We have many \foo{something}. As the string inside the curly braces contains odd number of character, we want to add a space after the string. The result will be \foo{something }. We only want odd number of characters only. For example, in \foo{string} the string inside the curly braces has even number of character, then we do not add a space.
What I have tried:
a dot means a character, two dots means two characters
asterisk means repeat 0 or more times
Then, [..]*. means 1, 3, 5 ... characters.
Then, \foo{[..]*.} will be my matching sequence.
Then, :%s/\foo{[..]*.}/\foo{[..]*. }/g will add the space we want.
But we failed.
This cmd should do:
:%s/\\foo{\zs[^}]\+\ze}/\=substitute(submatch(0), '$', len(submatch(0))%2?' ':'','g')/
:%s/\\\w\+{\([^}]\{2}\)*[^}]\zs\ze}/ /g
Explanation:
Find pairs of characters (\([^}]\{2}\)*) followed by another character ([^}]) in between a macro \\\w\+{...}. Then do a substitution adding an additional space.
Glory of details:
\\\w\+{...} find macros of the pattern \foo{...}
Look for character that does not match an ending curly brace, [^}]
Look for pairs of non-}'s, \([^}]\{2}\)*
Find an odd length string by finding pairs and then finding one more, \([^}]\{2}\)*[^}]
Use \zs and \ze to set where the start and end of the match
Use a space in the replacement portion of the substitution to add additional space and thereby making the string even in length
Fore more help see:
:h :s
:h :range
:h /\[]
:h /\(
:h /\zs
:h /\w
:h /star
:h /\+
This should work
:%s/\v[(\{(\w{2})*) ]#<!}/ }/g
break down:
%s " run substitute on every line
\v " very magic mode
[(\{(\w{2})*) ] " matches every `{` followed by an equal number of word
characters and space, since you don't want to add a space again everytime
you run it.
#<!} " negative lookahead, matches all } not following the pattern before
Update
To solve the precise problem as you can read in the comments, the following command helps.
g/^\\foo/s/\v(\{([^}]{2})*)#<!}/ }/g
Changes:
g/^\\foo/... " only run it on lines starting with \foo
[^}] " it does now match all characters between { and } not only word
characters.
Pitfalls: won't work correct with multiple braces on the line ( \foo{"hello{}"} and \foo{string} {here it would also match} f.e. will fail). if one of these is a problem. just tell me
Related
I have some txt in vi:
|NC_004718|29751nt|SARS
|NC_045512|29903nt|Severe
|NC_004718|29751nt|SARS
|NC_045512|29903nt|Severe
|NC_004718|29751nt|SARS
now I want to replace remove everything after NC_004718, my expected output is:
NC_004718
NC_045512
NC_004718
NC_045512
NC_004718
How to do it? Thanks.
I would recommend using a substitution with regular expression to match the entire string and to capture what you would like to keep in parentheses. That way you can then replace the entire string with just the match.
:%s/^|\([^|]\+\)|.\+/\1/
To break down what is happening:
% means that you want to apply the command to each line within the file.
s means that you are doing substitution command (on each line). The s command has a syntax of s/<regular expression pattern>/<replacement>/<flags>
The regular eression pattern in the above command is ^|\([^|]\+\)|.\+.
^ means match from the line start.
| matches the character |.
\([^|]\+\) matches all characters except for the character |. Note that the real regular expression is actually ([^|]+), the additional \ characters are there because Vim needs to know that they are intended to be special characters for processing and not exact characters it needs to match. Also note that the parentheses are there to capture the match into a group (see below).
| again matches the actual character |.
.\+ matches all characters until the end of the line. Note that the . is considered special character by default but + still needs a preceding \.
The replacement text is only \1. This denotes that Vim should replace the text with whatever was captured in the first group (i.e. the first set of parentheses).
There are no flags with this command so there is nothing after the last /.
For example,
:g/NC_\d\+/normal! ygnV]p
:g/regex/ to match lines
normal! to execute Normal mode commands
ygn to yank the text previously matched by :g
V to select the whole line
]p or p to replace the line with the match
If you have only lines like those you have shown try:
:%norm xf|D
I sometimes need to write the whole alphabet abcd…z and I hate typing it letter by letter in Vim's insert mode. Does there exist any method to do this more efficiently?
