Preconditions:
[numbers]
[vip]111,222[vip]
[standard]333[standard]
[numbers]
What I want:
Find everything between [numbers]
Problem:
When this text is in one line the solution is simple
(?<=\[numbers\])(.*?)(?=\[numbers\])
But it is possible to search when new line are like in preconditions?
in most regex varieties the dot (.) stands for anything ON THE LINE.
You can scan across end-of-lines by using an expression for "anything", for example:
(?<=\[numbers\]) ( [\d\D]*? ) (?=\[numbers\])
as [\d\D] stands for 'anything that is a digit, or anything that is NOT a digit'
You don't really need a regex for the presented problem, with awk it's enough to set the record separator:
awk 1 RS='\\[numbers\\]\n' ORS=''
Output:
[vip]111,222[vip]
[standard]333[standard]
Related
I'm needing to write some regex that takes a number and removes any trailing zeros after a decimal point. The language is Actionscript 3. So I would like to write:
var result:String = theStringOfTheNumber.replace( [ the regex ], "" );
So for example:
3.04000 would be 3.04
0.456000 would be 0.456 etc
I've spent some time looking at various regex websites and I'm finding this harder to resolve than I initially thought.
Regex:
^(\d+\.\d*?[1-9])0+$
OR
(\.\d*?[1-9])0+$
Replacement string:
$1
DEMO
Code:
var result:String = theStringOfTheNumber.replace(/(\.\d*?[1-9])0+$/g, "$1" );
What worked best for me was
^([\d,]+)$|^([\d,]+)\.0*$|^([\d,]+\.[0-9]*?)0*$
For example,
s.replace(/^([\d,]+)$|^([\d,]+)\.0*$|^([\d,]+\.[0-9]*?)0*$/, "$1$2$3");
This changes
1.10000 => 1.1
1.100100 => 1.1001
1.000 => 1
1 >= 1
What about stripping the trailing zeros before a \b boundary if there's at least one digit after the .
(\.\d+?)0+\b
And replace with what was captured in the first capture group.
$1
See test at regexr.com
(?=.*?\.)(.*?[1-9])(?!.*?\.)(?=0*$)|^.*$
Try this.Grab the capture.See demo.
http://regex101.com/r/xE6aD0/11
Other answers didn't consider numbers without fraction (like 1.000000 ) or used a lookbehind function (sadly, not supported by implementation I'm using). So I modified existing answers.
Match using ^-?\d+(\.\d*[1-9])? - Demo (see matches). This will not work with numbers in text (like sentences).
Replace(with \1 or $1) using (^-?\d+\.\d*[1-9])(0+$)|(\.0+$) - Demo (see substitution). This one will work with numbers in text (like sentences) if you remove the ^ and $.
Both demos with examples.
Side note: Replace the \. with decimal separator you use (, - no need for slash) if you have to, but I would advise against supporting multiple separator formats within such regex (like (\.|,)). Internal formats normally use one specific separator like . in 1.135644131 (no need to check for other potential separators), while external tend to use both (one for decimals and one for thousands, like 1.123,541,921), which would make your regex unreliable.
Update: I added -? to both regexes to add support for negative numbers, which is not in demo.
If your regular expressions engine doesn't support "lookaround" feature then you can use this simple approach:
fn:replace("12300400", "([^0])0*$", "$1")
Result will be: 123004
I know I am kind of late but I think this can be solved in a far more simple way.
Either I miss something or the other repliers overcomplicate it, but I think there is a far more straightforward yet resilient solution RE:
([0-9]*[.]?([0-9]*[1-9]|[0]?))[0]*
By backreferencing the first group (\1) you can get the number without trailing zeros.
It also works with .XXXXX... and ...XXXXX. type number strings. For example, it will convert .45600 to .456 and 123. to 123. as well.
More importantly, it leaves integer number strings intact (numbers without decimal point). For example, it will convert 12300 to 12300.
Note that if there is a decimal point and there are only zeroes after that it will leave only one trailing zeroes. For example for the 42.0000 you get 42.0.
If you want to eliminate the leading zeroes too then youse this RE (just put a [0]* at the start of the former):
[0]*([0-9]*[.]?([0-9]*[1-9]|[0]?))[0]*
I tested few answers from the top:
^(\d+\.\d*?[1-9])0+$
(\.\d*?[1-9])0+$
(\.\d+?)0+\b
All of them not work for case when there are all zeroes after "." like 45.000 or 450.000
modified version to match that case: (\.\d*?[1-9]|)\.?0+$
also need to replace to '$1' like:
preg_replace('/(\.\d*?[1-9]|)\.?0+$/', '$1', $value);
try this
^(?!0*(\.0+)?$)(\d+|\d*\.\d+)$
And read this
http://www.regular-expressions.info/numericranges.html it might be helpful.
