I'm looking for a regular expression that matches the first two specific fields (variable strings written in Perl). In a file a line without a comment # starts with any character, length unspecific followed by a whitespace and another nonspecific length string followed by a whitespace: name info data1 data2 data3.
The following works for matching the second field only but I want the first two fields to match exactly: /^[^#].*\s$INFO\s/ where $INFO="info". I tried variations of the above to no avail. My first attempt was this: /^[^#]$NAME\s$INFO\s/ which seemed logical to me if $NAME="name" for the above record.
My first attempt was this: /^[^#]$NAME\s$INFO\s/
This won't work because (implied from the question) the character before $NAME is either # or nothing. As such you just need to remove that first [^#]:
/^$NAME\s$INFO\s/
Which will match the string:
"$NAME $INFO <whatever or nothing>"
Although I'm not a regex expert, this may work (I also am not clear on the precise details of the question so I made some assumptions):
'$NAME=name #$INFO=info $DATA=data1 data2 data3'.replace(/#[\S]+/g,'').match(/\$[\S]+/g);
This returns an array. The first 2 elements are the 'fields' i.e. [0]='$NAME=name' AND [1]='$DATA=data1'
Hope that helps at all. And apologies to the gods for my regex.
Related
In R, I have a column of data in a data-frame, and each element looks something like this:
Bacteria;Bacteroidetes;Bacteroidia;Bacteroidales;Marinilabiaceae
What I want is the section after the last semicolon, and I've been trying to use 'sub' and also duplicating the existing column and create a new one with just the endings kept. In essence, I want this (the genus):
Marinilabiaceae
A snippet of the code looks like this:
mydata$new_column<- sub("([\\s\\S]*;)", "", mydata$old_column)
In this situation, I am using \\ rather than \ because of R's escape sequences. The sub replaces the parts I don't want and updates it to the new column. I've tested the Regex several times in places such as this: http://regex101.com/r/kS7fD8/1
However, I'm still struggling because the results are very bizarre. Now my new column is populated with the organism's domain rather than the genus: Bacteria.
How do I resolve this? Are there any good easy-to-understand resources for learning more about R's Regex formats?
Starting with your simple string,
string <- "Bacteria;Bacteroidetes;Bacteroidia;Bacteroidales;Marinilabiaceae"
You can remove everything up to the last semicolon with "^(.*);" in your call to sub
> sub("^(.*);", "", string)
# [1] "Marinilabiaceae"
You can also use strsplit with tail
> tail(strsplit(string, ";")[[1]], 1)
# [1] "Marinilabiaceae"
Your regular expression, ([\\s\\S]*;) wouldn't work primarily because \\s matches any space characters, and your string does not contain any spaces. I think it worked in the regex101 site because that regex tester defaults to pcre (php) (see "Flavor" in top-left corner), and R regex syntax is slightly different. R requires extra backslash escape characters in many situations. For reference, this R text processing wiki has come in handy for me many times before.
Make it Greedy and get the matched group from desired index.
(.*);(.*)
^^^------- Marinilabiaceae
Here is regex101 demo
Or to get the first word use Non-Greedy way
(.*?);(.*)
Bacteria -----^^^
Here is demo
To extract everything after the last ; to the end of the line you can use:
[^;]*?$
I'm really bad at regex, I have:
/(#[A-Za-z-]+)/
which finds words after the # symbol in a textbox, however I need it to ignore email addresses, like:
foo#things.com
however it finds #things
I also need it to include numbers, like:
#He2foo
however it only finds the #He part.
Help is appreciated, and if you feel like explaining regex in simple terms, that'd be great :D
/(?:^|(?<=\s))#([A-Za-z0-9]+)(?=[.?]?\s)/
#This (matched) regex ignores#this but matches on #separate tokens as well as tokens at the end of a sentence like #this. or #this? (without picking the . or the ?) And yes email#addresses.com are ignored too.
The regex while matching on # also lets you quickly access what's after it (like userid in #userid) by picking up the regex group(1). Check PHP documentation on how to work with regex groups.
You can just add 0-9 to your regex, like so:
/(#[A-Za-z0-9-]+)/
Don't think any more explanation is needed since you've been able to come this far by yourself. 0-9 is just like a-z (though numeric ofcourse).
In order to ignore emailaddresses you will need to provide more specific requirements. You could try preceding # with (^| ) which basically states that your value MUST be preceeded by either the start of the string (so nothing really, though at the start) or a space.
Extending this you can also use ($| ) on the end to require the value to be followed by the end of the string or a space (which means there's no period allowed, which is requirement for a valid emailaddress).
Update
$subject = "#a #b a#b a# #b";
preg_match_all("/(^| )#[A-Za-z0-9-]+/", $subject, $matches);
print_r($matches[0]);
I am trying to figure out the best regex to simply match only the last two strings in a url.
For instance with www.stackoverflow.com I just want to match stackoverflow.com
The issue i have is some strings can have a large number of periods for instance
a-abcnewsplus.i-a277eea3.rtmp.atlas.cdn.yimg.com
should also return only yimg.com
The set of URLS I am working with does not have any of the path information so one can assume the last part of the string is always .org or .com or something of that nature.
What regular expresion will return stackoverflow.com when run against www.stackoverflow.com and will return yimg.com when run against a-abcnewsplus.i-a277eea3.rtmp.atlas.cdn.yimg.com
under the condtions above?
You don't have to use regex, instead you can use a simple explode function.
So you're looking to split your URL at the periods, so something like
$url = "a-abcnewsplus.i-a277eea3.rtmp.atlas.cdn.yimg.com";
$url_split = explode(".",$url);
And then you need to get the last two elements, so you can echo them out from the array created.
