I'm new into vim, I have hug text file as follow:
ZK792.6,ZK792.6(let-60),cel-miR-62(18),0.239
UTR3,IV:11688688-11688716,0.0670782
ZC449.3b,ZC449.3(ZC449.3),cel-miR-62(18),0.514
UTR3,X:5020692-5020720,0.355907
First, I would like to get delete all rows with even numbers (2,4,6...).
Second, I would like to remove (18) from entire file. as a example:
cel-miR-62(18) would be cel-miR-62.
Third: How can I get delete all parentheses including it's inside?
Would someone help me with this?
For the first one:
:g/[02468]\>/d
where :g matches all lines by the regex between the slashes and runs d (delete line) on the matching lines. The regex is quite easy to read, the only interesting symbol there is perhaps the \>, which matches end of a word.
For the second question:
:%s/\V(18)//g
where % is the specification meaning "all lines of the file", s is the substitute command, \V sets the "very nomagic" mode of regexes (not sure what your default is, you might not need this) and the final g makes vim substitute all occurrences on each line (with an empty string, the one between slashes). Make sure that :set gdefault? prints nogdefault (the default setting of gdefault), otherwise, drop the final g from the substitute command.
To remove every even line (or every other line):
:g/^/+d
To remove every instance of (18):
:%s/(18)//g
Remove all the parenthetical content:
:%s/(.\\{-})//g
Note: the pattern in third answer is a non-greedy match.
Related
I am trying to clean up some code and I am trying to find a good way of achieving the following:
I am a #decent
guy
and I want:
I am a guy
I tried using
:g/#/d
but the whole line gets deleted and I only want to delete until the end of line. What is the best way to achieve this in vim?
Thank you.
That won't because the usage of that command:
:[range]g/pattern/cmd
defaults to range being the whole line, and you are not doing any substitution anyway.
Use:
:%s/#.\+\n//g
instead.
# Matches a literal #.
.\+\n Matches everything until the end of line, and a new line.
// Replaces the entire match with nothing.
With :global you would want something like
:global/#/normal! f#D | join
or
:global/#/substitute/#.*// | join
Try this instead:
:s/ # .*\n/ /
Explanation:
You were using the wrong command, as they may look similar to new users.
:[range]g/VRE/act Globally apply the "act"ion (one letter command) to all lines (in range, default all file) matching the VRE (pattern)
:[range]s/VRE/repl/f Substitute within lines (in range, default current line) the matching VRE (pattern) with the "repl"acement using optional "f"lags
Now about the pattern, I think this candidate cover most cases (all but comments at the beginning of a line and comments without space after pound sign)
# litteral space, then hash tag, then space again
.* dot for any character, star to mean the previous may occur many times or even be absent
$ dollar at end to stay at "end of line", but \n to catch en EOL here
press d + shift 4 or d + $, which means delete to end of the line
d means delete
shift 4 or $ means cursor to end of the line
Please let me know, How I can remove the last word from each line using vim commands?
Before :
abcdef 123
xyz 1256
qwert 2
asdf 159
after :
abcdef
xyz
qwert
asdf
Similarly please let me know how to remove the second word from each line using vim command?
Use the following regex (in ex mode):
%s/\s*\w\+\s*$//
This tells it to find optional whitespace, followed by one or more word characters, followed by optional whitespace, followed by end of line—then replace it with nothing.
The question's been answered already, but here's what I'd more likely end up doing:
Record a macro:
qq to record a macro into register "q"
$ to go to the end of the line
daw to delete a word
q to stop recording
Then select the rest of the lines:
j to go down a line
vG to select to the end of the file
And apply the macro:
:norm #q
Some similar alternatives:
:%norm $daw
qq$dawjq (note the added j) then 999#q to replay the macro many times. (Macro execution stops at the first "error" -- in this case, you'd probably hit the bottom of the file, j would not work, and the macro would stop.)
The key for this is the :substitute command; it is very powerful (and often used in vi / Vim).
You need to come up with a regular expression pattern that matches what you want to delete. For the last word, that's whitespace (\s), one or more times \+ (or any number (*), depending on how you want to treat single-word lines), followed by word characters (\w\+), anchored to the end of the line ($). Note that word has a special meaning in Vim; you may want to use a different atom (e.g. \S). Voila:
:%s/\s\+\w\+$//
For the second word, you can make use of the special \zs and \ze atoms that assert for matches, but do not actually match: Anchored at the start (^), match a word, then start the match for a second one:
:%s/^\w\+\s\+\zs\w\+\s\+//
Soon, you'll also want to reorder things, not just remove them. For that, you need to know capturing groups: \(...\). The text matched by those can then be referred to in the replacement part. For example, to swap the first and second words:
:%s/^\(\w\+\s\+\)\(\w\+\s\+\)/\2\1/
For details, have a look at the help, especially :help :substitute and :help pattern.
