I trying to match the following using regular expression but I struggling in matching the round bracket.
[??(Z)Z-axis Down Position Stroke]
Can anyone kindly advise ?
My Current expression as shown.
[[][[a-zA-Z0-9_?. - ]{1,30}[]]
You can add a backslash \ before any character to escape it. Try this regex:
\[\?\?\(Z\)Z-axis Down Position Stroke\]
When writing regex, I find regex101.com to be really helpful. It's a free website that evaluates your regex and lets you specify test cases etc, then breaks those down and tells you about the various matching conditions in them. Worth a look if you're learning regular expressions.
Edit: Also, it's necessary to escape the brackets, parentheses, and question marks because those all have special meaning in regex.
Related
I am trying to use lookbehinds in a regular expression and it doesn't seem to work as I expected. So, this is not my real usage, but to simplify I will put an example. Imagine I want to match "example" on a string that says "this is an example". So, according to my understanding of lookbehinds this should work:
(?<=this\sis\san\s*?)example
What this should do is find "this is an", then space characters and finally match the word "example". Now, it doesn't work and I don't understand why, is it impossible to use '+' or '*' inside lookbehinds?
I also tried those two and they work correctly, but don't fulfill my needs:
(?<=this\sis\san\s)example
this\sis\san\s*?example
I am using this site to test my regular expressions: http://gskinner.com/RegExr/
Many regular expression libraries do only allow strict expressions to be used in look behind assertions like:
only match strings of the same fixed length: (?<=foo|bar|\s,\s) (three characters each)
only match strings of fixed lengths: (?<=foobar|\r\n) (each branch with fixed length)
only match strings with a upper bound length: (?<=\s{,4}) (up to four repetitions)
The reason for these limitations are mainly because those libraries can’t process regular expressions backwards at all or only a limited subset.
Another reason could be to avoid authors to build too complex regular expressions that are heavy to process as they have a so called pathological behavior (see also ReDoS).
See also section about limitations of look-behind assertions on Regular-Expressions.info.
Hey if your not using python variable look behind assertion you can trick the regex engine by escaping the match and starting over by using \K.
This site explains it well .. http://www.phpfreaks.com/blog/pcre-regex-spotlight-k ..
But pretty much when you have an expression that you match and you want to get everything behind it using \K will force it to start over again...
Example:
string = '<a this is a tag> with some information <div this is another tag > LOOK FOR ME </div>'
matching /(\<a).+?(\<div).+?(\>)\K.+?(?=\<div)/ will cause the regex to restart after you match the ending div tag so the regex won't include that in the result. The (?=\div) will make the engine get everything in front of ending div tag
What Amber said is true, but you can work around it with another approach: A non-capturing parentheses group
(?<=this\sis\san)(?:\s*)example
That make it a fixed length look behind, so it should work.
You can use sub-expressions.
(this\sis\san\s*?)(example)
So to retrieve group 2, "example", $2 for regex, or \2 if you're using a format string (like for python's re.sub)
Most regex engines don't support variable-length expressions for lookbehind assertions.
Yesterday, I needed to add a file path to a regular expression creating a pattern like this:
"some_pattern/%s/more_pattern" % some_path
In the beginning the regular expression did not match, because some_path contained
several regex specific symbols like ? or .. As a quick fix I replaced them with [?]{1} and . with \..
However, I asked myself if there isn't a more reliable or better way to clean a string from regex specific symbols.
Is such functionality supported in the Python Standard library?
If not, do you know a regular expression to identify all regex symbols and clean them through a substitute?
Just apply re.escape and you'll be fine.
re.escape(string)
Return string with all non-alphanumerics
backslashed; this is useful if you want to match an arbitrary literal
string that may have regular expression metacharacters in it.
I have an HTML to LaTeX parser tailored to what it's supposed to do (convert snippets of HTML into snippets of LaTeX), but there is a little issue with filling in variables. The issue is that variables should be allowed to contain the LaTeX reserved characters (namely # $ % ^ & _ { } ~ \). These need to be escaped so that they won't kill our LaTeX renderer.
