I have about 256 lines in a text file that look like /*0*/L"", I want to remove the last , and then put the remaining as a function argument code.append(/*0*/L""); I tried doing it with vim but I don't have much experience in it. how can we place something within something else in vi or vim?
:%s#\v(/\*0\*/L""),#code.append(\1);#
:%s : substitute all lines
# : alternative separator
\v : use very magic mode (see :h magic)
(/\*0\*/L""), : capture the regex, excluding the trailing comma
\1 : insert first captured group
this line would do the substitution on all lines in your buffer, only if the line ending with comma. No matter you had /*0*/L"", or /*123*/L"",
%s/\v(.*),$/code.append(\1)/
if you want to shrink the sub on certain pattern, change the .* part in the above cmd to fit your needs.
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
In the code below, I want to replace/remove the , from the line above .VSS(VSS).
It is at multiple places in the file. I have basic knowledge of gvim and I could not figure out how to just search and then pipe it with replace.
ANTENNABWP7THVT ANTENNABWP7THVT_spr_gate156 (
**.I(LTIELO_NET),
.VSS(VSS),**
.VDD(VDD));
Matching each line that contains .VSS(VSS) and doing something with it can be done with :global. You then want to address the line above it; that's a :help :range: .-1 (or short -1). And removal of a (all with the /g flag) comma can be done with plain :substitute. Taken together:
:global/\.VSS(VSS)/-1substitute/,//
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.
I have a text file with about 2,000 lines of text and I want to remove a large portion of each line of text.
The text is in this format:
Important Text that I care about. Stuff I want to delete
I am unsure as to how to delete all of the text after the . in each line.
Can someone give me a quick command that would do this?
With substitutions:
:%s/\..*/./
With :normal command:
:%norm f.lD
Various additional :normal solutions:
:%norm )Dx
:%norm $T.D
:%norm f.C.
:%norm 0/\. /e<C-v><CR>D
Use the Substitution Ex Command to Trim All Lines
This is very similar to both answers, yet I think there is value in presenting it.
Like the other answers, I just used the ex substitution command:
:%s/[^.]*$//
Explanation of substitution:
% indicates a range for all lines.
[^.] is a character class of all non-period characters
* is a quantifier indicating 0 or more matches.
$ is an anchor which communicates to VIM that we want this pattern to match at the end of the line.
Addendum
The solution assumes each line will have a period, otherwise the command will not work as expected as #Qeole has indicated. Qeole's solution addresses non-periods lines appropriately.
Use search and replace "vim feature" combined with regex:
:%s/\..*$//g
with the cursor at the first character of first line.
fS<Ctrl-V>G$d
I always wanted to know, how you can substitute within given parameters.
If you have a line like this:
123,Hello,World,(I am, here), unknown
and you wnat to replace World with Foobar then this is an easy task: :%s/World/Foobar/
Now I wonder how I can get rid of a , which is wihtin the ( ).
Theoretically I just have to find the first occurance of ( then substitute the , with a blank until ).
Try lookahead and lookbehind assertions:
%s/([^)]*\zs,\ze.*)//
(\zs and \ze tell where pattern starts and end)
or
%s/\(([^)]*\)\#<=,\(.*)\)\#=//
The first one is more readable, the second one uses \( ... \) groupings with parentheses inside groups which makes it look like obfuscated, and \#<= which apart from being a nice ASCII-art duck is the lookbehind operator, and \#= that is the lookahead operator.
References: :help pattern (more detail at :help /\#=, :help /\ze, etc.)
You use the GUI and want to try those commands? Copy them into the clipboard and run :#+ inside Gvim.
Modifying slightly the answer of #Tom can give you a quite good and "a bit" more readable result :
%s/\(.*\)(\(.*\),\(.*\))\(.*\)/\1(\2\3)\4/
That way you will have : in \1 will store what is at the left outside of the parenthesis, \4 what is at the right outside of the parenthesis and \2 and \3 what is inside the parenthesis, respectively on the left (\2) and on the right (\3).
With that you can easily swap your elements if your file is organised as column.
You can also select the text you want to change (either with visual or visual-block modes) and enter the : to start the replace command. vi will automatically start the command with :'<,'> which applies the command to the selected area.
Replacing a , can be done with:
:'<,'>s/,/ /g
For your example, this is the same thing as suggested by #ubuntuguy
%s/\(.*\)(\(.*\),\(.*\)/\1(\2\3
This will do the exact replacement you want.
Yet another approach, based on the fact that actually you want to substitute only the first occurrence of , inside the parenthesis:
:%s#\((.\{-}\),#\1 #
Explanation:
:%s for substitution in the whole file (don't use % if you want to work only with the current line)
we can use # or : as a delimiter to make the command more readable
in (.\{-} we ask to find any symbol (dot) after the left parenthesis and the rest stands for 0 or more occurrence (as few as possible) of this symbol. This expression is put inside \(...\) to be able to refer to this group as \1 in the future.