I have a number of floats/strings that look as follows:
12339.0
133339
159.0
dfkkei
something
32439
Some of them have trailing .0. How can I show all the numbers without the trailing .0 as a regular repression, including the items that are not a number? I tried something like that, hoping it would exclude all .0 from the capture group, but it doesn't work: (.*)(:?.0)?
https://regex101.com/r/sC6jO2/1
You may use a simpler regex:
\.0+$
And replace with an empty string, see regex demo.
The regex matches a . (\.) followed with 1 or more zeros (0+) up to the end of string ($).
If you plan to match two groups as in your initial attempt, use
^(.*?)(?:\.0+)?$
See this regex demo
Here,
^ - start of string
(.*?) - Group 1 capturing any 0+ chars other than a newline, as few as possible (=lazily), up to a
(?:\.0+)? - optional sequence of . + one or more zeros
$ - at the end of the string.
Related
How can I replace all digits in URL after third slash, on characters #### with regexp?
In this case, the number of # must correspond to the number of replaced digits.
Numbers can be in more than one slash section.
Also, the location of the digits is not fixed, but exactly after the third slash
Examples:
/path/to/something/1234/end
/path/to/something/12/1234/end
To:
/path/to/something/####/end
/path/to/something/##/####/end
I tried to use an expression, but it does not give the desired result:
"(?<=/)\\d+(?=/|$), #####"
This regexp is needed to implement the grok pattern in Logstash (gsub function).
P.s. Why after third slash? Because because the numbers can be at the beginning, but they do not need to be changed (/path/to_1/something/1234/end)
You can use
(?:\G(?!^)|^((?:/[^/]*){3}/))(\D*)\d
as regex and $1$2# as replacement.
See the regex demo.
Details:
(?:\G(?!^)|^((?:/[^/]*){3}/)) - end of the previous match (\G(?!^)) or (|) start of string + three occurrences of / and then zero or more non-slash shars and then a slash char captured into Group 1 (^((?:/[^/]*){3}/))
(\D*) - Group 2: any zero or more non-digits
\d - a digit
The replacement is a concatenation of Group 1 + Group 2 values and a # char.
I am trying to find regex which would find below matches. I would replace these with blank. I am able to create regex for few of these conditions individually, but I am not able to figure out how to create one regex for all of these
Strings:
song1 artist (SiteWithMp3Keyword.com).mp3
02.song2 | siteWithdownloadKeyword.in 320 Kbps
song3 [SitewithDjKeyword.in] 128kbps.mp3
Output
song1 artist.mp3
song2
song3.mp3
Criteria for match:
Case Insensitive
Find Strings with particular keyword and remove whole word, even if inside any braces
Find kpbs keyword and remove it along with any number before it (128/320)
if string ends in .mp3, keep it as it is.
Remove junk characters (like | ) and replace _ with space.
Remove number if present at start of string, like 001_ 02. etc.
Trim whitespaces before and after remaining string
Example Regex for 2.
\S+(mp3|dj|download)\S+
https://regex101.com/r/nxp4d3/1
Try this regex ....
Find:^[0-9. ]*(song\d+ (\w+ )?).*?(\.mp3 ?)?$
Replace with:$1$3
P.S , if this code doesn't solve your problem, please share a sample of your real data, so someone well better understand you,
Thanks...
For the example data, you might use:
^\h*(?:\d+\W*)?(\w+(?:\h+\w+)*).*?(\.mp3)?\h*$
The pattern matches:
^ Start of string
\h* Match optional leading spaces
(?:\d+\W*)? Match 1+ digits followed by optional non word characters
(\w+(?:\h+\w+)*) Capture group 1, match word characters optionally repeated with a space in between
.*? Match any character except a newline, as least as possible
(\.mp3)? Optionally capture .mp3 in group 2
\h* Match optional trailing spaces
$ End of string
Regex demo
Replace with capture group 1 and group 2
$1$2
I am trying to use replace in Sublime using regular expressions but I'm stuck. I tried various combinations but don't seem to be getting there.
This is the input and my desired output:
Input: N_BBP_c_46137_n
Output : BBP
I tried combinations of:
[^BBP]+\b
\*BBP*+\g
But none of the above (and many others) don't seem to work.
