The command J joins lines.
The command gJ joins lines removing spaces
Is there also a command to Join lines adding a separator between the lines?
Example:
Input:
text
other text
more text
text
What I want to do:
- select these 4 lines
- if there are spaces at start and/or EOL remove them
- join lines adding a separator '//' between them
Output:
text//other text//more text//text
You can use :substitute for that, matching on \n:
:%s#\s*\n\s*#//#g
However, this appends the separator at the end, too (because the last line in the range also has a newline). You could remove that manually, or specify the c flag and quit the substitution before the last one, or reduce the range by one and :join the last one instead:
:1,$-1s#\s*\n\s*#//#g|join
I wrote a plugin "Join", could do what you wanted, and more.
https://github.com/sk1418/Join
Except for all features provided by the build-in :join command, Join can:
Join lines with separator (string)
Join lines with or without trimming the leading/trailing whitespaces
Join lines with negative count (backwards join)
Join lines in reverse
Join lines and keep joined lines (without removing joined lines)
Join lines with any combinations of above options
check the homepage for details and examples/screenshots.
There are few ways to do it, but I would recommend going by simplest route possible - recording a macro or doing multi-step command, for example by:
Appending to all lines excluding last by
Using substitution (:1,$-1s#$#//#)
Appending (:1,$-1norm A//)
And then join using visual selection (vGgJ) or by any other method.
Unless you're doing this operation very often you most likely forget any complex commands or existence of specialized plugin in your config, thus my recommendation of using generic, often used sub steps.
Another substitution, for the sake of diversity:
:%s:\n\ze.://
Will list 50 items per line seperated by spaces:
seq 0 70 | xargs -L 50 | sed 's/ /,/g'
Output:
0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49
50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70
Related
I have a large pipe-delimited text file that should have one 3-column record per line. Many of the records are split up by line breaks within a column.
I need to do a find/replace to get three, and only three, pipes per line/record.
Here's an example (I added the line breaks (\r\n) to demonstrate where they are and what needs to be replaced):
12-1234|The quick brown fox jumped over the lazy dog.|Every line should look similar to this one|\r\n
56-7890A|This record is split\r\n
\r\n
on to multiple lines|More text|\r\n
09-1234AS|\r\n
||\r\n
\r\n
56-1234|Some text|Some more text\r\n
|\r\n
76-5432ABC|A record will always start with two digits, a dash and four digits|There may or may not be up to three letters after the four digits|\r\n
The caveat is that I need to retain those mid-record line breaks for the target system. They need to be replaced with \.br\. So the final result of the above should look like this:
12-1234|The quick brown fox jumped over the lazy dog.|Every line should look similar to this one|\r\n
56-7890A|This record is split\.br\\.br\on multiple lines|More text|\r\n
09-1234AS|\.br\||\.br\\r\n
56-1234|Some text|Some more text\.br\|\r\n
76-5432ABC|A record will always start with two digits, a dash and four digits|There may or may not be up to three letters after the four digits|\r\n
As you can see the mid-record line breaks have all been replaced with \.br\ and the end-of-line line breaks have been retained to keep each three-column/pipe record on its own line. Note the last record's text, explaining how each line/record begins. I included that in case that would help in building a regex to properly identify the beginning of a record.
I'm not sure if this can be done in one find/replace step or if it needs to be (or just should be) split up into a couple of steps.
I had the thought to first search for |\r\n, since all records end with a pipe and a CRLF, and replace those with dummy text !##$. Then search for the remaining line breaks with \r\n, which will be mid-column line breaks and replace those with \.br\, then replace the dummy text with the original line breaks that I want to keep |\r\n.
That worked for all but records that looked like the third record in the first example, which has several line breaks after a pipe within the record. In such a large file as I am working with it wasn't until much later that I found that the above process I was using didn't properly catch those instances.
You can use
(?:\G(?!^(?<!.))|^\d{2}-\d+[A-Z]*\|[^|]*?(?:\|[^|]*?)?)\K\R+
Replace with \\.br\\. See the regex demo. Details:
(?:\G(?!^(?<!.))|^\d{2}-\d+[A-Z]*\|[^|]*?(?:\|[^|]*?)?) - either the end of the previous match (\G(?!^(?<!.))) or (|) start of a line, two digits, 0, one or more digits, zero or more letters, a |, then any zero or more chars other than |, as few as possible, and then an optional sequence of | and any zero or more chars other than |, as few as possible (see ^\d{2}-\d+[A-Z]*\|[^|]*?(?:\|[^|]*?)?)
\K - omit the text matched
\R+ - one or more line breaks.
See the Notepad++ demo:
If you need to remove empty lines after this, use Edit > Line Operations > Remove Empty Lines.