I know about the ga command which gives me the ascii code of the character where the cursor is … but don't know anything about how to mix it with my standard solution to type numbers from 1 to (for example) 5000: a1ESCqqyyp^Aq4998#q …
Using set nrformats+=alpha:
ia<Esc>qqylp<C-a>q24#q
Step by step:
ia<Esc> " Start with 'a'
qqylp<C-a>q " #q will duplicate the last character and increment it
24#q " Append c..z
If your shell does brace expansion this is a pretty elegant solution:
:r !printf '\%s' {a..z}
:read! reads the output of an external command into the current buffer. In this case, it reads the output of the shell's printf applied to {a..z} after it's been expanded by the shell.
How about this command:
:put =join(map(range(char2nr('a'),char2nr('z')),'nr2char(v:val)'),'')
Collect the ASCII values of the characters in the range from a to z, then map them over the nr2char() function and insert the result into the current buffer with :put =.
When you leave out the enclosing join( … ,'') you get the characters on a separate line each.
See
:h nr2char(),
:h char2nr(),
:h :put,
and look up range(), map(), join() and friends in the list-functions table.
First, set nrformats+=alpha.
Then:
ia<ESC>Y25p<CTRL-V>}g<CTRL-A>k26gJ
Which means:
ia insert the initial a
Y25p yank the a and duplicate it on 25 lines
<CTRL-V> go into visual block mode
} go to the last character at the end of the current paragraph
g<CTRL-A> incrementally increase each alphabetic character (see help v_g_CTRL-A)
k go up one line
26gJ join 26 lines without inserting or removing any spaces
Which leads to:
abcdefghijklmnopqrstuvwxyz
I have found a shorter solution (you don't need to change nrformats beforehand) while solving http://www.vimgolf.com/challenges/5ebe8a63d8085e000c2f5bd5
iabcdefghijklm<Esc>yiwg??P
which means:
iabcdefghijklm<Esc> insert first half of the alphabet
yiw copy it
g?? ROT13 encode (shift by 13 letters) to get the second half
P paste the first half
You might try using Vim abbreviations or a full-fledged snippet manager plugin like UltiSnips. It might take a few moments to set up, and you'd have to type that alphabet one more time to define it as an abbreviation or snippet, but after that you'd be able to insert the alphabet or any other common chunk of text much more easily.
Is there an easy way using a macro or ~10 line function (no plugin!) to center some text between the first and last word (=sequence of non-blank characters) on a line? E.g. to turn
>>> No user serviceable parts below. <<<
into
>>> No user serviceable parts below. <<<
by balancing the spaces +/-1? You can assume no tabs and the result should not contain tabs, but note that the first word may not start in column 1. (EDIT: ... in fact, both delimiter words as well as the start and end of the text to center may be on arbitrary columns.)
source this function:
fun! CenterInSpaces()
let l = getline('.')
let lre = '\v^\s*\S+\zs\s*\ze'
let rre = '\v\zs\s*\ze\S+\s*$'
let sp = matchstr(l,lre)
let sp = sp.matchstr(l,rre)
let ln = len(sp)
let l = substitute(l,lre,sp[:ln/2-1],'')
let l = substitute(l,rre,sp[ln/2:],'')
call setline('.',l)
endf
note
this function might NOT work in all cases. I just wrote it quick for usual case. this is not a plugin after all
the codes lines could be reduced by combining function calls. but i think it is clear in this way, so I just leave it like this.
if it worked for you, you could create a map
it works like this: (last two lines I typed #: to repeat cmd call)
You can use the :s command with the \= aka sub-replace-expression.
:s#\v^\s*\S+\zs(\s+)(.{-})(\s+)\ze\S+\s*$#\=substitute(submatch(1).submatch(3),'\v^(\s*)(\1\s=)$','\1'.escape(submatch(2),'~&\').'\2','')#
Overview
Capture the text (including white-space) between the >>> and <<< marks. Divide up the white-space on both sides of the text in half and substitute in the non-white-space text in between. This white-space balancing act is done via the regex engine's backtracking because math is hard. Lets go shopping!
