Lets say, I have wrote this piece of LaTeX in vim:
A_{\vec B}_
^
Cursor
The cursor is at the marked position. Now I want to add a \vec in front of the A.
I tried <Esc>Bi\vec <Esc>WW but I want B and W to jump over spaces which are inside of curled braces.
Any ideas how to do that?
Edit: Maybe I should've wrote, that I wanted to map this to a shortcut. A and B are just Examples for Strings which do not include spaces.
SOLUTION: I think I have found, what I was searching. With ? you can do a backward search with a regular expression.
You can't make B or W treat a braced string as a Word, but you can use % to jump from one end of a braced string to the other.
There are several ways to achieve a similar job. W and B only stop at whitespaces, as far as I know, so you can jump twice between words with 2B, or set insert mode at the beginning with I, or search for a letter, with f, F, t or T. Also 0 and $ to go to the beginning or end directly. There are many possibilities. I hope one of them can be useful instead.
FAi\vecESC$
Does it assuming that B is really at the end of the line
Related
I'm trying to figure out how to dt or df the last occurrence of a character in a string.
For example, let's say I have the following line:
foo not.relevant.text.bar
If I f df. I expectedly get foo relevant.text.bar but I would like to get foo bar. Using f 3df. is not an option as I don't know how many of that character will be in the string. Additionally, I may want to get foo .bar (f 3dt.), or if the line ends with a dot, I may want to get foo .. I want to always find the last one regardless of how many there are.
Is this possible without a regex? I suppose I could use a regex but I was hoping there was a simple vim command that I'm missing. I find myself trying to do something like this often.
one way without using regex, without counting "dot" (could be other letters)... see if others have better way..
foo[I]not.relevant.text.bar ([I] is cursor)
you could try:
lmm$T.d`m
or in this format, may look better?
lmm$T.d`m
this will do the job. you could create a mapping if you use that often.
EDIT
I add a GIF animation to show it works. :)
note
I typed #= in normal mode after moving my cursor to the right starting point (by f(space)), to display the keys I pressed in command line.
You can use my JumpToLastOccurrence plugin. It provides ,f / ,F / ,t / ,T commands that do just that.
I would use f df...
It is not necessarily shorter to type, but I find it easier to use "repeat last command" than counting in advance the number of word/sentence I want to delete.
Then you can adjust the number of . you type to adjust the length of the string you want to delete.
For your example: ET.dB
foo not.relevant.text.bar
And it works, as long as the cursor is anywhere within the text following "foo".
Strip Path from Path+Filename: ET/dB
I use it for stripping a pathname of all but the trailing filename.
Strip the path from /some/long/path/filename.ext leaving only the filename.
Just as long as:
The cursor is anywhere within the bold word
There are no spaces in that word
E Go to the end (since there are no spaces - also works if not the last thing on the line)
T/ Find the last / (stop just after it, so it will be deleted, as well)
dB Delete to the beginning of the word
In visual mode:
$F.d^
The $ goes to the end of the current line, F searches backward for a period and d^ deletes till the beginning of the line.
What is the shortest combo?
P.S. I usually do ACtrl+wCtrl+wCtrl+wEsc.
In other words, I have to jump into Insert mode.
P.P.S. d3b doesn't work.
This sequence of commands,
$3bD
is the shortest way to do it I can think of at the moment but it leaves out an ugly trailing space.
$3gelD
is another way that gets rid of the trailing space.
You can also think outside the box and play with spaces instead of words:
$3F D
$ 3b D which translates to
$ to the end of a line
3 three of whatever is next
b backward three [one without modifier] words
D delete the characters under the cursor until the end of the line
In addition, you can also use the normal command in ex mode to achieve the same
:norm! $3bD
$3bD
Or
$3bDx
...if you also want to delete the trailing space.
Go to the end of the line, go back 3 words, then delete to the end of the line: $ 3b D
or
Go to the end of the line, delete back 3 words, then delete the extra character: $ 3db x
As others have mentioned, $3bD is the shortest, and $3bDx will suffice to remove the remaining whitespace, assuming it's exactly one space or tab.
In order to account for all whitespace, $3gelD can be used. This option has the benefit of cleaning up properly in most cases, but it fails when a line contains exactly 3 words because the 3ge motion jumps back to the previous line.
An alternative approach is based on daw. Since 3daw acts forward, either $daw.. or $3b3daw must be used. For 3 words, $daw.. is shorter and probably more desirable, but $3b3daw is more Vim-esque and can easily be extended to more words.
b, ge, and w can be replaced with B, gE, and W in order to remove WORDs instead of words.
Using vim I would like to replace all characters up to a certain one with another character, say a blank space - without affecting the layout/number of characters in the line. Here's an example:
Before:
real(kind=R12), intent(out) :: my_var
After replacing , intent(out) with blanks (i.e. starting from ,, and going up to )):
real(kind=R12) :: my_var
I know about r to replace one character, and about nr to replace n characters, but I would like to know whether I can accomplish my task without first having to count the characters I want to replace.
Thanks a lot for your replies!
Visual mode is probably the shortest way here:
vt:r
v enter visual mode
t: select till :
r (note space after r) replace selected region with spaces.
In command mode type 'df?' to delete up to that (?) character. Then 'i' to go back to insert.
For example if the following sentence is in your view:
The wizard quickly jinxed the gnomes before they vaporized.
and you enter dfs
You will be left with:
before they vaporized.
I know about r to replace one character
Did you know that R will keep you in that replace mode? So you could hit R and then hold Space until you've replaced everything you want.