To turn N_BBP_c_46137_n into BBP and according to the comment just want that entire long name such as N_BBP_ to be replaced by only BBP* you might also use a capture group to keep BBP.
\bN_(BBP)_\S*
\bN_ Match N preceded by a word boundary
(BBP) Capture group 1, match BBP (or use [A-Z]+ to match 1+ uppercase chars)
_\S* Match _ followed by 0+ times a non whitespace char
In the replacement use the first capturing group $1
Regex demo
You may use
(N_)[^_]*(_c_\d+_n)
Replace with ${1}some new value$2.
Details
(N_) - Group 1 ($1 or ${1} if the next char is a digit): N_
[^_]* - any 0 or more chars other than _
-(_c_\d+_n) - Group 2 ($2): _c_, 1 or more digits and then _n.
See the regex demo.
Remove trailing zeros to a number with 4 decimals
Sample expected output:
1.7500 -> 1.75
1.1010 -> 1.101
1.0000 -> 1
I am new with REGEX so I just tried this one first but not working:
REPLACE ALL OCCURRENCES OF REGEX '^\.[0]\d{0,3}' IN lv_rate WITH space.
Need help for the right regex to use. Thanks!
EDIT: SHIFT lv_rate RIGHT DELETING TRAILING '0' is not an option.
Try replacing on the following regex pattern:
\.?0+$
Use empty string as the replacement. This will match an optional decimal point, followed by trailing zeroes until the end of the string. See the demo below to see this pattern working.
Demo
This answer assumes that all inputs would always have a decimal component. If not, then we would need to add additional logic.
If you want to remove trailing zeros to a number with 4 decimals, one option is to use a capturing group and use group 1 in the replacement.
^(\d+(?=\.\d{4}$)(?:\.\d*[1-9])?)\.?0+$
In parts
^ Start of string
( Capture group 1
\d+ Match 1+ digits
(?=\.\d{4}$) Assert what is on the right is a . and 4 digits
(?:\.\d*[1-9])? Optionally match digits until the last digit 1-9
) Close group 1
\.?0+ Match an optional . and 1 or more times a zero
$ End of string
Regex demo
Given any URL, like:
https://stackoverflow.com/v1/summary/1243PQ/details/P1/9981
How do I extract the numeric or alphanumeric part of the URL? I.e. the following strings from the url given above:
1. v1
2. 1243PQ
3. P1
4. 9981
To rephrase, a regex to extract strings from a string (URL) which have at least 1 digit and 0 or more alphabet characters, separated by '/'.
I tried to capture a repeating group (^[a-zA-Z0-9]+)+ and ([a-zA-Z]{0,100}[0-9]{1,100})+ but it didn't work. In hindsight intuition does say this shouldn't work. I am unsure how do I match patterns over a group and not just a single character.
If I understand what you really want:
Extracting parts with only numbers or with numbers following alphabets
then; I can suggest this regex:
\b[a-zA-Z]*[0-9]+[a-zA-z]*\b
Regex Demo
I use \b to assert position of a word boundary or a part.
As numbers are required and alphabets can comes before or after that I use above regex.
If following alphabets are not required then I can suggest this regex:
\b[a-zA-z0-9]*[0-9]+[a-zA-Z0-9]*\b
Regex Demo
I believe this should work for you:
(\d*\w+\d+\w*)
EDIT: actually, this should be sufficient
(\w+\d+\w*)
or
(\w*\d+\w*)
Well, you could do this:
(\w*\d+\w*) with the g (global) regex option
On the example URL, it would look like this:
const regex = /(\w*\d+\w*)/g;
const url = 'https://stackoverflow.com/v1/summary/1243PQ/details/P1/9981';
console.log(url.match(regex))
Try \/[a-zA-Z]*\d+[a-zA-Z0-9]*
Explanation:
\/ - match / literally
[a-zA-Z]* - 0+ letters
\d+ - 1+ digits - thanks to this, we require at least one digits
[a-zA-Z0-9]* - 0+ letters or digits
Demo
It will captrure together with / at the beginning, so you need to trim it.