I often find myself adding numbers on the fly to a list of numbers.
p.e.
38
12 x
215 x
98 x
03 x
23
What I want to do is to select a visual block of numbers (x in the above example)
and increase or decrease the numbers with another number.
I tried to do it using two macro's (I suppose one macro isn't possible):
#a to increase the number --> 5#a --> to increase every number with 5 (#a = '^Aj')
#x to decrease the number --> 5#x --> to decrease every number with 5 (#x = '^Xj')
but...
I don't know
1) how to use the macro only in my selection (without counting lines)
2) how to change the increase/decrease number on the fly without creating a whole new macro.
3) How to resolve this: when I add 100 to the above numbers, the numbers 12, 98 and 03 are moved 1 character to the right.
Another solution is to create a function but it is i.m.o. to complex to add every time a value in an input box for a few numbers I have to change.
Once you have all your lines selected, you can do:
:'<,'>norm 5<C-v><C-a> <-- inserts ^A
to add 5 to every number.
The alignment problem can't be avoided AFAIK and yes, vimscript is probably the right tool for the job.
Increment the numbers isn't to bad. You have a handful of options, but I personally suggest using Tim Pope's speeddating plugin. It will provide a nice <c-a> visually mode mapping.
However if you want a quick and dirty mapping here you go:
xnoremap <silent> <c-a> :<c-u>exe "'<,'>norm! ".min([col("'<"),col("'>")]).'<bar>'.v:count1."\<c-a>"<cr>
Since you mentioned alignment you may also want to look at godlygeek's Tabular plugin. Drew over at vimcasts did a screencast on using tabular.
If you decide to make your own mapping/function/plugin I would also suggest you look into the following:
:h :s
:h /\%V
:h sub-replace-expression
:h printf(
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 incoming input entries.
Like these
750
1500
1
100
25
55
And There is an lookup table like given below
25
7
5
75
So when I will receive my first entry, in this case its 750. So this will look up into lookup entry table will try to match with a string which having max match from left to right.
So for 750, max match case would be 75.
I was wondering, Is that possible if we could write a regex for this kind of scenario. Because if I choose using startsWith java function It can get me output of 7 as well.
As input entries will be coming from text file one by one and all lookup entries present in file different text file.
I'm using java language.
May I know how can I write a regex for this flavor..?
This doesn't seem like a regex problem at first, but you could actually solve it with a regex, and the result would be pretty efficient.
A regex for your example lookup table would be:
/^(75?|5|25)/
This will do what you want, and it will avoid the repeated searches of a naive "check every one" approach.
The regex would get complicated,though, as your lookup table grew. Adding a couple of terms to your lookup table:
25
7
5
75
750
72
We now have:
/^(7(50?|2)?|5|25)/
This is obviously going to get complicated quickly. The trick would be programmatically constructing the appropriate regex for arbitrary data--not a trivial problem, but not insurmountable either.
That said, this would be an..umm...unusual thing to implement in production code.
I would be hesitant to do so.
In most cases, I would simply do this:
Find all the strings that match.
Find the longest one.
(?: 25 | 5 | 75? )
There is free software that automatically makes full blown regex trie for you.
Just put the output regex into a text file and load it instead.
If your values don't change very much, this is a very fast way to do a lookup.
If it does change, generate another one.
Whats good about a full blown trie here, is that it never takes more than 8
steps to match.
The one I just did http://imgur.com/a/zwMhL
App screenshot
Even a 175,000 Word Dictionary takes no more than 8 steps.
Internally the app initially makes a ternary tree from the input
then converts it into a full blown regex trie.
I'm in need of a regex, which takes a minimum and a maximum number to determine valid input, And I want the maximum and minimum to be dynamic.
I have been trying to get this done using this link
https://stackoverflow.com/a/13473595/1866676
But couldn't get it to work. Can someone please let me know how to do this.
Let's say I want to make a html5 input box, and I Want it to only receive numbers from 100 to 1999
What would a regex for this like this look like?
First off, while it is possible to do this, I think if there is a simpler way to choose a number range such as <input type="number" min="1" max="100">, that way would be preferred.
Having said that, here's how the kind of regex you requested works:
ones: ^[0-9]$ // just set the numbers -- matches 0 to 9
tens: ^[1-3]?[0-9]$ //set max tens and max ones -- matches 0 to 39
tens where max does not end in 9 ^[1-2]?[0-9]$|^[3][0-4]$ // 0 to 34
only tens: ^[1][5-9]$|^[2-3][0-9]$|^[4][0-5]$ // 15 to 45
Here, lets pick an arbitrary number 1234 to 2345
^[1][2][3][4-9]$|
^[1][2][4-9][0-9]$|
^[1][3-9][0-9][0-9]$|
^[2][0-2][0-9][0-9]$|
^[2][3][0-3][0-9]$|
^[2][3][4][0-5]$
https://regex101.com/r/pP8rQ7/4
Basically the ending of the middle series always needs to be a straight range that can reach 9 unless we are dealing with the ones place, and if it cant, you have to build it upwards toward the middle each time we have a value that can't start in 0 and then once we reach a value that cant end in 9 break early and set it in the next condition.
Notice the pattern, as each place solidifies. Also keep in mind that when dealing with going from lower to higher places, optional operators ? should be used.
Its a bit complex, but its nowhere near impossible to design a custom range with a bit of thought.
