I have a sequence of digits like below. I want to combine digits to group of 4. Can someone give a vim regex to do that?
Input : 1234 56 7890 1234
The output should be: 1234 5678 9012 34
Two pass:
s/ //g
s/\(.\{4\}\)/\1 /g
I would do this in two steps:
(1) remove the blanks right of digit groups
:s/\(\d\+\) /\1/g
(2) grouping:
:s/\(\d\{4}\)/\1 /g
In case of many lines record a macro
or do these steps for an marked area.
You can do it in one pass, but it looks awful:
:s/\(\d\) *\(\d\) *\(\d\) *\(\d\) */\1\2\3\4 /g
While this works, it's rather asinine. It's faster to do one of those nice two-pass solutions than to spend 2 minutes working out one Monsteregex™ that does it in one pass. Plus, the two-pass solutions are easier to understand.
:s/\(\d\)\s*\(\d\)\s*\(\d\)\s*\(\d\)\s*/\1\2\3\4 /g
works, but I prefers Autocracy's solution.
Related
Regex: [0-9]{6-8}([0-9]{4})
Test Strings:
sfad 123456781234 afd sadfa fdads
sfd 12345671234 24312 fasdfa dsfafd
221 1234561234 safd safd23 34
Expected:
I need the end part 1234 captured in a group on each line.
Actual: No matches. :(
I would like to make [0-9]{6-8} to match the least possible characters so all these 3 strings would match. How do I make this lazy as it seems to be greedy now.
I need only regex solutions as this is part of a bigger solution. Here's a link to play with it: https://regex101.com/r/eF5pA9/1
[0-9]{6,8}([0-9]{4}) with g modifier matches all three.
Thanks #anubhava and #Anderson Pimentel.
I need to validate string with 2 groups which are separated with one space with next rules:
Each group needs to be at least 2 character long but less or equal to 15
Both groups together can't be more than 20 chars long (not counting space)
Groups can only contain letters (that's simple, it's [a-zA-Z])
Following these rules, here are some examples
Firstname Lastname (Valid)
Somename T (Invalid, 2nd one is <2)
Somethingsomettt Here (Invalid, first one is > 15)
Somethingsome Somethingsome (Invalid, total > 20)
It'd be simple [a-zA-Z]{2,15} [a-zA-Z]{2,15} if it wasn't for that 2+2<=total<=20 condition.
Is it even possible to limit it this way? If it is - how?
UPDATE
Just for the sake of it, resulting regex was supposed to be ^(?=[a-zA-Z ]{5,21}$)[a-zA-z]{2,15} [a-zA-Z]{2,15}$, #vks was closest one to it. Nevertheless, thanks #popovitsj and #Avinash Raj too.
^(?=.{5,21}$)[a-zA-Z]{2,15} [a-zA-Z]{2,15}$
Try this.See demo.
http://regex101.com/r/nA6hN9/30
This can be done with lookahead. Something like this:
^(?=.{1,20}$)[a-zA-z]{2,14} [a-zA-Z]{2,14}$
You could try the below regex which uses negative lookahead,
(?!^.{22,})^[a-zA-Z]{2,15} [a-zA-Z]{2,15}$
DEMO
I have a text with some lines (200+) in this format:
10684 - The jackpot ? discuss Lev 3 --- ? ---
10755 - Garbage Heap ? discuss Lev 5 --- ? ---
I hant to retrieve the first number (10684 or 10755) only if number after "Lev" is greater than 3.
I'm able to get the first number with this regex: ([0-9]+) - but without the 'level' restrictions.
How this could be made?
Thanks in advance.
(\d+) - .*?Lev (?:[4-9]|[1-9]\d+)
The first \d+ matches line number as you have done.
The next .*? is a lazy quantifier, which will not consume too many characters. And the following expression will guide it to the right place. (lazy quantifier is usually more efficient)
The second parenthesis, (?:[4-9]|[1-9]\d+), matches either single digital numbers greater than 3 or two digital numbers without leading zero.
Alright stackoverflow doesn't properly show my image. Take this link : http://regexr.com?36n5l
Example Output:
Regular expressions doesn't recognize numbers as numbers (only strings). You can do this though:
([0-9]+) - .*Lev (?:[4-9][^0-9]|[1-9][0-9]+)
Basically, we use the alternation operator (|) to accept only a single digit greater than 3 (enforced by checking that the following character is not a digit) or a multi-digit number not beginning with a zero.
In case that level number might be the end of the line, though, you might have to do this:
([0-9]+) - .*Lev (?:[4-9](?:[^0-9]|$)|[1-9][0-9]+)
(I'm assuming whatever regex engine you're using can't handle lookaround assertions. In the future, try to always include what language you're using when you're asking a regex question.)
Ah, I just read your edit that the number is always less than 10. Well, that's much easier then:
([0-9]+) - .*Lev [4-9]
A lookahead is really the best thing because it will leave just the number:
/\d+(?=.*Lev (0*[4-9]|[1-9]\d))/
A bit of Awk trickery:
awk -F '\? +discuss +Lev' '$2>3 { split($1,a,/ */); print a[1] }' file
In bash use this:
var=">3"
perl -lne '/(\d+) - .*Lev (\d+)/; print $1 if $2'"$var"
This is a good solution to be able to pass the condition by parameter.
I keep getting into situations where I end up making two regular expressions to find subtle changes (such as one script for 0-9 and another for 10-99 because of the extra number)
I usually use [0-9] to find strings with just one digit and then [0-9][0-9] to find strings with multiple digits, is there a better wildcard for this?
ex. what expression would I use to simultaneously find the strings
6:45 AM and 10:52 PM
You can specify repetition with curly braces. [0-9]{2,5} matches two to five digits. So you could use [0-9]{1,2} to match one or two.