However, I'd still go with Thor's answer. Visual mode allows you to use the efficient text navigation methods in vim without having to count out characters.
But if you disagree, there's always EasyMotion.
You can use regular expression here (use (.*?) to reference all values up to a token).
For instance:
The regex: (.*?)foo will get rid of everything up to foo.
For example if I have some code like:
foo = bar("abc", "def", true, callback);
Is there a nice command to move true to the 1st or 2nd position leaving the commas intact?
P.S as a bonus my friend want to know if this works in Emacs too.
In Vim if you place the cursor at the start of the first word and do dWWP then it will have the desired effect. Here is a breakdown:
dW delete the current word, including the comma and the following whitespace
W move to the start of the next word
P insert the deleted text before the cursor
This will work if there are further parameters after the pair to be swapped - it will need to be modified if there are only two parameters or you want to swap the last two parameters, since it will paste the text after the closing bracket.
Alternatively you could use a regex substitution:
:%s/(\([^,]\+\),\s*\([^,)]\+\)/(\2, \1/
This will find the first two arguments after the open bracket and swap them.
update:
A search of vim.org found the swap parameters plugin, which should do exactly what you want and can handle situations that either of the above methods cannot.
I don't know the answer for vi, but in Emacs, transpose-sexps (C-M-t) will swap two arguments either side of the cursor. Actually transpose-words (M-t) was my first guess, but that leaves the quotes behind.
You need a transpose emacs command. But its limited to not guessing that its transposing in lists, it only considers text (it can't guess the 1st, 2nd word of list). Try this.
Keep your cursor at after comma of true. Use M-x transpose-words. By default it will transpose with next word from the point. Shortcut is M-t.
You can use C-u 2 M-t for transpose with next second word.
Now coming to your question. If you want to move true, to backward 1 word, use C-u -1 M-t, and for backward 2 words C-u -2 M-t.
Am not a VIM guy. So sorry bout that.
If you want to do this as a refactoring, not just as text manipulation, I'd suggest looking into Xrefactory, a refactoring tool for Emacsen (free for C/Java, commercial for C++).
Transposing previous (Ctrl-t p) and next (Ctrl-t n) argument ... add the
following into your .vimrc file:
map <C-t>p ?,\\|(<CR>wd/,\\|)<CR>?,\\|(<CR>"_dw?,\\|(<CR>a, <C-c>?,<CR>P/,<CR>w
map <C-t>n ?,\\|(<CR>wv/,<CR>d"_dw/\\,\\|)<CR>i, <C-r>"<C-c>?,<CR>?,\\|(<CR>w
My previous question seems to be a bit ambiguous, I will rephrase it:
I have a file like this:
copythis abc
replacethis1 xyz
qwerty replacethis2
hasfshd replacethis3 fslfs
And so on...
NOTE: replacethis1, replacethis2, replacethis3, ... could be any words
How do I replace "replacethis1","replacethis2","replacethis3",.. word by "copythis" word by using minimum vim commands.
One way I can do is by these steps:
delete "replacethis1","replacethis2","replacethis3",.. by using 'dw'
copy "copythis" using 'yw'
move cursor to where "replacethis1" was and do 'p'; move cursor to where "replacethis2" was and do 'p' and so on...
Is there a better way to do this in VIM (using less number of vim commands)?
Since you changed your question, I'd do it this way:
Move to the first "replacethis1" and type cw (change word), then type "copythis" manually.
Move to the next "replacethis", hit . (repeat last operation)
Move to the next "replacethis", hit .,
and so on, and so on.
If "copythis" is a small word, I think this is the best solution.
The digit needs to be included, and there could be more than one instance per line:
:%s/replacethis\d/copythis/g
Given that "replacethis[1-3]" can be arbitrary unrelated words, the quickest/simplest way to do this globally would be:
:%s/replacethis1\|replacethis2\|replacethis3/copythis/g
(Note that you need to use \| to get the pipes to function as "or". Otherwise, vim will look for the literal | character.)
I've been struggling with this for a long time too, I think I just worked out the cleanest way:
Use whichever command is cleanest to put copythis into register r:
/copythis
"rye
Then go to the replacement and replace it with the contents of r:
/replacethis
cw<CTRL-R>r<ESC>
Then you can just n.n.n.n.n.n.n. for the rest of them, or if they're wildly different just go to the beginning of each and hit .
The key is replacing and pasting in one step so you can use . later.
:%s/copythis/replacethis/g
To replace all occurrences of copythis with replacethis. Or you can specify a range of line numbers like:
:8,10 s/copythis/replacethis/g
Note, the /g on the end will tell it to replace all occurrences. If you leave that off it will just do the first one.
create this mapping:
:map z cwcopythis^[
( ^[ is the escape character, you can type it in vim using Ctrl+V Ctrl+[ )
go to each word you want to replace and press z
if u need to do essentially the same action multiple times - swap 1st word of one line with second word of the next line, I say you could record a macro and call it whenever you need to
Have you tried string replacement?
%s/replacethis/copythis
A host of other parameters are possible to fine-tune the replacement. Dive into the Vim help for more details. Some more examples here.
You can remap e.g. the m key in normal mode to delete the word under the cursor and paste the buffer: :nnoremap m "_diwP.
Then you can just copy the desired word, move the cursor anywhere onto the to-be-replaced word and type m.
EDIT: Mapping to m is a bad idea since it is used to mark locations. But you can use e.g. ; anyway.