I'm new to Regular Expressions, and I have been trying to figure out how to code this: I need to find numbers greater than 25000 where the same line also has the number " 19" somewhere on that line (that's a space then 19). The problem is that the numbers have commas in them. I tried a few options:
This finds lines with any numbers over 25000:
^.*(25,|26,|27,|28,|29,|30,|31,|32,|33,|34,|35,|36,|37,|38,|39,|40,|41,|42,|43,|44,|45,|46,|47,|48,|49,|50,|51,|52,|53,|54,|55,|56,|57,|58,|59,|60,|61,|62,|63,|64,|65,|66,|67,|68,|69,|70,|71,|72,|73,|74,|75,|76,|77,|78,|79,|80,|81,|82,|83,|84,|85,|86,|87,|88,|89,|90,|91,|92,|93,|94,|95,|96,|97,|98,|99,|100,|101,|102,|103,|104,|105,|106,|107,|108,|109,|110,|111,|112,|113,|114,|115,|116,|117,|118,|119,|120,|121,|122,|123,|124,).*$
This finds line with both " 19" and 26, (but not with the comma behind the 26)
^.*( 19.*26).*$
Any help is appreciated!
Numbers over 25000 can be represented as follows :
\d{6,}|2[5-9]\d{3}|[3-9]\d{4}
That is, in english :
numbers of 6 digits or more
numbers of 5 digits starting with 2 and another digit equal or greater than 5
numbers of 5 digits starting with a digit greater than 2
So the complete regex would look like this :
.*(\d{6,}|2[5-9]\d{3,}|[3-9]\d{4,}).* 19.*
Which is said number somewhere in the line, followed by 19 somewhere in the line.
Here is a test run on regex101 for you to test with your data.
I also second the comment that this isn't a job for regular expressions, which as you can see work on characters rather than numbers.
I would try something like this:
^(([0-9,]*([3-9][0-9]|2[5-9]),?[0-9]{3})\s?)$
That should handle the numeric part. You didn't really explain if the " 19" would come before or after that, and what would delimit that from the numeric part, but just insert (\s19) wherever that bit needs to go.
example
Thanks everyone. The following RegEx worked for me:
^.* 19.(25,|26,|27,|28,|29,|30,|31,|32,|33,|34,|35,|36,|37,|38,|39,|40,|41,|42,|43,|44,|45,|46,|47,|48,|49,|50,|51,|52,|53,|54,|55,|56,|57,|58,|59,|60,|61,|62,|63,|64,|65,|66,|67,|68,|69,|70,|71,|72,|73,|74,|75,|76,|77,|78,|79,|80,|81,|82,|83,|84,|85,|86,|87,|88,|89,|90,|91,|92,|93,|94,|95,|96,|97,|98,|99,|100,|101,|102,|103,|104,|105,|106,|107,|108,|109,|110,|111,|112,|113,|114,|115,|116,|117,|118,|119,|120,|121,|122,|123,|124,).$
This finds lines that have " 19" first in the line then a number greater than 25K later in the line, when the numbers have commas in them. I couldn't use the shortcut "number ranges" that were suggested because there are other numbers on the lines without commas that are over 25K that I don't want to flag. Maybe there's any easier way that my brute force method, but if not, at least this works. Thanks again!
Related
source txt file:
34|Gurla Mandhata|7694|25243|2788|Nalakankar Himalaya|30°26'19"N
81°17'48"E|Dhaulagiri|1985|6 (4)|China
command input:
:%s/\(\d\+\)\(\d\d\d\)/\1,\2/g
command output:
34|Gurla Mandhata|7,694|25,243|2,788|Nalakankar Himalaya|30°26'19"N
81°17'48"E|Dhaulagiri|1,985|6 (4)|China
Desired output:
34|Gurla Mandhata|7,694|25,243|2,788|Nalakankar Himalaya|30°26'19"N
81°17'48"E|Dhaulagiri|1985|6 (4)|China
Basically 1985 is supposed to be 1985 and not 1,985. I tried to put a \? so every time the pattern matches it stops and a °+ after so it has to detect a ° to match the pattern, but no success. It just replaces the ° and everything before that, complete mess.
My knowledge of regular expressions however combined with the substitute is weak and I'm stuck here.
EDIT
the first 3 numbers represent heights of mountains, those 3 need to change with a (,) and the last number ( 1985 ) represents a year, which must not be changed.
Mathematical solutions are not going to work as loophole since there are mountains with a height off less than 1900
You haven't told us what is the difference between 1985 and other numbers, so I assumed that your "small" numbers are less than 2000.
You almost got it:
:%s/(\d*[2-90])(\d\d\d)/\1,\2/g
Alternatively if that isn't what you want, you can use c flag (:h s_flags):
:%s/\(\d\+\)\(\d\d\d\)/\1,\2/gc
this line will leave the last 3 columns untouched, just do substitution on the content before it:
%s/\v(.*)((\|[^|]*){3}$)/\=substitute(submatch(1),'\v(\d+)(\d{3})','\1,\2','g').submatch(2)/g
Note that the above line will change 1000000 into 1000,000 instead of 1,000,000. Vim's printf() doesn't support %'d, it is pity. If you do have number > 1m, we can find other solutions.
update
I solved it myself, by using 3 seperate commands; one for every number string in the file:
%s/^\(\d*|[^|]*|\)\(\d\+\)\(\d\d\d\)|/\1\2,\3|/g
:%s/^\(\d*|[^|]*|\d\+,*\d*|\)\(\d\+\)\(\d\d\d\)|/\1\2,\3|/g
:%s/^\(\d*|[^|]*|\d\+,*\d*|\d\+,*\d*|\)\(\d\+\)\(\d\d\d\)|/\1\2,\3|/g
In case you want to use perl:
:%!perl -F'\|' -lane 'for(#F[2..4]) { s/(\d+)(\d{3})/\1,\2/;} print join "|", #F'
I have files that are split in two blocks, the first half contains the information I need, the second half always starts with a 4 digit number (between 1400 and 1900). I need to delete the second block, hence my question:
How do I delete all lines in a file after (and including) the first 4 digit number?
