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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 use linux and I'm trying to use sed for this. I download a CSV from an institutional site providing some data to be analyzed. There are several thousand lines per CSV, and many columns per row (I haven't counted them, but I think the number is useless). The fields are separated by semicolons and quoted, so the format per line is:
"Field 1";"Field 2";"Field 3"; .... ;"Field X";
Each correct line ends with semicolon and '\n'. The problem is that, from time to time, there's some field that incorrectly has a newline, and the solution is to delete the newline character, so the two lines go back to be together into only one. Example of an incorrect line:
"Field 1";"Field 2";"Fi
eld 3";"Field X";
I've found that there can be a \n right after the opening quote or somewhere in the between the quotes.
I've found a way to manage this last case, where the newline is right after the quote:
sed ':a;N;$!ba;s/";"\n/";"/g' file.csv
but not for "any number of alphabet characters after the quote not ending in semicolon". I have a pattern file (to be used with -f) with these lines:
:a;N;$!ba;s/";"\n/";"/g
:a;N;$!ba;s/\([A-z]\)\n/\1/g
:a;N;$!ba;s/\([:alpha:]\)\n/\1/g
The first line of the pattern file works, but I've tried combinations of the second and third and I always get an empty file.
If current line doesn't end with a semicolon, read and append next line to pattern space and remove line break.
sed '/[^;]$/{N;s/\n//}' file
I have a csv file where every line but the first starts with a number and looks like this:
subject,parameter1,parameter2,parameter3
1,blah,blah,blah
3,blah,blah,blah
2,blah,blah,blah
44,blah,blah,blah
12,blah,blah,blah
14,blah,blah,blah
11,blah,blah,blah
10,blah,blah,blah
11,blah,blah,blah
13,blah,blah,blah
3,blah,blah,blah
...
I would like to delete all lines except the first that start, say, with the numbers 1,6,12.
I was trying something like this:
:g!/^[1 6 12]\|^subject/d
But the 12 is interpreted as "1 or 2" so this also erases the lines that start with 2..
What am I missing, and what should be the most efficient way to do this?
Btw instead of 1, 6, 12, my list contains many multiple single and 2-digit numbers.
The character class [1 6 12] means "any single character that is in this class,
i.e. any one of ' ', 1, 2, 6 (the repeated 1 is ignored).
You could use
:g!/^1,\|^6,\|^12,\|^subject/d
which is close to your original syntax - but it works (tested with vim on Mac OS X).
Note - it is important to include the comma, so that the line starting with 1 doesn't "protect" 11, 12345, etc.
You might want to do this differently though - using grep.
Put all the "white listed" numbers in a file, one per line, like so:
^subject
^1,
^2,
^6,
^12,
then do
grep -f whitelist csvFile
and the output will be your "edited" file (which you can pipe to a new file).
If you are even more interested in "efficiency", you could make your text file (let's continue to call it whitelist) just
subject
1
2
6
12
and use the following command:
cat whitelist | xargs -I {} grep "^"{}"," cvsFile
This needs a bit of explaining.
xargs - take the input one line at a time
-I {} - and insert that line in the command that follows, at the {}
This means that the grep command will be run n times (once per line in the whitelist file), and each time the regular expression that is fed into grep will be the concatenation of
"^" - start of line
{} - contents of one line of the input file (whitelist)
"," - comma that follows the number
So this is a compact way of writing
grep "^subject," csvFile; grep "^1," csvFile; grep "^2," csvFile;
etc.
It has the advantage that you can now generate your whitelist any way you want - as long as it ends up in a file, one line at a time, you can use it; the disadvantage is that you are essentially running grep n times. If your files get very large, and you have a large number of items in your white list, that may start to be a problem; but since your OS is likely to put the file into cache after the first read-through, it is really quite fast. The use of the ^ anchor makes the regular expression very efficient - as soon as it doesn't find a match it goes on to the next line.
Use a global match:
:v/^\(subject\|1\|6\|12\),/ delete
For every line that does not match that regular expression, delete it.
It yields:
subject,parameter1,parameter2,parameter3
1,blah,blah,blah
12,blah,blah,blah
EDIT: Just now I realised that you were already using the global match. You error was in the character class. It matches any character inside it regardless of repeated ones, in your case numbers one, two, six and a space. You must separate them in different branches, like I did before.
a "functional" alternative:
:g/./if index([1,12,6],str2nr(split(getline("."),",")[0]))<0|exec 'normal! dd'|endif
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!
A several line document has a header/title section and then about 10 listings under each. I need to put the header/title info in with each of the listings so that they can be properly uploaded into a website (using comma and pipe delimiters). It looks like this:
SectionName1 and TitleName1
1111 - The SubSectionName A
222 - The SubSectionName B
3333 - The SubSectionName C
SectionName2 and TitleName2
444 - The SubSectionName D
55555 - The SubSectionName E
66 - The SubSectionName F
Repeating several hundred times. What I need is to produce something like:
SectionName1,TitleName1,1111,SubSectionNameA
SectionName1,TitleName1,222,SubSectionNameB
SectionName1,TitleName1,3333,SubSectionNameC
SectionName2,TitleName2,444,SubSectionNameD
SectionName2,TitleName2,55555,SubSectionNameE
SectionName2,TitleName2,66,SubSectionNameF
I realize there can multiple approaches to this solution, but I'm having a difficult time pulling the trigger on any one method. I understand submatches, joins and getline but I am not good at practical use of them in this scenario.