Notes:
using \v or very magic mode to reduce escaping as this command is long enough
already
use # as an alternative separator instead of the usual / for :s/pat/sub/ in hopes to make it slightly more readable
Matching Pattern
:s#\v^\s*\S+\zs(\s+)(.{-})(\s+)\ze\S+\s*$#...
:s with no range supplied only do the substitution on the current line.
^\s*\S+ match the starting white-space followed by non-white-space. >>> in this case.
(\s+)(.{-})(\s+) match white-space followed by the "text" followed by white-space
3 capture groups: 1) leading white-space, 2) the "text", and 3) trailing white-space. These will be later referenced by submatch(1), submatch(2), and submatch(3) respectively
.{-} is vim-speak for non-greedy matching or .*? in perl-speak
without the non-greedy matching the second capture group would include too much white-space at its end
\S+\s*$ match the non-white-space (i.e. <<<) and any trailing white-space
Use \zs and ze to designate the start and end of the match to be replaced
Replacement
\=substitute(submatch(1).submatch(3),'\v^(\s*)(\1\s=)$','\1'.escape(submatch(2),'~&\').'\2','')
\= tells vim that replacement will be a vim expression. Also allows the use of submatch() functions
substitute({str}, {pat}, {sub}, {flags}) Our expression will be a nested substitution
substitute(submatch(1).submatch(3), ...) do a substitute over the concatenation of leading and trailing white-spacing captured in submatch(1) and submatch(3)
The {pat} is ^(\s*)(\1\s=)$. Match some white-space followed by white-space of the same length as the first or 1 character longer. Capture both halves.
escape(submatch(2),'~&\') escape submatch(2) for any special characters. e.g. ~,&,\1, ...
The {sub} is '\1'.escape(submatch(2),'~&\').'\2'. Replace with the the escaped submatch(2) (i.e. the "text" we want to center) in between the halves of white-space, \1 and \2 from the {pat}
No {flag}'s are needed so ''
Usage
If you use this often I would suggest creating a command and putting it in ~/.vimrc.
command! -range -bar -nargs=0 CenterBetween <line1>,<line2>s#\v^\s*\S+\zs(\s+)(.{-})(\s+)\ze\S+\s*$#\=substitute(submatch(1).submatch(3),'\v^(\s*)(\1\s=)$','\1'.submatch(2).'\2','')#`
Otherwise use this once and then repeat the last substitution via & on each needed line.
For more help see
:h :s/
:h :s/\=
:h sub-replace-\=
:h submatch(
:h substitute(
:h escape(
:h /\v
:h /\S
:h /\{-
:h /\zs
:h &
EDIT by Kent
Don't be jealous, your answer has it too. ^_^
I didn't change the command, just cp/paste to my vim. only add |noh at the end to disable highlighting.
If execute this command, it looks like:
I don't know of any good way. I usually do it in a semi-automatic way, by using :center on a line of text that only contains the parts that are to be centered and then move the result into the line containing the surrounding parts.
If nobody else has a better answer, perhaps boxes can help if you need to do this kind of thing a lot.
I have a line in a source file: [12 13 15]. In vim, I type:
:%s/\([0-90-9]\) /\0, /g
wanting to add a coma after 12 and 13. It works, but not quite, as it inserts an extraspace [12 , 13 , 15].
How can I achieve the desired effect?
Use \1 in the replacement expression, not \0.
\1 is the text captured by the first \(...\). If there were any more pairs of escaped parens in your pattern, \2 would match the text capture between the pair starting at the second \(, \3 at the third \(, and so on.
\0 is the entire text matched by the whole pattern, whether in parentheses or not. In your case this includes the space at the end of your pattern.
Also note that [0-90-9] is the same as [0-9]: each [...] collection matches just one character. It happens to work anyway, because in your data ‘a digit followed by a space’ matches in the same places as ‘2 digits followed by a space’. (If you actually needed to only insert commas after 2 digits, you could write [0-9][0-9].)