I have a bash script that outputs two CSV columns. I need to prepend the three-digit number of those rows of the second column that contain them with "f. " and keep the rest of the rows intact. I have tried different ways so far but each has failed in one way or another.
What I've tried mainly has been to use regular expressions with either the first or second column to separate the desired rows from the rest, but I can't separate and prepend at the same time without cancelling out or messing up the process somehow. Some of the commands I've used so far have been: $ sed $ cut as well as (nested) for loops, read-while loops, if/else and if/else/elif statements, etc. What follows is one such (failed) solution:
for var1 in "^.*_[^f]_.*"
do
sed -i "" "s:$MSname::" $pathToCSV"_final.csv"
for var2 in "^.*_f_.*"
do
sed -i "" "s:$MSname:f.:" $pathToCSV"_final.csv"
done
done
And these are some sample rows:
abc_deg0014_0001_a_1.tif,British Library 1 Front Board Outside
abc_deg0014_0002_b_000.tif,British Library 1 Front Board Inside
abc_deg0014_0003_f_001r.tif,British Library 1 001r
abc_deg0014_0004_f_001v.tif,British Library 1 001v
…
abc_deg0014_0267_f_132r.tif,British Library 1 132r
abc_deg0014_0268_f_132v.tif,British Library 1 132v
abc_deg0014_0269_y_999.tif,British Library 1 Back Board Inside
abc_deg0014_0270_z_1.tif,British Library 1 Back Board Outside
Here $MSname = British Library 1 (since with different CSVs the "British Library 1" part can change to other words that I need to remove/replace and that's why I use parameter expansion).
The desired result:
abc_deg0014_0002_b_000.tif,Front Board Inside
abc_deg0014_0003_f_001r.tif,f. 001r
…
abc_deg0014_0268_f_132v.tif,f. 132v
abc_deg0014_0269_y_999.tif,Back Board Inside
If you look closely, you'll notice these rows are also differentiated from the rest by "f" in their first column (the rows that shouldn't get the "f. " in front of their second column are differentiated by "a", "b", "y", and "z", respectively, in the first column).
You are not using var1 or var2 for anything, and even if you did, looping over variables and repeatedly running sed -i on the same output file is extremely wasteful. Ideally, you would like to write all the modifications into a single sed script, and process the file only once.
Without being able to guess what other strings than "British Library 1" you have and whether those require different kinds of actions, I would suggest something along the lines of
sed -i '/^[^,]*_f_[^,_]*,/s/,British Library 1 /,f. /
s/,British Library 1 /,/' "${pathToCSV}_final.csv"
Notice how the sed script in single quotes can be wrapped over multiple physical lines. The first line finds any lines where the last characters between underscores in the first comma-separated column is f, and replaces ",British Library 1 " with ",f. ". (I made some adjustments to the spacing here -- I hope they make sense for you.) On the following line, we simply replace any (remaining) occurrences of ",British Library 1 " with just a comma; the idea is that only the lines which didn't match the regex on the previous line will still contain this string, and so we don't have to do another regex match.
This can easily be extended to cover more patterns in the same sed script, rather than repeatedly looping over the file and rewriting one pattern at a time. For example, if your next task is to replace Windsor Palace A with either a. or nothing depending on whether the penultimate underscore-separated subfield in the first field contains a, that should be obvious enough:
sed -i '/^[^,]*_f_[^,_]*,/s/,British Library 1 /,f. /
s/,British Library 1 /,/
/^[^,]*_a_[^,_]*,/s/,Windsor Palace A /,a. /
s/,Windsor Palace A /,/' "${pathToCSV}_final.csv"
In some more detail, the regex says
^ beginning of line
[^,]* any sequence of characters which are not a comma
_f_ literal characters underscore, f, underscore
[^,_]* any sequence of characters which are not a comma or an underscore
, literal comma
You should be able to see that this will target the last pair of underscores in the first column. It's important to never skip across the first comma, and near the end, not allow any underscores after the ones we specifically target before we finally allow the comma column delimiter.
Finally, also notice how we always use double quotes around variables which contain file names. There are scenarios where you can avoid this but you have to know what you are doing; the easy and straightforward rule of thumb is to always put double quotes around variables. For the full scoop, see When to wrap quotes around a shell variable?