The program that handles the conversion and everything is written in Python, so I tried to find a nice solution. My first idea was to simply do a .replace(), but replace doesn't allow you to match only if the first is not a \. My second attempt was a regex, but I failed miserably at that.
The regex I came up with is ([^\][#\$%\^&_\{\}~\\]). I hoped that this would match any of the reserved characters, but only if it didn't have a \ in front. Unfortunately, this matches ever single character in my input text. I've also tried different variations on this regex, but I can't get it to work. The variations mainly consisted of removing/adding slashes in the second part of the regex.
Can anyone help with this regex?
EDIT Whoops, I seem to have included the slashes as well. Shows how awake I was when I posted this :) They shouldn't be escaped in my case, but it's relatively easy to remove them from the regexes in the answers. Thanks all!
The [^\] is a character class for anything not a \, that is why it is matching everything. You want a negative lookbehind assertion:
((?<!\)[#\$%\^&_\{\}~\\])
(?<!...) will match whatever follows it as long as ... is not in front of it. You can check this out at the python docs
The regex ([^\][#\$%\^&_\{\}~\\]) is matching anything that isn't found between the first [ and the last ], so it should be matching everything except for what you want it to.
Moving around the parenthesis should fix your original regex ([^\\])[#\$%\^&_\{\}~\\].
I would try using regex lookbehinds, which won't match the character preceding what you want to escape. I'm not a regex expert so perhaps there is a better pattern, but this should work (?<!\\)[#\$%\^&_\{\}~\\].
If you're looking to find special characters that aren't escaped, without eliminating special chars preceded by escaped backslashes (e.g. you do want to match the last backslash in abc\\\def), try this:
(?<!\\)(\\\\)*[#\$%\^&_\{\}~\\]
This will match any of your special characters preceded by an even number (this includes 0) of backslashes. It says the character can be preceded by any number of pairs of backslashes, with a negative lookbehind to say those backslashes can't be preceded by another backslash.
The match will include the backslashes, but if you stick another in front of all of them, it'll achieve the same effect of escaping the special char, anyway.
I recently learned a little Python and I couldnt find a good list of the RegEx's (don't know if that is the correct plural tense...) with complete explanations even a rookie will understand :)
Anybody know a such list?
Vide:
Well, for starters - hit up the python docs on the re module. Good list of features and methods, as well as info about special regex characters such as \w. There's also a chapter in Dive into Python about regular expressions that uses the aforementioned module.
Check out the re module docs for some basic RegEx syntax.
For more, read Introduction To RegEx, or other of the many guides online. (or books!)
You could also try RegEx Buddy, which helps you learn regular expressions by telling you what they do an parsing them.
The Django Book http://www.djangobook.com/en/2.0/chapter03/ chapter on urls/views has a great "newbie" friendly table explaining the gist of regexes. combine that with the info on the python.docs http://docs.python.org/library/re.html and you'll master RegEx in no time.
an excerpt:
Regular Expressions
Regular expressions (or regexes) are a compact way of specifying patterns in text. While Django URLconfs allow arbitrary regexes for powerful URL matching, you’ll probably only use a few regex symbols in practice. Here’s a selection of common symbols:
Symbol Matches
. (dot) Any single character
\d Any single digit
[A-Z] Any character between A and Z (uppercase)
[a-z] Any character between a and z (lowercase)
[A-Za-z] Any character between a and z (case-insensitive)
+ One or more of the previous expression (e.g., \d+ matches one or more digits)
? Zero or one of the previous expression (e.g., \d? matches zero or one digits)
* Zero or more of the previous expression (e.g., \d* matches zero, one or more than one >digit)
{1,3} Between one and three (inclusive) of the previous expression (e.g., \d{1,3} matches >one, two or three digits)
But it's turtles all the way down!
I am battling regular expressions now as I type.