I know it's not what the original question is looking for, but anyone who is looking to format money and would only like to remove two consecutive trailing zeros, like so:
£30.00 => £30
£30.10 => £30.10 (and not £30.1)
30.00€ => 30€
30.10€ => 30.10€
Then you should be able to use the following regular expression which will identify two trailing zeros not followed by any other digit or exist at the end of a string.
([^\d]00)(?=[^\d]|$)
I'm a bit late to the party, but here's my solution:
(((?<=(\.|,)\d*?[1-9])0+$)|(\.|,)0+$)
My regular expression will only match the trailing 0s, making it easy to do a .replaceAll(..) type function.
Breaking it down, part one: ((?<=(\.|,)\d*?[1-9])0+$)
(?<=(\.|,): A positive look behind. Decimal must contain a . or a , (commas are used as a decimal point in some countries). But as its a look behind, it is not included in the matched text, but still must be present.
\d*?: Matches any number of digits lazily
[1-9]: Matches a single non-zero character (this will be the last digit before trailing 0s)
0+$: Matches 1 or more 0s that occur between the last non-zero digit and the line end.
This works great for everything except the case where trailing 0s begin immediately, like in 1.0 or 5.000. The second part fixes this (\.|,)0+$:
(\.|,): Matches a . or a , that will be included in matched text.
0+$ matches 1 or more 0s between the decimal point and the line end.
Examples:
1.0 becomes 1
5.0000 becomes 5
5.02394900022000 becomes 5.02394900022
Is it really necessary to use regex? Why not just check the last digits in your numbers? I am not familiar with Actionscript 3, but in python I would do something like this:
decinums = ['1.100', '0.0','1.1','10']
for d in decinums:
if d.find('.'):
while d.endswith('0'):
d = d[:-1]
if d.endswith('.'):
d = d[:-1]
print(d)
The result will be:
1.1
0
1.1
10
I have this very long cfg file, where I need to find the latest occurrence of a line starting with a specific string. An example of the cfg file:
...
# format: - search.index.[number] = [search field]:element.qualifier
...
search.index.1 = author:dc.contributor.*
...
search.index.12 = language:dc.language.iso
...
jspui.search.index.display.1 = ANY
...
I need to be able to get the last occurrence of the line starting with search.index.[number] , more specific: I need that number. For the above snippet, that number would be 12.
As you can see, there are other lines too containing that pattern, but I do not want to match those.
I'm using Groovy as a programming/scripting language.
Any help is appreciated!
Have you tried:
def m = lines =~ /(?m)^search\.index\.(\d+)/
m[ -1 ][ 1 ]
Try this as your expression :
^search\.index\.(\d+)/
And then with Groovy you can get your result with:
matcher[0][0]
Here is an explanation page.
I don't think you should go for it but...
If you can do a multi-line search (anyway you have to here), the only way would be to read the file backward. So first, eat everything with a .* (om nom nom)(if you can make the dot match all, (?:.|\s)* if you can't). Now match your pattern search\.index\.(\d+). And you want to match this pattern at the beginning of a line: (?:^|\n) (hoping you're not using some crazy format that doesn't use \n as new line character).
So...
(?:.|\s)*(?:^|\n)search\.index\.(\d+)
The number should be in the 1st matching group. (Test in JavaScript)
PS: I don't know groovy, so sorry if it's totally not appropriate.
Edit:
This should also work:
search\.index\.(\d+)(?!(?:.|\s)*?(?:^|\n)search\.index\.\d+)
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Extract info inside all parenthesis in R (regex)
I inported data from excel and one cell consists of these long strings that contain number and letters, is there a way to extract only the numbers from that string and store it in a new variable? Unfortunately, some of the entries have two sets of brackets and I would only want the second one? Could I use grep for that?
the strings look more or less like this, the length of the strings vary however:
"East Kootenay C (5901035) RDA 01011"
or like this:
"Thompson-Nicola J (Copper Desert Country) (5933039) RDA 02020"
All I want from this is 5901035 and 5933039
Any hints and help would be greatly appreciated.