//this will return the second to last element, yimg
echo $url_split[count($url_split)-2];
//this will echo the period
echo ".";
//this will return the last element, com
echo $url_split[count($url_split)-1];
So in the end you'll get yimg.com as the final output.
Hope this helps.
I don't know what did you try so far, but I can offer the following solution:
/.*?([\w]+\.[\w]+)$/
There are a couple of tricks here:
Use $ to match till the end of the string. This way you'll be sure your regex engine won't catch the match from the very beginning.
Use grouping inside (...). In fact it means the following: match word that contains at least one letter then there should be a dot (backslashed because dot has a special meaning in regex and we want it 'as is' and then again series of letters with at least one of letters).
Use reluctant search in the beginning of the pattern, because otherwise it will match everything in a greedy manner, for example, if your text is :
abc.def.gh
the greedy match will give f.gh in your group, and its not what you want.
I assumed that you can have only letters in your host (\w matches the word, maybe in your example you will need something more complicated).
I post here a working groovy example, you didn't specify the language you use but the engine should be similar.
def s = "abc.def.gh"
def m = s =~/.*?([\w]+\.[\w]+)$/
println m[0][1] // outputs the first (and the only you have) group in groovy
Hope this helps
if you needed a solution in a Perl Regular Expression compatible way that will work in a number of languages, you can use something like that - the example is in PHP
$url = "a-abcnewsplus.i-a277eea3.rtmp.atlas.cdn.yimg.com";
preg_match('|[a-zA-Z-0-9]+\.[a-zA-Z]{2,3}$|', $url, $m);
print($m[0]);
This regex guarantees you to fetch the last part of the url + domain name. For example, with a-abcnewsplus.i-a277eea3.rtmp.atlas.cdn.yimg.com this produces
yimg.com
as an output, and with www.stackoverflow.com (with or without preceding triple w) it gives you
stackoverflow.com
as a result
A shorter version
/(\.[^\.]+){2}$/
Before somebody points me to that question, I know that one can't parse html with regex :) And this is not what I am trying to do.
What I need is:
Input: a string containing html.
Output: replace all opening tags
***<tag>
So if I get
<a><b><c></a></b></c>, I want
***<a>***<b>***<c></a></b></c>
as output.
I've tried something like:
(<[~/].+>)
and replace it with
***$1
But doesn't really seem to work the way I want it to. Any pointers?
Clarification: it's guaranteed that there are no self closing tags nor comments in the input.
You just have two problems: ^ is the character to exclude items from a character class, not ~; and the .+ is greedy, so will match as many characters as possible before the final >. Change it to:
(<[^/].+?>)
You can also probably drop the parentheses and replace with $0 or $&, depending on the language.
Try using: (<[^/].*?>) and replace it with ***$1
This question already has answers here:
Regular expression to match a line that doesn't contain a word
(34 answers)
Closed 6 years ago.
Ok, so this is something completely stupid but this is something I simply never learned to do and its a hassle.
How do I specify a string that does not contain a sequence of other characters. For example I want to match all lines that do NOT end in '.config'
I would think that I could just do
.*[^(\.config)]$
but this doesn't work (why not?)
I know I can do
.*[^\.][^c][^o][^n][^f][^i][^g]$
but please please please tell me that there is a better way
You can use negative lookbehind, e.g.:
.*(?<!\.config)$
This matches all strings except those that end with ".config"
Your question contains two questions, so here are a few answers.
Match lines that don't contain a certain string (say .config) at all:
^(?:(?!\.config).)*$\r?\n?
Match lines that don't end in a certain string:
^.*(?<!\.config)$\r?\n?
and, as a bonus: Match lines that don't start with a certain string:
^(?!\.config).*$\r?\n?
(each time including newline characters, if present.
Oh, and to answer why your version doesn't work: [^abc] means "any one (1) character except a, b, or c". Your other solution would also fail on test.hg (because it also ends in the letter g - your regex looks at each character individually instead of the entire .config string. That's why you need lookaround to handle this.
(?<!\.config)$
:)
By using the [^] construct, you have created a negated character class, which matches all characters except those you have named. Order of characters in the candidate match do not matter, so this will fail on any string that has any of [(\.config) (or [)gi.\onc(])
Use negative lookahead, (with perl regexs) like so: (?!\.config$). This will match all strings that do not match the literal ".config"
Unless you are "grepping" ... since you are not using the result of a match, why not search for the strings that do end in .config and skip them? In Python:
import re
isConfig = re.compile('\.config$')
# List lst is given
filteredList = [f.strip() for f in lst if not isConfig.match(f.strip())]
I suspect that this will run faster than a more complex re.
As you have asked for a "better way": I would try a "filtering" approach. I think it is quite easy to read and to understand:
#!/usr/bin/perl
while(<>) {
next if /\.config$/; # ignore the line if it ends with ".config"
print;
}
As you can see I have used perl code as an example. But I think you get the idea?
added:
this approach could also be used to chain up more filter patterns and it still remains good readable and easy to understand,
next if /\.config$/; # ignore the line if it ends with ".config"
next if /\.ini$/; # ignore the line if it ends with ".ini"
next if /\.reg$/; # ignore the line if it ends with ".reg"
# now we have filtered out all the lines we want to skip
... process only the lines we want to use ...
I used Regexpal before finding this page and came up with the following solution when I wanted to check that a string doesn't contain a file extension:
^(.(?!\.[a-zA-Z0-9]{3,}))*$ I used the m checkbox option so that I could present many lines and see which of them did or did not match.
so to find a string that doesn't contain another "^(.(?!" + expression you don't want + "))*$"
My article on the uses of this particular regex