With awk, you can look at the firth field to see whether it matches "3digits + 1 letter" then print with f. in this case and just remove fields 2,3 and 4 in the other case. For example:
awk -F'[, ]' '{
if($5 ~ /.?[[:digit:]]{3}[a-z]$/) {
printf("%s,f. %s\n",$1,$5)}
else {
printf("%s,%s %s %s\n",$1,$5,$6,$7)
}
}' test.txt
On the example you provide, it gives:
abc_deg0014_0001_a_1.tif,Front Board Outside
abc_deg0014_0002_b_000.tif,Front Board Inside
abc_deg0014_0003_f_001r.tif,f. 001r
abc_deg0014_0004_f_001v.tif,f. 001v
abc_deg0014_0267_f_132r.tif,f. 132r
abc_deg0014_0268_f_132v.tif,f. 132v
abc_deg0014_0269_y_999.tif,Back Board Inside
abc_deg0014_0270_z_1.tif,Back Board Outside
I have a text file that looks like the following:
Chanelle
Jettie
Winnie
Jen
Shella
Krysta
Tish
Monika
Lynwood
Danae
2649
2466
2890
2224
2829
2427
2816
2648
2833
2453
I need to make it look like this
Chanelle 2649
Jettie 2466
... ...
I tried a lot on sublime editor but couldn't figure out the regex to do that. Can somebody demonstrate if it can be done.
I tested the following in Notepad++ but it should work universally.
Use this as the search string:
(?:(\s+[A-Za-z]+)(\r?\n))((?:\s*[A-Za-z]*\r?\n)+)\s+(\d+)
and this as the replacement:
$1 $4$2$3
Running a replace with it once will do one line at a time, if you run it multiple times it'll continue to replace lines until there are no matching lines left.
Alternatively, you can use this as the replacement if you want to have the values aligned by tabs, but it's not going to match in all cases:
$1\t\t$4$2$3
While the regex answer by SeinopSys will work, you don't need a regex to do this - instead, you can take advantage of Sublime's multiple cursors.
Place your cursor at the beginning of line 1, then hold down Shift↓ to select all the names.
Hit CtrlShiftL (Selection -> Split into Lines) to split the selection into lines.
CtrlC to copy.
Place your cursor on line 11 (the first number line) and press CtrlShift↓ (Windows/OS X) or AltShift↓ (Linux) to place a cursor at the beginning of each number line.
Hit CtrlV to paste the names before the numbers.
You can now delete the names at the top and you're all set. Alternatively, you could use CtrlX to cut the names in step 3.
I have a file with long lines and need to see/ copy what the values are in a specic location(s) for the whole file but copy the rest of the line.
If the text width is small enough, ~184 columns, I can use :set colorcolumnnum to highlight the value. However over 184 characters it gets a bit unwieldy scrolling.
I tried :g/\%1237c/y Z, for one of the positions I needed, but that yanked the entire line.
eg for a smaller sample :g/\%49c/y Z will yank all of line 1 and 2 but I want to yank, or copy, the character at that column ie = on line 1 and x on line 2.
vim: filetype=help foldmethod=indent foldclose=all modifiable noreadonly
Table of Contents *sfcontents* *vim* *regex* *sfregex*
*sfsearch* - Search specific commands
|Ampersand-replaces-previous-pattern|
|append-a-global-search-to-a-register|
*sfHelp* Various Help related commands
There are two problems with your :g command:
For each matching line, the cursor is positioned on the first column. So even though you've matched at a particular column, that position is lost.
The \%c atom actually matches byte indices (what Vim somewhat confusingly names "columns"), so your measurement will be off for Tab and non-ASCII characters. Use the virtual column atom \%v instead.
Instead of :global, I would use :substitute with a replace-expression, in the idiom described at how to extract regex matches using vim:
:let t=[] | %s/\%49v./\=add(t, submatch(0))[-1]/g | let ## = join(t, "\n")
Alternatively, if you install my ExtractMatches plugin, I'd be that short command invocation:
:YankMatchesToReg /\%50v./
Let's say I have the following text in Vim:
file1.txt
file2.txt
file3.txt
renamed1.txt
renamed2.txt
renamed3.txt
I want a transformation as follows:
file1.txt renamed1.txt
file2.txt renamed2.txt
file3.txt renamed3.txt
What I have in mind is something like the following:
:1,3 s/$/ <the text that is 4 lines below this line>
I'm stuck with how to specify the <the text that is 4 lines below this line> part.
I have tried something like .+4 (4 lines below the current line) but to no avail.
You can do it with blockwise cut & paste.
1) insert space at the start of each "renamed" line, e.g. :5,7s/^/ /
2) Use blockwise visual selection (ctrl-v) to select all the "file" lines, and press d to delete them
3) use blockwise visual selection again to select the space character at the start of all the renamed lines, and press p. This will paste the corresponding line from the block you deleted to the start of each line.
:1,3:s/\ze\n\%(.*\n\)\{3}\(.*\)/ \1
explained:
\ze - end of replaced part of match - the string matched by the rest of the pattern will not be consumed
\n - end of current line
\%(.*\n\)\{3} - next 3 lines
\(.*\) - content of 4th line from here
This will leave the later lines where they are.
I would make a macro for this really. Delete the lower line, move up, paste, Join lines, then run the macro on the others. The other method I think would be appropriate is a separate script to act as a filter.