I have an XML file where I only care about the size attribute of a certain element.
First I used
global!/<proto name="geninfo"/d
to delete all lines that I don't care about. That leaves me a whole bunch of lines that look like this:
<proto name="geninfo" pos="0" showname="General information" size="174">
I want to delete everything but the value for "size."
My plan was to use substitute to get rid of everything not matching 'size="[digit]"', the remove the string 'size' and the quotes but I can't figure out how to substitute the negation of a string.
Any idea how to do it, or ideas on a better way to achieve this? Basically I want to end up with a file with one number (the size) per line.
You can use matching groups:
:%s/^.*size="\([0-9]*\)".*$/\1/
This will replace lines that contain size="N" by just N and not touch other lines.
Explanation: this will look for a line that contains some random characters, then somewhere the chain size=", then digits, then ", then some more random characters, then the end of the line. Now what I did is that I wrapped the digits in (escaped) parenthesis. That creates a group. In the second part of the search-and-replace command, I essentially say "I want to replace the whole line with just the contents of that first group" (referred to as \1).
:v:size="\d\+":d|%s:.*size="\([^"]\+\)".*:\1:
The first command (until the | deletes every line which does not match the size="<SOMEDIGIT(S)>" pattern, the second (%s... removes everything before and after size attr's " (and " will also be removed).
HTH
Related
The list name is disk and its below:
disks
['5000cca025884d5\n', '5000cca025a1ee6\n']
The file name is p and its below:
c0t5000CCA025884D5Cd0 solaris
/scsi_vhci/disk#g5000cca025884d5c
c0t5000CCA025A1EE6Cd0
/scsi_vhci/disk#g5000cca025a1ee6c
c3t50060E8007DB981Ad1
/pci#400/pci#1/pci#0/pci#8/SUNW,emlxs#0/fp#0,0/ssd#w50060e8007db981a,1
c3t50060E8007DB981Ad2
/pci#400/pci#1/pci#0/pci#8/SUNW,emlxs#0/fp#0,0/ssd#w50060e8007db981a,2
c3t50060E8007DB981Ad3
/pci#400/pci#1/pci#0/pci#8/SUNW,emlxs#0/fp#0,0/ssd#w50060e8007db981a,3
c3t50060E8007DB981Ad4
i want to search elements of a list in file
There are a couple of things to look at here:
I haven't actually used re.match() before, but I can see the first issue: Your list of disks has a newline character after every entry, so that will mess up matches. Also, re.match() only matches from the start of the line. Your lines start with numbers, so you need to search during the line, using re.search(). Finally, you should make it case insensitive; one option to d this is to make everything lowercase just as your disks list is.
try adapting your loop as so:
#.strip() will get rid of new lines and .lower() will make the string lowercase
for line in q:
if re.search(disks[0].strip(),line.lower()):
print line
If that doesn't fix it, I would try making it print out disks[0].strip() and line for every iteration of the loop (not just when it matches the if clause) to make sure it's reading in what you think it is.
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 am a beginner at Vim and I've been reading about substitution but I haven't found an answer to this question.
Let's say I have some numbers in a file like so:
1
2
3
And I want to get:
(1)
(2)
(3)
I think the command should resemble something like :s:\d\+:........ Also, what's the difference between :s/foo/bar and :s:foo:bar ?
Thanks
Here is an alternative, slightly less verbose, solution:
:%s/^\d\+/(&)
Explanation:
^ anchors the pattern to the beginning of the line
\d is the atom that covers 0123456789
\+ matches one or more of the preceding item
& is a shorthand for \0, the whole match
Let me address those in reverse.
First: there's no difference between :s/foo/bar and :s:foo:bar; whatever delimiter you use after the s, vim will expect you to use from then on. This can be nice if you have a substitution involving lots of slashes, for instance.
For the first: to do this to the first number on the current line (assuming no commas, decimal places, etc), you could do
:s:\(\d\+\):(\1)
The \(...\) doesn't change what is matched - rather, it tells vim to remember whatever matched what is inside, and store it. The first \(...\) is stored in \1, the second in \2, etc. So, when you do the replacement, you can reference \1 to get the number back.
If you want to change ALL numbers on the current line, change it to
:s:\(\d\+\):(\1):g
If you want to change ALL numbers on ALL lines, change it to
:%s:\(\d\+\):(\1):g
You can do what you want with:
:%s/\([0-9]\)/(\1)/
%s means global search and replace, that is do the search/replace for every line in the file. the \( \) defines a group, which in turn is referenced by \1. So the above search and replace, finds all lines with a single digit ([0-9]), and replaces it with the matched digit surrounded by parentheses.
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).