There are many possible regular expressions to do this. Here is one:
x=c("East Kootenay C (5901035) RDA 01011","Thompson-Nicola J (Copper Desert Country) (5933039) RDA 02020")
> gsub('.+\\(([0-9]+)\\).+?$', '\\1', x)
[1] "5901035" "5933039"
Lets break down the syntax of that first expression '.+\\(([0-9]+)\\).+'
.+ one or more of anything
\\( parentheses are special characters in a regular expression, so if I want to represent the actual thing ( I need to escape it with a \. I have to escape it again for R (hence the two \s).
([0-9]+) I mentioned special characters, here I use two. the first is the parentheses which indicate a group I want to keep. The second [ and ] surround groups of things. see ?regex for more information.
?$ The final piece assures that I am grabbing the LAST set of numbers in parens as noted in the comments.
I could also use * instead of . which would mean 0 or more rather than one or more i in case your paren string comes at the beginning or end of a string.
The second piece of the gsub is what I am replacing the first portion with. I used: \\1. This says use group 1 (the stuff inside the ( ) from above. I need to escape it twice again, once for the regex and once for R.
Clear as mud to be sure! Enjoy your data munging project!
Here is a gsubfn solution:
library(gsubfn)
strapplyc(x, "[(](\\d+)[)]", simplify = TRUE)
[(] matches an open paren, (\\d+) matches a string of digits creating a back-reference owing to the parens around it and finally [)] matches a close paren. The back-reference is returned.
Sorry in advance that this might be a little challenging to read...
I'm trying to parse a line (actually a subject line from an IMAP server) that looks like this:
=?utf-8?Q?Here is som?= =?utf-8?Q?e text.?=
It's a little hard to see, but there are two =?/?= pairs in the above line. (There will always be one pair; there can theoretically be many.) In each of those =?/?= pairs, I want the third argument (as defined by a ? delimiter) extracted. (In the first pair, it's "Here is som", and in the second it's "e text.")
Here's the regex I'm using:
=\?(.+)\?.\?(.*?)\?=
I want it to return two matches, one for each =?/?= pair. Instead, it's returning the entire line as a single match. I would have thought that the ? in the (.*?), to make the * operator lazy, would have kept this from happening, but obviously it doesn't.
Any suggestions?
EDIT: Per suggestions below to replace ".?" with "[^(\?=)]?" I'm now trying to do:
=\?(.+)\?.\?([^(\?=)]*?)\?=
...but it's not working, either. (I'm unsure whether [^(\?=)]*? is the proper way to test for exclusion of a two-character sequence like "?=". Is it correct?)
Try this:
\=\?([^?]+)\?.\?(.*?)\?\=
I changed the .+ to [^?]+, which means "everything except ?"
A good practice in my experience is not to use .*? but instead do use the * without the ?, but refine the character class. In this case [^?]* to match a sequence of non-question mark characters.
You can also match more complex endmarkers this way, for instance, in this case your end-limiter is ?=, so you want to match nonquestionmarks, and questionmarks followed by non-equals:
([^?]*\?[^=])*[^?]*
At this point it becomes harder to choose though. I like that this solution is stricter, but readability decreases in this case.
One solution:
=\?(.*?)\?=\s*=\?(.*?)\?=
Explanation:
=\? # Literal characters '=?'
(.*?) # Match each character until find next one in the regular expression. A '?' in this case.
\?= # Literal characters '?='
\s* # Match spaces.
=\? # Literal characters '=?'
(.*?) # Match each character until find next one in the regular expression. A '?' in this case.
\?= # Literal characters '?='
Test in a 'perl' program:
use warnings;
use strict;
while ( <DATA> ) {
printf qq[Group 1 -> %s\nGroup 2 -> %s\n], $1, $2 if m/=\?(.*?)\?=\s*=\?(.*?)\?=/;
}
__DATA__
=?utf-8?Q?Here is som?= =?utf-8?Q?e text.?=
Running:
perl script.pl
Results:
Group 1 -> utf-8?Q?Here is som
Group 2 -> utf-8?Q?e text.
EDIT to comment:
I would use the global modifier /.../g. Regular expression would be:
/=\?(?:[^?]*\?){2}([^?]*)/g
Explanation:
=\? # Literal characters '=?'
(?:[^?]*\?){2} # Any number of characters except '?' with a '?' after them. This process twice to omit the string 'utf-8?Q?'
([^?]*) # Save in a group next characters until found a '?'
/g # Repeat this process multiple times until end of string.
