \d matches a digit.
If you want to allow some text after the extracted match, add something like (?:/\w+)? to allow for a slash and one single word token as an optional expression after the main match. (Maybe adorn that with \s* on both sides of the slash, too.)
I've been trying for about two hours now to write a regular expression which matches a single character that's not preceded or followed by the same character.
This is what I've got: (\d)(?<!\1)\1(?!\1); but it doesn't seem to work! (testing at https://regex101.com/r/whnj5M/6)
For example:
In 1111223 I would expect to match the 3 at the end, since it's not preceded or followed by another 3.
In 1151223 I would expect to match the 5 in the middle, and the 3 at the end for the same reasons as above.
The end goal for this is to be able to find pairs (and only pairs) of characters in strings (e.g. to find 11 in 112223 or 44 in 123544) and I was going to try and match single isolated characters, and then add a {2} to it to find pairs, but I can't even seem to get isolated characters to match!
Any help would be much appreciated, I thought I knew RegEx pretty well!
P.S. I'm testing in JS on regex101.com because it wouldn't let me use variable length lookbacks in Python on there, and I'm using the regex library to allow for this in my actual implementation.
Your regex is close, but by using simply (\d) you are consuming characters, which prevents the other match from occurring. Instead, you can use a positive lookahead to set the capture group and then test for any occurrences of the captured digit not being surrounded by copies of itself:
(?=.*?(.))(?<!\1)\1(?!\1)
By using a lookahead you avoid consuming any characters and so the regex can match anywhere in the string.
Note that in 1151223 this returns 5, 1 and 3 because the third 1 is not adjacent to any other 1s.
Demo on regex101 (requires JS that supports variable width lookbehinds)
The pattern you tried does not match because this part (\d)(?<!\1) can not match.
It reads as:
Capture a digit in group 1. Then, on the position after that captured
digit, assert what is captured should not be on the left.
You could make the pattern work by adding for example a dot after the backreference (?<!\1.) to assert that the value before what you have just matched is not the same as group 1
Pattern
(\d)(?<!\1.)\1(?!\1)
Regex demo | Python demo
Note that you have selected ECMAscript on regex101.
Python re does not support variable width lookbehind.
To make this work in Python, you need the PyPi regex module.
Example code
import regex
pattern = r"(\d)(?<!\1.)\1(?!\1)"
test_str = ("1111223\n"
"1151223\n\n"
"112223\n"
"123544")
matches = regex.finditer(pattern, test_str)
for matchNum, match in enumerate(matches, start=1):
print(match.group())
Output
22
11
22
11
44
#Theforthbird has provided a good explanation for why your regular explanation does not match the characters of interest.
Each character matched by the following regular expression is neither preceded nor followed by the same character (including characters at the beginning and end of the string).
r'^.$|^(.)(?!\1)|(?<=(.))(?!\2)(.)(?!\3)'
Demo
Python's re regex engine performs the following operations.
^.$ match the first char if it is the only char in the line
| or
^ match beginning of line
(.) match a char in capture group 1...
(?!\1) ...that is not followed by the same character
| or
(?<=(.)) save the previous char in capture group 2...
(?!\2) ...that is not equal to the next char
(.) match a character and save to capture group 3...
(?!\3) ...that is not equal to the following char
Suppose the string were "cat".
The internal string pointer is initially at the beginning of the line.
"c" is not at the end of the line so the first part of the alternation fails and the second part is considered.
"c" is matched and saved to capture group 1.
The negative lookahead asserting that "c" is not followed by the content of capture group 1 succeeds, so "c" is matched and the internal string pointer is advanced to a position between "c" and "a".
"a" fails the first two parts of the assertion so the third part is considered.
The positive lookbehind (?<=(.)) saves the preceding character ("c") in capture group 2.
The negative lookahead (?!\2), which asserts that the next character ("a") is not equal to the content of capture group 2, succeeds. The string pointer remains just before "a".
The next character ("a") is matched and saved in capture group 3.