I believe that should be doable using notepad++ and regular expressions, but I'm new to regex and have no idea how...
I know it's a noob-ish question but nonetheless, help would be greatly appreciated.
The Regex 1(?:[4-8]\d\d|900)(?:.|[\r\n])+\z will select a text starting with 1400-1900 till the end of a file.
I have text with 6 numbers typically stored in one line
SomeData\n0.00 0.00 0.00 31,570.07 0.00 31,570.07\nSomeData
SomeData\n0.00 0.00 0.00 485,007.24 0.00 485,007.24\nSomeData
This regex worked fine on it:
\n[0-9,.-]* [0-9,.-]* [0-9,.-]* [0-9,.-]* [0-9,.-]* [0-9,.-]*\n
I noticed that every once in a while I get this:
SomeData\n0.00 0.00 10,921,594\n.89\n-\n9,563,271.0\n6\n0.00 1,358,323.83\nSomeData
Note how the linebreaks are randomly inserted after a sign or between numbers as if the system stored the values without filtering linebreaks.
I am struggling to get this extracted. I tried various expressions but my more successful one was [0-9,.-][\n]{0,1}[0-9,.-][ ]{0,1} to match an individual number.
What expression can I use to match both variations of the number formats preferably already stripping out the inconstant line breaks?
Update: Going with
[-\n]{0,2}[0-9,]+[\n.0-9]{3,4}[\n ]{0,1}
Please let me know if I there's a better way
One way would be to write an exact representation of what constitutes a number, so in your case [-+]?[0-9]+[0-9,]*(?:\.[0-9]+)? would do the trick. This helps, because then your search can know when a number starts and when one ends (because of rules like: a sign always is at the start a dot cannot appear multiple times, etc.). Then you want to match pairs of six delimited by either a new line or space so wrap it in a capture group and limit by 6: (...[ \n]*){6,6}. This helps because then the regex engine can figure out by backtracking what to consider a number by knowing how many it should match. Then you want to allow new lines in pretty much any position, so place the new line in each character group. You might also want to anchor the numbers on both sides, but this is not necessary, because now the regex engine will try to identify valid tuples of 6 numbers. End result is:
SomeData\n([-+]?[0-9\n]+[0-9,\n]*(?:\.[0-9\n]+)?[ \n]){6,6}SomeData
This will find tuples of 6 numbers no matter where the enters are. Here is an example: https://regex101.com/r/jD5nT8/1
I need help forming regex to limit user input to only numerics and only up to 10 occurrences.
I have regex that is working to keep input to numerics only, but I cannot limit it to up to 10.
Here is what I have:
^(0|[1-9][0-9]*)$
I am okay accepting negative numbers, decimals, and 0's. Any advice?
^\s*([0-9)+){0,10}\s*$
This basically says I want to 0 to 10 things, where each thing is all digits. I added the \s* on either side to allow the user to have put spaces before or after their numbers. This would accept things like
10 1231231 1231 1231 23112 123123
If what you really want is just a single number, that is only up to 10 digits, it is even easier:
^\s[0-9]{1,10}\s$
The regex you're looking for is this:
/(?=^[-+]?\d*\.?\d+$)^.{1,10}$/
Keep in mind that this regex will allow maximum length of input to 10 which includes optional + or - sign at start and a decimal point ..
you can try this:
^(?<=\s)(\-?[\d]{1,10}(?=\s))$
This fails as it has 11 digits
Debug.WriteLine(Regex.IsMatch("12345678901", #"^\d{1,10}$").ToString());
Posted the above answer before you clarified you want up to 10 set of numbers delimited by space.
Tested the accepted answer and in .NET Regex it fails for me.
Even fixing the syntax error it still does not parse by space
Give this a try
Debug.WriteLine(Regex.IsMatch(" 12345 678901 12 ", #"^\s*([+-]?\d+)(\s+[+-]?\d+){1,9}\s*$").ToString());
I have a list of numbers that I want to find at least 3 of...
here is an example
I have a large list of numbers in a sql database in the format of (for example)
01-02-03-04-05-06
06-08-19-24-25-36
etc etc
basically 6 random numbers between 0 and 99.
Now I want to find the strings where at least 3 of a set of given numbers occurs.
For example:
given: 01-02-03-10-11-12
return the strings that have at least 3 of those numbers in them.
eg
01-05-06-09-10-12 would match
03-08-10-12-18-22 would match
03-09-12-18-22-38 would not
I am thinking that there might be some algorithm or even regular expression that could match this... but my lack of computer science textbook experience is tripping me up I think.
No - this is not a homework question! This is for an actual application!
I am developing in ruby, but any language answer would be appreciated
You can use a string replacement to replace - with | to turn 01-02-03-10-11-12 into 01|02|03|10|11|12. Then wrap it like this:
((01|02|03|10|11|12).*){3}
This will find any of the digit pairs, then ignore any number of characters... 3 times. If it matches, then success.