Tested in a Perl script:
use warnings;
use strict;
while ( <DATA> ) {
printf qq[Group -> %s\n], $1 while m/=\?(?:[^?]*\?){2}([^?]*)/g;
}
__DATA__
=?utf-8?Q?Here is som?= =?utf-8?Q?e text.?= =?utf-8?Q?more text?=
Running and results:
Group -> Here is som
Group -> e text.
Group -> more text
Thanks for everyone's answers! The simplest expression that solved my issue was this:
=\?(.*?)\?.\?(.*?)\?=
The only difference between this and my originally-posted expression was the addition of a ? (non-greedy) operator on the first ".*". Critical, and I'd forgotten it.
I saw other questions dealing with the finding the n-th occurrence of a word/pattern, but I couldn't find how you would actually substitute the n-th occurrence of a pattern in vim. There's the obvious way of hard coding all the occurrences like
:s/.*\(word\).*\(word\).*\(word\).*/.*\1.*\2.*newWord.*/g
Is there a better way of doing this?
For information,
s/\%(\(pattern\).\{-}\)\{41}\zs\1/2/
also works to replace 42th occurrence. However, I prefer the solution given by John Kugelman which is more simple -- even if it will not limit itself to the current line.
You can do this a little more simply by using multiple searches. The empty pattern in the :s/pattern/repl/ command means replace the most recent search result.
:/word//word//word/ s//newWord/
or
:/word//word/ s/word/newWord/
You could then repeat this multiple times by doing #:, or even 10#: to repeat the command 10 more times.
Alternatively, if I were doing this interactively I would do something like:
3/word
:s//newWord/r
That would find the third occurrence of word starting at the cursor and then perform a substitution.
Replace each Nth occurrence of PATTERN in a line with REPLACE.
:%s/\(\zsPATTERN.\{-}\)\{N}/REPLACE/
To replace the nth occurrence of PATTERN in a line in vim, in addtion to the above answer I just wanted to explain the pattern matching i.e how it is actually working for easy understanding.
So I will be discussing the \(.\{-}\zsPATTERN\)\{N} solution,
The example I will be using is replacing the second occurrence of more than 1 space in a sentence(string).
According to the pattern match code->
According to the zs doc,
\zs - Scroll the text horizontally to position the cursor at the start (left
side) of the screen.
.\{-} 0 or more as few as possible (*)
Here . is matching any character and {} the number of times.
e.g ab{2,3}c here it will match where b comes either 2 or 3 times.
In this case, we can also use .* which is 0 or many as many possible.
According to vim non-greedy docs, "{-}" is the same as "*" but uses the shortest match first algorithm.
\{N} -> Matches n of the preceding atom
/\<\d\{4}\> search for exactly 4 digits, same as /\<\d\d\d\d>
**ignore these \<\> they are for exact searching, like search for fred -> \<fred\> will only search fred not alfred.
\( \) combining the whole pattern.
PATTERN here is your pattern you are matching -> \s\{1,} (\s - space and {1,} as explained just above, search for 1 or more space)
"abc subtring def"
:%s/\(.\{-}\zs\s\{1,}\)\{2}/,/
OUTPUT -> "abc subtring,def"
# explanation: first space would be between abc and substring and second
# occurence of the pattern would be between substring and def, hence that
# will be replaced by the "," as specified in replace command above.
This answers your actual question, but not your intent.
You asked about replacing the nth occurrence of a word (but seemed to mean "within a line"). Here's an answer for the question as asked, in case someone finds it like I did =)
For weird tasks (like needing to replace every 12th occurrence of "dog" with "parrot"), I like to use recursive recordings.
First blank the recording in #q
qqq
Now start a new recording in q
qq
Next, manually do the thing you want to do (using the example above, replace the 12th occurrence of "dog" with "parrot"):
/dog
nnnnnnnnnnn
delete "dog" and get into insert
diwi
type parrot
parrot
Now play your currently empty "#q" recording
#q
which does nothing.
Finally, stop recording:
q
Now your recording in #q calls itself at the end. But because it calls the recording by name, it won't be empty anymore. So, call the recording:
#q
It will replay the recording, then at the end, as the last step, replay itself again. It will repeat this until the end of the file.
TLDR;
qq
q
/dog
nnnnnnnnnnndiwiparrot<esc>
#q
q
#q
Well, if you do /gc then you can count the number of times it asks you for confirmation, and go ahead with the replacement when you get to the nth :D