[0-9]{1,2}:[0-9]{2} (AM|PM)
I personally prefer to use \d for digits, thus
\d{1,2}:\d{2} (AM|PM)
[0-9] 1 or 2 times followed by : followed by 2 [0-9]:
[0-9]{1,2}:[0-9]{2}\s(AM|PM)
or to be valid time:
(?:[1-9]|1[0-2]):[0-9]{2}\s(?:AM|PM)
If you are looking for a time patten, you'd do something like:
\d{1,2}:\d{1,2} (AM|PM)
Or for more specific time regex
[0-1]{0,1}[0-9]{1,2}:[0-5][0-9] (AM|PM)
Much like the other answers, except the AM/PM is not captured, which should be more efficient
\d{1,2}:\d{1,2}\s(?:AM|PM)
if I have a file containing:
1 ABC
2 123XYZ
3 6:45 AM
4 123DHD
5 ABC
6 10:52 PM
7 CDE
and run the following
$>grep -P '6:45\sAM|10:52\sPM' temp
6:45 AM
10:52 PM
$>.
should do the trick (-P is a perl regx)
EDIT:
Perhaps I misunderstood, the other answers are very good if I were looking to just find a time, but you seem to be after specific times. the others would match ANY time in HH:MM format.
overall, I believe the items you are after would be the | pipe character which is used in this case to allow alternative phrases and the {n,m} match n-m times {1,2} would match 1-2 times, etc.
It can be able to check all type of time formats :
e.g. 12:05PM, 3:19AM, 04:25PM, 23:52PM
my $time = "12:52AM";
if ($time =~ /^[01]?[0-9]\:[0-5][0-9](AM|PM)/) {
print "Right Time Dude...";
}
else { print "Wrong time Dude"; }
This is the regex you want.
/^[01]?[0-9]\:[0-5][0-9](AM|PM)/
Having this string as input:
Sat, 6 May 2017 02:08:08 +0000
I did this regEx to get combinations of one or two digits:
[0-9]*:[0-9]*:[0-9]*
I want to create a regex that will match any of these values
7-5
6-6 ((0-99) - (0-99))
6-4
6-3
6-2
6-1
6-0
0-6
1-6
2-6
3-6
4-6
the 6-6 example is a special case, here are some examples of values:
6-6 (23-8)
6-6 (4-25)
6-6 (56-34)
Is it possible to make one regex that can do this?
If so, is it possible to further extend that regex for the 6-6 special case such that the the difference between the two numbers within the parentheses is equal to 2 or -2?
I could easily write this with procedural code, but i'm really curious if someone can devise a regex for this.
Lastly, if it could be further extended such that the individual digits were in their own match groups I'd be amazed. An example would be for 7-5, i could have a match group that just had the value 7, and another that had the value 5. However for 6-6 (24-26) I'd like a match group that had the first six, a match group for the second 6, a match group for the 24 and a match group for the 26.
This may be impossible, but some of you can probably get this part of the way there.
Good luck, and thanks for the help.
NO. The answer is "We can't," and the reason is because you're trying to use a hammer to dig a hole.
The problem with writing one long "clever" (this word causes a knee-jerk reaction in many people who are far more anti-regex than I) regex is that, six months from now, you'll have forgotten those clever regex features that you used so heavily, and you'll have written six months worth of code related to something else, and you'll get back to your impressive regex and have to tweak one detail, and you'll say, "WTF?"
This is what (I understand) you want, in Perl:
# data is in $_
if(/7-5|6-[0-4]|[0-4]-6|6-6 \((\d{1,2})-(\d{1,2})\)/) {
if($1 and $2 and abs($1 - $2) == 2) {
# we have the right difference
}
}
Some might say that the given regex is a bit much, but I don't think it's too bad. If the \d{1,2} bit is a little too obscure you could use \d\d? (which is what I used at first, but didn't like the repetition).
You can do it like this:
7-5|6-[0-4]|[0-5]-6|6-6 \(\d\d?-\d\d?\)
Just add parens to get your match groups.
Off the top of my head (there may be some errors but the principle should be good):
\d-\d|6-6 (\d+-\d+)
And like with any regexp, you can surround what you want to extract with parentheses for match groups:
(\d)-(\d)|(6)-(6) ((\d)+-(\d+))
In the 6-6 case, the first two parentheses should get the sixes, and the second two should get the multi-digit values that come afterwards.
Here is one that will match only the numbers you want and let you get each digit by name:
p = r'(?P<a>[0-4]|6|7)-(?P<b>[0-4]|6|5) *(\((?P<c>\d{1,2})-(?P<d>\d{1,2})\))?'
To get each digit you could use:
values = re.search(p, string).group('a', 'b', 'c', 'd')
Which will return a four element tuple with the values you are looking for (or None if no match was found).
One problem with this pattern is that it will patch the stuff in the parenthesis whether or not there was a match to '6-6'. This one will only match the final parenthesis if 6-6 is matched:
p = r'(?P<a>[0-4]|(?P<tmp_a>6)|7)-(?P<b>(?(tmp_a)(?P<tmp_b>6)|([0-4]|5)))(?(tmp_b) *(\((?P<c>\d{1,2})-(?P<d>\d{1,2})\))?)'
I don't know of any way to look for a difference between the numbers in the parenthesis; regex only knows about strings, not numerical values . . .
(I am assuming python syntax here; the perl syntax is slightly different, though perl supports the python way of doing things.)