To remove the second word from the start of a line, use the following:
:%s/^\(\s*\w\+\s\+\)\w\+\s*/\1/
Update
To treat special characters as part of the word, you have to use the \S (which matches all non-whitespace characters) instead of \w (which matches only word characters [0-9A-Za-z_]). Then, the command would be:
:%s/^\(\s*\S\+\s\+\)\S\+\s*/\1/
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.
I'm learning the power of g and want to delete all lines containing an expression, to the end of the sentence (marked by a period). Like so:
There was a little sheep. The sheep was black. There was another sheep.
(Run command to find all sentences like There was and delete to the next period).
The sheep was black.
I've tried:
:g/There was/d\/\. in an attempt to "delete forward until the next period" but I get a trailing characters error.
:g/There was/df. but get a df. is not an editor command error.
Any thoughts?
The action associated with g must be able to act on the line without needing position information from the pattern match that g implies. In the command you are using, the delete forward command needs a starting position that is not being provided.
The problem is that g only indicates a line match, not a specific character position for it's pattern match. I did the following and it did what I think you want:
:g/There was/s/There was[^.]*[.]//
This found lines that matched the pattern There was, and performed a substitution of the regular expression There was[^.]*[.] with the empty string.
This is equivalent to:
:1,$s/There was[^.]*[.]//g
I'm not sure what the g is getting you in your use case, except the automatic application to the entire file line range (same as 1,$ or %). The g in this latter example has to do with applying the substitution to all patterns on the same line, not with the range of lines affected by the substitution command.
I'd just use a regex:
%s/There was\_.\{-}\.\s\?//ge
Note how \_. allows for cross-line sentences
You can use :norm like this:
:g/There was/norm 0weldf.
This finds lines with "There was" then executes the normal commands 0weldf..
0: go to beginning of line
w: go to next word (in this case, "was")
e: go the end of the word (so cursor is on the 's' of "was")
l: move one character to the right (so we don't delete any of "was")
df.: delete until the next '.', inclusive.
If you want to keep the period use dt. instead of df..
If you don't want to delete from the beginning of the line and instead want to do sentences, the :%s command is probably more appropriate here. (e.g. :%s/\(There was\)[^.]*\./\1/g or %s/\(There was\)[^.]*\./\1./g if you want to keep the period at the end of the sentence.
Use search and replace:
:%s/There was[^.]*\.\s*//g
It happens sometimes that I have to look into various log and trace files on Windows and generally I use for the purpose VIM.
My problem though is that I still can't find any analog of grep -v inside of VIM: find in the buffer a line not matching given regular expression. E.g. log file is filled with lines which somewhere in a middle contain phrase all is ok and I need to find first line which doesn't contain all is ok.
I can write a custom function for that, yet at the moment that seems to be an overkill and likely to be slower than a native solution.
Is there any easy way to do it in VIM?
I believe if you simply want to have your cursor end up at the first non-matching line you can use visual as the command in your global command. So:
:v/pattern/visual
will leave your cursor at the first non-matching line. Or:
:g/pattern/visual
will leave your cursor at the first matching line.
you can use negative look-behind operator #<!
e.g. to find all lines not containing "a", use /\v^.+(^.*a.*$)#<!$
(\v just causes some operators like ( and #<! not to must have been backslash escaped)
the simpler method is to delete all lines matching or not matching the pattern (:g/PATTERN/d or :g!/PATTERN/d respectively)
I'm often in your case, so to "clean" the logs files I use :
:g/all is ok/d
Your grep -v can be achieved with
:v/error/d
Which will remove all lines which does not contain error.
It's probably already too late, but I think that this should be said somewhere.
Vim (since version about 7.4) comes with a plugin called LogiPat, which makes searching for lines which don't contain some string really easy. So using this plugin finding the lines not containing all is ok is done like this:
:LogiPat !"all is ok"
And then you can jump between the matching (or in this case not matching) lines with n and N.
You can also use logical operations like & and | to join different strings in one pattern:
:LP !("foo"|"bar")&"baz"
LP is shorthand for LogiPat, and this command will search for lines that contain the word baz and don't contain neither foo nor bar.
I just managed a somewhat klutzy procedure using the "g" command:
:%g!/search/p
This says to print out the non-matching lines... not sure if that worked, but it did end up with the cursor positioned on the first non-matching line.
(substitute some other string for "search", of course)
You can search with following line and press n to jump to the first non-matching line
^\(.*all is ok\)\#!.*$
Breakdown of operators:
^ -> means start of the line
\( and \) -> To match a whole string multiple times, it must be grouped into one item. This is done by putting "\(" before it and "\)" after it.
\#! -> Matches with zero width if the preceding atom does NOT match at the current position.
.* -> Matches any character repeated 1 or more times
$ -> end of the line
Here is sample animation how it works. For simplicity I searched for word apple.
You can iterate through the non-matches using g and a null substitution:
:g!/pattern/s/^//c
If you reply "n" each time you wont even mark the file as changed.
You need ctrl-C to escape from the circle (or keep going to bottom of file).