"I have a line in a source file:..."
then you type :%s/... this will do the substitution on all lines, if it matched. or that is the single line in your file?
If it is the single line, you don't have to group, or [0-9], just :%s/ \+/,/g will do the job.
The fine answers already point interesting solutions, but here's another one,
making use of the \zs, which marks the start of the match. In this pattern:
/[0-9]\zs /
The searched text is /[0-9] /, but only the space counts as a match. Note
that you can use the class \d to simplify the digit character class, so the
following command shall work for your needs:
:s/\d\d\zs /, /g ; matches only the space, replace by `, '
You said you have multiple lines and these changes are only to certain lines.
You can either visually select the lines to be changed or use the :global
command, which searches for lines matching a pattern and applies a command to
them. Now you'd need to build an expression to match the lines to be changed
in a less precise as possible way. If the lines that begins with optional
spaces, a [ and two digits are the only lines to be matched and no other
ones, then this would work for you:
:g/\s*[\d\d/s/\d\d\zs /, /g
Check the help for pattern.txt for \ze and similar and
:global.
Homework: use the help to understand \zs and see how this works:
:s/\d\d\zs\ze /,/g
I have the following characters being repeated at the end of every line:
^[[00m
How can I remove them from each line using the Vim editor?
When I give the command :%s/^[[00m//g, it doesn't work.
You could use :%s/.\{6}$// to literally delete 6 characters off the end of each line.
The : starts ex mode which lets you execute a command. % is a range that specifies that this command should operate on the whole file. The s stands for substitute and is followed by a pattern and replace string in the format s/pattern/replacement/. Our pattern in this case is .\{6}$ which means match any character (.) exactly 6 times (\{6}) followed by the end of the line ($) and replace it with our replacement string, which is nothing. Therefore, as I said above, this matches the last 6 characters of every line and replaces them with nothing.
I would use the global command.
Try this:
:g/$/norm $xxxxxx
or even:
:g/$/norm $5Xx
I think the key to this problem is to keep it generic and not specific to the characters you are trying to delete. That way the technique you learn will be applicable to many other situations.
Assuming this is an ANSI escape sequence, the ^[ stands for a single <Esc> character. You have to enter it by pressing Ctrl + V (or Ctrl + Q) on many Windows Vim installations), followed by Esc. Notice how this is then highlighted in a slightly different color, too.
It's easy enough to replace the last six characters of every line being agnostic to what those characters are, but it leaves considerable room for error so I wouldn't recommend it. Also, if ^[ is an escape character, you're really looking for five characters.
Escape code
Using ga on the character ^[ you can determine whether it's an escape code, in which case the status bar would display
<^[> 27, Hex 1b, Octal 033
Assuming it is, you can replace everything using
:%s/\%x1b\[00m$//gc
With \%x1b coming from the hex value above. Note also that you have to escape the bracket ([) because it's a reserved character in Vim regex. $ makes sure it occurs at the end of a line, and the /gc flags will make it global and confirm each replacement (you can press a to replace all).
Not escape code
It's a simple matter of escaping then. You can use either of the two below:
:%s/\^\[\[00m$//gc
:%s/\V^[[00m\$//gc
If they are all aligning, you can do a visual-block selection and delete it then.
Otherwise, if you have a sequence unknown how to input, you can visually select it by pressing v, then mark and yank it y (per default into register "). Then you type :%s/<C-R>"//g to delete it.
Note:
<C-R>" puts the content of register " at the cursor position.
If you yanked it into another register, say "ay (yank to register a - the piglatin yank, as I call it) and forgot where you put it, you can look at the contents of your registers with :reg.
<C-R> is Vim speak for Ctrl+R
This seems to work fine when the line is more than 5 chars long:
:perldo $_ = substr $_, 0, -5
but when the line is 5 or less chars long it does nothing.
Maybe there is a easy way in perl to delete the last 5 chars of a string, but I don't really know it:)
Use this to delete:
:%s/^[[00m//gc