The negative lookahead (?!\3), which asserts that the following character ("t") does not equal the content of capture group 3, succeeds, so "a" is matched and the string pointer advances to just before "t".
The same steps are performed when evaluating "t" as were performed when evaluating "a". Here the last token ((?!\3)) succeeds, however, because no characters follow "t".
I'm looking for a clean way to identify occurrences of [variableName] followed by the exact string .add(.
A variable name is a string which contains one or more characters from a-z, A-Z, 0-9 and an underscore.
One more thing is that it cannot start with any of the characters from 0-9, but I don't mind ignoring this condition because there are no such cases in the text that I need to parse anyway.
I've been following several tutorials, but the farthest I got was finding all occurrences of what I've referred to above as "variableName":
import re
txt = "The _rain() in+ Spain5"
x = re.split("[^a-zA-Z0-9_]+", txt)
print(x)
What is the right way to do it?
You may use
re.findall(r'\w+(?=\.add\()', txt, flags=re.ASCII)
The regex matches:
\w+ - 1+ word chars (due to re.ASCII, it only matches [A-Za-z0-9_] chars)
(?=\.add\() - a positive lookahead that matches a location immediately followed with .add( substring.
I have two scenarios to match . Length should be exactly 16.
Pattern should contain A-F,a-f,0-9 and '-' in 1st case.
AC-DE-48-23-45-67-AB-CD
ACDE48234567ABCD
I have tried with r'^([0-9A-Fa-f]{16})$|(([0-9A-Fa-f]{2}\-){7}[0-9A-Fa-f]{2})$'this , which is working fine . Looking for better expression .
You can simplify the regex by considering the string to be a group of two hex digits followed by an optional -, followed by 6 similar groups (i.e. if the first group had a -, the subsequent ones must too), followed by a group of 2 hex digits:
^[0-9A-Fa-f]{2}(-?)([0-9A-Fa-f]{2}\1){6}[0-9A-Fa-f]{2}$
Use of the re.I flag allows you to remove the a-f from the character classes:
^[0-9A-F]{2}(-?)([0-9A-F]{2}\1){6}[0-9A-F]{2}$
You can also simplify slightly further by replacing 0-9 by \d in the character classes (although personally I find 0-9 easier to read):
^[\dA-F]{2}(-?)([\dA-F]{2}\1){6}[\dA-F]{2}$
Demo on regex101
Sample python code:
import re
strs = ['AC-DE-48-23-45-67-AB-CD',
'ACDE48234567ABCD',
'AC-DE48-23-45-67-AB-CD',
'ACDE48234567ABC',
'ACDE48234567ABCDE']
for s in strs:
print(s + (' matched' if re.match(r'^[0-9A-F]{2}(-?)([0-9A-F]{2}\1){6}[0-9A-F]{2}$', s, re.I) else ' didn\'t match'))
Output
AC-DE-48-23-45-67-AB-CD matched
ACDE48234567ABCD matched
AC-DE48-23-45-67-AB-CD didn't match
ACDE48234567ABC didn't match
ACDE48234567ABCDE didn't match
I'm trying to find exact matches of strings in Lua including, special characters. I want the example below to return that it is an exact match, but because of the - character it returns nil
index = string.find("test-string", "test-string")
returns nil
index = string.find("test-string", "test-")
returns 1
index = string.find("test-string", "test")
also returns 1
How can I get it to do full matching?
- is a pattern operator in a Lua string pattern, so when you say test-string, you're telling find() to match the string test as few times as possible. So what happens is it looks at test-string, sees test in there, and since - isn't an actual minus sign in this case, it's really looking for teststring.
Do as Mike has said and escape it with the % character.
I found this helpful for better understanding patterns.
You can also ask for a plain substring match that ignores magic characters:
string.find("test-string", "test-string",1,true)
you need to escape special characters in the pattern with the % character.
so in this case you are looking for
local index = string.find('test-string', 'test%-string')