I would like to determine a pattern for the following example file: b410cv11_test.ext. I want to be able to do a search for files that match the pattern of the example file aforementioned. Where do I start (so lost and confused) and what is the best way of arriving at a solution that best matches the file pattern? Thanks in advance.
Further clarification of question:
I would like the pattern to be as follows: must start with 'b', followed by three digits, followed by 'cv', followed by two digits, then an underscore, followed by 'release', followed by .'ext'
Now that you have a human readable description of your file name, it's quite straight forward to translate it into a regular expression (at least in this case ;)
must start with
The caret (^) anchors a regular expression to the beginning of what you want to match, so your re has to start with this symbol.
'b',
Any non-special character in your re will match literally, so you just use "b" for this part: ^b.
followed by [...] digits,
This depends a bit on which flavor of re you use:
The most general way of expressing this is to use brackets ([]). Those mean "match any one of the characters listed within. [ASDF] for example would match either A or S or D or F, [0-9] would match anything between 0 and 9.
Your re library probably has a shortcut for "any digit". In sed and awk you could use [[:digit:]] [sic!], in python and many other languages you can use \d.
So now your re reads ^b\d.
followed by three [...]
The most simple way to express this would be to just repeat the atom three times like this: \d\d\d.
Again your language might provide a shortcut: braces ({}). Sometimes you would have to escape them with a backslash (if you are using sed or awk, read about "extended regular expressions"). They also give you a way to say "at least x, but no more than y occurances of the previous atom": {x,y}.
Now you have: ^b\d{3}
followed by 'cv',
Literal matching again, now we have ^b\d{3}cv
followed by two digits,
We already covered this: ^b\d{3}cv\d{2}.
then an underscore, followed by 'release', followed by .'ext'
Again, this should all match literally, but the dot (.) is a special character. This means you have to escape it with a backslash: ^\d{3}cv\d{2}_release\.ext
Leaving out the backslash would mean that a filename like "b410cv11_test_ext" would also match, which may or may not be a problem for you.
Finally, if you want to guarantee that there is nothing else following ".ext", anchor the re to the end of the thing to match, use the dollar sign ($).
Thus the complete regular expression for your specific problem would be:
^b\d{3}cv\d{2}_release\.ext$
Easy.
Whatever language or library you use, there has to be a reference somewhere in the documentation that will show you what the exact syntax in your case should be. Once you have learned to break down the problem into a suitable description, understanding the more advanced constructs will come to you step by step.
To avoid confusion, read the following, in order.
First, you have the glob module, which handles file name regular expressions just like the Windows and unix shells.
Second, you have the fnmatch module, which just does pattern matching using the unix shell rules.
Third, you have the re module, which is the complete set of regular expressions.
Then ask another, more specific question.
I would like the pattern to be as
follows: must start with 'b', followed
by three digits, followed by 'cv',
followed by two digits, then an
underscore, followed by 'release',
followed by .'ext'
^b\d{3}cv\d{2}_release\.ext$
Your question is a bit unclear. You say you want a regular expression, but could it be that you want a glob-style pattern you can use with commands like ls? glob expressions and regular expressions are similar in concept but different in practice (regular expressions are considerably more powerful, glob style patterns are easier for the most common cases when looking for files.
Also, what do you consider to be the pattern? Certainly, * (glob) or .* (regex) will match the pattern. Also, _test.ext (glob) or ._test.ext (regexp) pattern would match, as would many other variations.
Can you be more specific about the pattern? For example, you might describe it as "b, followed by digits, followed by cv, followed by digits ..."
Once you can precisely explain the pattern in your native language (and that must be your first step), it's usually a fairly straight-forward task to translate that into a glob or regular expression pattern.
if the letters are unimportant, you could try \w\d\d\d\w\w\d\d_test.ext which would match the letter/number pattern, or b\d\d\dcv\d\d_test.ext or some mix of the two.
When working with regexes I find the Mochikit regex example to be a great help.
/^b\d\d\dcv\d\d_test\.ext$/
Then use the python re (regex) module to do the match. This is of course assuming regex is really what you need and not glob as the others mentioned.