In order to merge the odd number line and the even number line in two methods.
One use command :s, the other use command :g and :s.
It's our homework and I could not get appropriate answer from the google.
And I had worked out the first one, which means I can solve it with command :s:
:%s/\(^.*$\)\n\(^.*$\)/\1 \2
And how could I use command :d and :s to solve it?
BEFORE:
1 aa
2 bb
3 abc
4 abc
5 an apple
6 is a bug
7 mazic
8 homework!
9 try a time
10 dodo
AFTER:
1 aa bb
2 abc abc
3 an apple is a bug
4 mazic homework!
5 try a time dodo
thanks to everyone and I have leant about how to solve it before the lesson.hah
:g/\(^.*$\)\n\(^.*$\)/s//\1 \2
What you can do here is :
Move the cursor to the line number to which you want to append the next line and then type below command in normal mode.
:s/\n/ /
Another way is go to the particular line and press SHIFT+V and then type below command:
:'<,'>s/\n/, /
Note that when you are in visual mode and press : then :'<,'> will automatically get typed. You just need to type regex ahead of that.
In both the above commands, g is not needed as it will not do any impact because only one \n will be there for each line.
You don't need to use :substitute here, there's a special command :join.
You can use the Ex command with :global, using ^ as the pattern to match all lines:
:global/^/join
Or use the shorter normal mode variant J:
:%normal! J
Related
I am attempting to remove .nc1 at the end of a line. I receive .nc1 in batches as a steel fabricator. We run into issues with our files where, line 5 in the example below, has an unnecessary .nc1 extension at the end. Problem I have, is that I cannot simply replace the value as it appears in line 2 as well.
In the example photo I have attached, I am looking to remove line 5 .nc1 extension and keep line 2 as is, .nc1 extension removal will be applied in a batch editing to all of my .nc1 files via find/replace.
ST
** BB233.nc1
F88
BB233
BB233.nc1
1000
A992
1
W21X201
Change to this
ST
** BB233.nc1
F88
BB233
BB233
1000
A992
1
W21X201
I was looking into Positive and/or Negative lookahead/lookbehind but didnt have much luck in making it work. I am a novice/lack thereof when it comes to using RegEx.
Match .nc1 only at the end of lines starting with whitespace, capturing the part you want to keep and putting it back, effectively deleting .nc1
Search: ^(\s+.*)\.nc1$
Replace: $1
I have a source code file where I want to comment out (using single-line comments) certain blocks that have a specific opening and ending pattern. For example:
1
2
BEGIN
3
4
END
5
6
I want the output to be of the form:
1
2
//BEGIN
//3
//4
//END
5
6
I have tried to use sed, and I can match the whole block, but what I can't figure out is how to get each line of that match and perform another sed operation to replace the start of the line with the two forward slashes.
Edit: the BEGIN and END can also be on the same line.
Answers using any other tool are also welcome. Also, kindly provide an explanation for your code for beginners like me!
source txt file:
34|Gurla Mandhata|7694|25243|2788|Nalakankar Himalaya|30°26'19"N
81°17'48"E|Dhaulagiri|1985|6 (4)|China
command input:
:%s/\(\d\+\)\(\d\d\d\)/\1,\2/g
command output:
34|Gurla Mandhata|7,694|25,243|2,788|Nalakankar Himalaya|30°26'19"N
81°17'48"E|Dhaulagiri|1,985|6 (4)|China
Desired output:
34|Gurla Mandhata|7,694|25,243|2,788|Nalakankar Himalaya|30°26'19"N
81°17'48"E|Dhaulagiri|1985|6 (4)|China
Basically 1985 is supposed to be 1985 and not 1,985. I tried to put a \? so every time the pattern matches it stops and a °+ after so it has to detect a ° to match the pattern, but no success. It just replaces the ° and everything before that, complete mess.
My knowledge of regular expressions however combined with the substitute is weak and I'm stuck here.
EDIT
the first 3 numbers represent heights of mountains, those 3 need to change with a (,) and the last number ( 1985 ) represents a year, which must not be changed.
Mathematical solutions are not going to work as loophole since there are mountains with a height off less than 1900
You haven't told us what is the difference between 1985 and other numbers, so I assumed that your "small" numbers are less than 2000.
You almost got it:
:%s/(\d*[2-90])(\d\d\d)/\1,\2/g
Alternatively if that isn't what you want, you can use c flag (:h s_flags):
:%s/\(\d\+\)\(\d\d\d\)/\1,\2/gc
this line will leave the last 3 columns untouched, just do substitution on the content before it:
%s/\v(.*)((\|[^|]*){3}$)/\=substitute(submatch(1),'\v(\d+)(\d{3})','\1,\2','g').submatch(2)/g
Note that the above line will change 1000000 into 1000,000 instead of 1,000,000. Vim's printf() doesn't support %'d, it is pity. If you do have number > 1m, we can find other solutions.
