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 directory with a bunch of text files, all of which follow this structure:
...
- Some random number of list items of random text
- And even more of it
PATTERN_A (surrounded by empty lines)
- Again, some list items of random text
- Which does look similar as the first batch
PATTERN_B (surrounded by empty lines)
- And even more some random text
....
And I need to run a replace operation (let's say, I need to prepend CCC at the beginning of the line, just after the dash) on only those "list items", which are between PATTERN_A and PATTERN_B. The problem is they aren't really much different from the text above PATTERN_A, or below PATTERN_B, so an ordinary regex can't really catch them without also affecting the remaining text.
So, my question would be, what tool and what regex should I use to perform that replacement?
(Just in case, I'm fine with Vim, and I can collect those files in a QuickFix for a further :cdo, for example. I'm not that good with awk, unfortunately, and absolutely bad with Perl :))
Thanks!
If I have understood your questions, you can do so quite easily with a pattern-range selection and the general substitution form with sed (stream editor). For example, in your case:
$ sed '/PATTERN_A/,/PATTERN_B/s/^\([ ]*-\)/\1CCC/' file
- Some random number of list items of random text
- And even more of it
PATTERN_A (surrounded by empty lines)
-CCC Again, some list items of random text
-CCC Which does look similar as the first batch
PATTERN_B (surrounded by empty lines)
- And even more some random text
(note: to substitute in place within the file add the -i option, and to create a backup of the original add -i.bak which will save the original file as file.bak)
Explanation
/PATTERN_A/,/PATTERN_B/ - select lines between PATTERN_A and PATTERN_B
s/^\([ ]*-\)/\1CCC/ - substitute (general form 's/find/replace/') where find is from beginning of line ^ capturing text between \(...\) that contains [ ]*- (any number of spaces and a hyphen) and then replace with \1 (called a backreference that contains all characters you captured with the capture group \(...\)) and appending CCC to its end.
Look things over and let me know if you have questions or if I misinterpreted your question.
With Perl also, you can get the results
> perl -pe ' { s/^(\s*-)/\1CCC/g if /PATTERN_A/../PATTERN_B/ } ' mass_replace.txt
...
- Some random number of list items of random text
- And even more of it
PATTERN_A (surrounded by empty lines)
-CCC Again, some list items of random text
-CCC Which does look similar as the first batch
PATTERN_B (surrounded by empty lines)
- And even more some random text
....
>
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./
I have a data file as follows.
1,14.23,1.71,2.43,15.6,127,2.8,3.06,.28,2.29,5.64,1.04,3.92,1065
1,13.2,1.78,2.14,11.2,100,2.65,2.76,.26,1.28,4.38,1.05,3.4,1050
1,13.16,2.36,2.67,18.6,101,2.8,3.24,.3,2.81,5.68,1.03,3.17,1185
1,14.37,1.95,2.5,16.8,113,3.85,3.49,.24,2.18,7.8,.86,3.45,1480
1,13.24,2.59,2.87,21,118,2.8,2.69,.39,1.82,4.32,1.04,2.93,735
Using vim, I want to reomve the 1's from each of the lines and append them to the end. The resultant file would look like this:
14.23,1.71,2.43,15.6,127,2.8,3.06,.28,2.29,5.64,1.04,3.92,1065,1
13.2,1.78,2.14,11.2,100,2.65,2.76,.26,1.28,4.38,1.05,3.4,1050,1
13.16,2.36,2.67,18.6,101,2.8,3.24,.3,2.81,5.68,1.03,3.17,1185,1
14.37,1.95,2.5,16.8,113,3.85,3.49,.24,2.18,7.8,.86,3.45,1480,1
13.24,2.59,2.87,21,118,2.8,2.69,.39,1.82,4.32,1.04,2.93,735,1
I was looking for an elegant way to do this.
Actually I tried it like
:%s/$/,/g
And then
:%s/$/^./g
But I could not make it to work.
EDIT : Well, actually I made one mistake in my question. In the data-file, the first character is not always 1, they are mixture of 1, 2 and 3. So, from all the answers from this questions, I came up with the solution --
:%s/^\([1-3]\),\(.*\)/\2,\1/g
and it is working now.
A regular expression that doesn't care which number, its digits, or separator you've used. That is, this would work for lines that have both 1 as their first number, or 114:
:%s/\([0-9]*\)\(.\)\(.*\)/\3\2\1/
Explanation:
:%s// - Substitute every line (%)
\(<something>\) - Extract and store to \n
[0-9]* - A number 0 or more times
. - Every char, in this case,
.* - Every char 0 or more times
\3\2\1 - Replace what is captured with \(\)
So: Cut up 1 , <the rest> to \1, \2 and \3 respectively, and reorder them.
This
:%s/^1,//
:%s/$/,1/
could be somewhat simpler to understand.
:%s/^1,\(.*\)/\1,1/
This will do the replacement on each line in the file. The \1 replaces everything captured by the (.*)
:%s/1,\(.*$\)/\1,1/gc
.........................
You could also solve this one using a macro. First, think about how to delete the 1, from the start of a line and append it to the end:
0 go the the start of the line
df, delete everything to and including the first ,
A,<ESC> append a comma to the end of the line
p paste the thing you deleted with df,
x delete the trailing comma
So, to sum it up, the following will convert a single line:
0df,A,<ESC>px
Now if you'd like to apply this set of modifications to all the lines, you will first need to record them:
qj start recording into the 'j' register
0df,A,<ESC>px convert a single line
j go to the next line
q stop recording
Finally, you can execute the macro anytime you want using #j, or convert your entire file with 99#j (using a higher number than 99 if you have more than 99 lines).
Here's the complete version:
qj0df,A,<ESC>pxjq99#j
This one might be easier to understand than the other solutions if you're not used to regular expressions!
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.