If you are more specific, we can craft an exact example, but this is generally how it is done:beginning-range|middle-range|end-range
You should only need beginning or end-ranges in certain cases like if the min or max does not end in 9. the ? means that the range that comes after it is optional. (so for example in the first case it lets us have both single and double numbers.
so for 100 - 1999 it's quite simple actually because you have lots of 9's and 0's
/^[1-9][0-9][0-9]$|^[1][0-9][0-9][0-9]$/
https://regex101.com/r/pP8rQ7/1
Note: Single values don't need ranges [n] I just added them for readability.
Edit: There used to be a regex range generator at: http://gamon.webfactional.com/regexnumericrangegenerator/. It appears to be offline now.
Essentially, you can't.
For every numeric range, there exists a regex that will match numbers in that range, therefore it is possible to write code that can generate a such regex. But such a regex is not a simple reformatting of the range ends.
However, such code would require colossal effort and complexity to write compared to code that simply checked the number using numeric methods.
With HTML 5 simply put a range input...
<form>
Quantity (between 100 and 1999):
<input type="number" name="quantity" min="100" max="1999">
</form>
with regex:
^([12345679])(\d)(\d)|^(1)(\d)(\d)(\d)
So if you need to create the regex dinamically it's possible but a bit tricky and complex
I am close, but I need some help to complete a regex. Here is the goal:
Should succeed:
10.05
3.00
50
Should fail:
55.99 (>50)
3.001 (can't have the "1" at the end)
0.50 (< 3)
.99 (< 3)
$50 (can't have "$")
5.2 (if decimal, must have 2 digits after)
Here's the regex I have so far, but it doesn't quite do all the above correctly:
^([1-4][0-9]|50|[3-9])+(\.[0-9][0-9])?$
Can anyone share the answer? Thanks!
^(50(\.00)?|([1-4][0-9]|[3-9])(\.[0-9][0-9])?)$
There were two issues. Firstly, you had allowed non-zero values after the decimal point, even if the value before it was 50. So I separated that out on the top level. Secondly, just remove the +. Because due to it, you can have much larger numbers (by chaining 50 and 43 together, for instance).
However, as Bergi mentioned in a comment, it would be better to just check the format, and do the range check separately (without regex). This would be the format check:
^\d+(\.\d\d)?$
I found a online utility that returns a regex for integers when input the lower and upper limits of the range you want. I used it for the part before the . with limits 3-50 and after the . with limits 0-99. Here is the result:
^0*([3-9]|[1-4][0-9]|50)(\.[0-9]{2})?$
A quick glance... just remove the +
^([1-4][0-9]|50|[3-9])(.[0-9][0-9])?$
You should remove the + before the potential cents. Also, you will need to handle 50$ as a special case, because it can only have .00 after it and not any cent amount.
Also, I changed the [0-9] to the shortcut for digits: \d
/^((0?[3-9]|[1-4]\d)(\.\d\d)?|50(\.00)?)$/
I know this is a long shot, but I have a huge text file and I need to add a given number to other numbers matching some criteria.
Eg.
identifying text 1.1200
identifying text 1.1400
and I'd like to transform this (by adding say 1.15) to
identifying text 2.2700
identifying text 2.2900
Normally I'd do this in Python, but it's on a Windows machine where I can't install too many things. I've got Vim though :)
Here is a simplification and a fix on hobbs' solution:
:%s/identifying text \zs\d\+\(.\d\+\)\=/\=(1.15+str2float(submatch(0)))/
Thanks to \zs, there is no need to recall the leading text. Thanks to str2float() a single addition is done on the whole number (in other words, 1.15 + 2.87 will give the expected result, 4.02, and not 3.102).
Of course this solution requires a recent version of Vim (7.3?)
You can do a capturing regex and then use a vimscript expression as a replacement, something like
:%s/\(identifying text \)\(\d\+\)\.\(\d\+\)/
\=submatch(1) . (submatch(2) + 1) . "." . (submatch(3) + 1500)
(only without the linebreak).
Your number format seems to be a fixed one, so it's easy to convert to int and come back (remove the dot) add 11500 and put the dot back.
:%s/\.//
:%normal11500^A " type C-V then C-a
:%s/....$/.&/
If you don't want to do that on all the lines but only the one which match 'identifying text' replace all the % by 'g/indentifying text/'
For integers you can just use n^A to add n to a number (and n^X to subtract it). I doubt whether that works for fractional numbers though.
Well this might not be a solution for vim but I think awk can help:
cat testscript | LC_ALL=C awk '{printf "%s %s %s %s %.3f\n", $1,$2,$3,$4,$5+1.567 }'
and the test
this is a number 1.56
this is a number 2.56
this is a number 3.56
I needed the LC_ALL=C for the correct conversion of the floating point separator, and maybe there is a more elegant solution for printing the beginning/ rest of the string. And the result looks like:
this is a number 3.127
this is a number 4.127
this is a number 5.127
Using macro
qa .......................... start record macro 'a'
/iden<Enter> ................ search 'ident*' press Enter
2w .......................... jump 2 words until number one (before dot)
Ctrl-a ...................... increases the number
2w .......................... jump to number after dot
1500 Ctrl-a ................. perform increases 1500 times
q ........................... stop record to macro 'a'
if you have 300 lines with this pattern just now making
300#a