update
I solved it myself, by using 3 seperate commands; one for every number string in the file:
%s/^\(\d*|[^|]*|\)\(\d\+\)\(\d\d\d\)|/\1\2,\3|/g
:%s/^\(\d*|[^|]*|\d\+,*\d*|\)\(\d\+\)\(\d\d\d\)|/\1\2,\3|/g
:%s/^\(\d*|[^|]*|\d\+,*\d*|\d\+,*\d*|\)\(\d\+\)\(\d\d\d\)|/\1\2,\3|/g
In case you want to use perl:
:%!perl -F'\|' -lane 'for(#F[2..4]) { s/(\d+)(\d{3})/\1,\2/;} print join "|", #F'
So first a sample of the actual data mangled (data is originally a mix of text and numbers, there's no significance to any of the data at this point and some of the patterns are just because I replaced most of the characters with 0s, 1s and Zs because the random number generator in my brain is broken):
011.0ZN1ZZ 001.F5ZS1Z 001.ZO5ZY0
014.5ZZZ1Z 001.1SZZOZ 001.ZLMZY0
016.01NM1SU54 001.EX0Z1Z 001.LIZZOZ
018.01NM1SS41 001.F83Z1Z 001.0011M1SU54
014.ZZ1YZZ 001.ZZZ1IZ 001.0011M1SS41
013.2EBSIZ 001.ZZZ11Z 001.0011SE4
01N.ZINSIZ 001.ZZZZ1Z P01.ZZZZ1Z
01N.01NSE4 001.LSZZHG N01.ZZZZ1Z
001.01ON5O 001.5Z21OL F01.ZZZZ1Z
001.NE5ZO1 001.ZOM05O D01.ZZZZ1Z
001.ZO5ZOZ 001.01NO1G Z01.ZZZZ1Z
001.ZO5ZOZ 001.01NO1G Z01.ZZZZ1Z
001.011ZOZ 001.01NZ0Y
Some additional comments.. I can clean up whitespace and deal with record length with no issues, so I'd like to simplify the question to this, I'm just including the above in case there's a solution to the simplified version that can't be easily extended to a more complex version.
1 7 13
2 8 14
3 9 15
4 10 16
5 11 17
6 12 18
19 25
20 26
21 27
22 28
23 29
24
So there will be a variable number of pages, but the same number of columns and rows on each page (although, in case it matters significantly, it's actually 12x3 instead of 6x3 but I wanted to keep it simple if possible), although the last page may be some empty rows/columns.
I'm using notepad++ but I have access to various gnutilities so if there's a solution that's way, way better than a regular expression I don't mind, although since I'll be using this a lot and use notepad++ a lot I'd appreciate a regex solution if it isn't too insane.
If you've got Git installed on your Windows machine, you may use Perl bundled with it from Git bash. Provided your input file is named data, try the following command (caution: it will orverwrite the input file):
echo >>data ; \
perl -i -lane'
$i=0;
push #{$c[$i++]}, $_ foreach #F;
if (/^\s*$/) {
push #l, #{$_} foreach #c;
print "#l\015";
#l=#c=();
}' data
The Perl command treats each line of input as space delimited fields and accumulates the fields in the #c matrix. When encounters an empty line (if (/^\s*$/) ...), it prints the matrix columns concatenated in a list.
The input file is changed in-place. A backup copy data.bak is created.
The input file may not end with an empty line so I add one with echo >>data. This makes the Perl script shorter and easier.
Another trick is the trailing \015 in print "#l\015";. This allows us to get Windows CRLF line endings in Unix-flavoured Git bash environment.
A demo can be found here: https://ideone.com/vnYoOd. But since Ideone forbids file read/write, the original command has been modified to make the code run there.
Hi I am new to this forum. I want to use SED to replace an expression on even lines of a file. My problem is that I cannot think f how to save the changes in the original file (i.e, how to overwrite the changes in the file). I have tried with :
sed -n 'n;p;' filename | sed 's/aaa/bbb/'
but this does not save the changes. I appreciate your help on this.
Try :
sed -i '2~2 s/aaa/bbb/' filename
The -i option tells sed to work in place, so not to write the edited version to stout and leave the original file be, but to apply the changes to the file. The 2~2 portion is the address for the lines sed should apply the commands. 2~2 means edit only even lines. 1~2 would edit only odd lines. 5~6 would edit every fifth line, starting at line 5 etc...
#Mithrandir's answer is an excellent, correct and complete one.
I will just add that the m~n addressing method is a GNU sed extension that may not work everywhere. For example, not all Macs have GNU sed, as well as *BSD systems may not have it either.
So, if you have a file like the following one:
$ cat f
1 ab
2 ad
3 ab
4 ac
5 aa
6 da
7 aa
8 ad
9 aa
...here is a more universal solution:
$ sed '2,${s/a/#A#/g;n}' f
1 ab
2 #A#d
3 ab
4 #A#c
5 aa
6 d#A#
7 aa
8 #A#d
9 aa
What does it do? The address of the command is 2,$, which means it will be applied to all lines between the second one (2) and the last one ($). The command in fact are two commands, treated as one because they are grouped by brackets ({ and }). The first command is the replacement s/a/#A#/g. The second one is the n command, which gets, in the current iteration, the next line, appends it to the current pattern space. So the current iteration will print the current line plus the next line, and the next iteration will process the next next line. Since I started it at the 2nd line, I am doing this process at each even line.
Of course, since you want to update the original file, you should call it with the -i flag. I would note that some of those non-GNU seds require you to give a parameter to the -i flag, which will an extension to be append to a file name. This file name is the name of a generated backup file with the old content. (So, if you call, for example, sed -i.bkp s/a/b/ myfile.txt the file myfile.txt will be altered, but another file, called myfile.txt.bkp, will be created with the old content of myfile.txt.) Since a) it is required in some places and b) it is accepted in GNU sed and c) it is a good practice nonetheless (if something go wrong, you can reuse the backup), I recommend to use it:
$ ls
f
$ sed -i.bkp '2,${s/a/#A#/g;n}' f
$ ls
f f.bkp
Anyway, my answer is just a complement for some specific scenarios. I would use #Mithrandir's solution, even because I am a Linux user :)
This might work for you:
sed -i 'n;s/aaa/bbb/' file
Use sed -i to edit the file in place.