Any help to get me mentally started would be greatly appreciated.
Let me propose the following quite general Ex command solving the
issue.1
:g/^\s*\h/d|let#"=substitute(#"[:-2],'\s\+and\s\+',',','')|ki|/\n\s*\h\|\%$/kj|
\ 'i,'js/^\s*\(\d\+\)\s\+-\s\+The/\=#".','.submatch(1).','/|'i,'js/\s\+//g
At the top level, this is the :global command that enumerates the lines
starting with zero or more whitespace characters followed by a Latin letter or
an underscore (see :help /\h). The lines matching this pattern are supposed
to be the header lines containing section and title names. The rest of the
command, after the pattern describing the header lines, are instructions to be
executed for each of those lines.
The actions to be performed on the headers can be divided into three steps.
Delete the current header line, at the same time extracting section
and title names from it.
:d|let#"=substitute(#"[:-2],'\s\+and\s\+',',','')
First, remove the current line, saving it into the unnamed register,
using the :delete command. Then, update the contents of that
register (referred to as #"; see :help #r and :help "") to be
result of the substitution changing the word and surrounded by
whitespace characters, to a single comma. The actual replacement is
carried out by the substitute() function.
However, the input is not the exact string containing the whole header
line, but its prefix leaving out the last character, which is
a newline symbol. The [:-2] notation is a short form of the
[0:-2] subscript expression that designates the substring from the
very first byte to the second one counting from the end (see :help
expr-[:]). This way, the unnamed register holds the section and the
title names separated by comma.
Determine the range of dependent subsection lines.
:ki|/\n\s*\h\|\%$/kj
After the first step, the subsection records belonging to the just
parsed header line are located starting from the current line (the one
followed the header) until the next header line or, if there is no
such line below, the end of buffer. The numbers of these lines are
stored in the marks i and j, respectively. (See :helpg ^A mark
is for description of marks.)
The marks are placed using the :k command that sets a specified mark
at the last line of a given range which is the current line, by
default. So, unlike the first line of the considered block, the last
one requires a specific line range to point out its location.
A particular form of range, denoting the next line where a given
pattern matches, is used in this case (see :help :range). The
pattern defining the location of the line to be found, is composed in
such a way that it matches a line immediately preceding a header (a
line starting with possible whitespace followed by an alphabetical
character), or the very last line. (See :help pattern for details
about syntax of Vim regular expressions.)
Transform the delineated subsection lines according to desired format,
prepending section and title names found in the corresponding header
line.
:'i,'js/^\s*\(\d\+\)\s\+-\s\+The/\=#".','.submatch(1).','/|'i,'js/\s\+//g
This step comprised of the two :substitute commands that are run
over the range of lines delimited by the locations labelled by the
marks i and j (see :help [range]).
The first substitution command matches the beginning of a subsection
line—an identifier followed by a hyphen and the word The, all
floating in a whitespace—and replaces it with the contents of the
unnamed register, holding the section and title names concatenated
with a comma, the matched identifier, and another comma. The second
substitution finalizes the transformation by squeezing all whitespace
characters on the line to gum the subsection name and the following
letter together.
To construct the replacement string in the first :substitute
command, the substitute-with-an-expression feature is used (see :help
sub-replace-\=). The substitution part of the command should start
with \= for Vim to interpret the remaining text not in a regular
way, but as an expression (see :help expression). The result of
that expression's evaluation becomes the substitution string. Note
the use of the submatch() function in the substitute expression to
retrieve the text of a submatch by its number.
1 The command is wrapped for better readability, its one-line
version is listed below for ease of copy-pasting into Vim command line. Note
that the wrapped command can be used in a Vim script without any change.
:g/^\s*\h/d|let#"=substitute(#"[:-2],'\s\+and\s\+',',','')|ki|/\n\s*\h\|\%$/kj|'i,'js/^\s*\(\d\+\)\s\+-\s\+The/\=#".','.submatch(1).','/|'i,'js/\s\+//g
Simplest/fastest way I can think of is a simple macro. Do once, rinse, repeat.
Assuming your cursor is initially on the first character of the first line (S of SectionName), this macro should work as long as the document is exactly in the same format as posted above.
f ctT,<Esc>yyjpjjpjddkkkddkkkJr,f ctS,<Esc>f xjJr,f ctS,f xjJr,f ctS,<Esc>f xjdd
well I think the question is not that clear. why in your demo input, after "-", the text was like:
55555 - The SubSectionName E
but in your expected output, it turned into:
55555,SubSectionNameE
all spaces were removed, this is ok, but why "The" was removed as well? is there any pattern for "the" ?
I wrote an awk oneliner, it removes all spaces in output, but leave those "The" there, you can change it to get the right output you need.
awk -F' and ' -vOFS="," 'NF>1{s=$1;t=$2;next;}$1{gsub(/\s+/,"");gsub(/-/,",");print s,t,$0} ' input
test on your example input:
kent$ cat v
SectionName1 and TitleName1
1111 - The SubSectionName A
222 - The SubSectionName B
3333 - The SubSectionName C
SectionName2 and TitleName2
444 - The SubSectionName D
55555 - The SubSectionName E
66 - The SubSectionName F
kent$ awk -F' and ' -vOFS="," 'NF>1{s=$1;t=$2;next;}$1{gsub(/\s+/,"");gsub(/-/,",");print s,t,$0} ' v
SectionName1,TitleName1,1111,TheSubSectionNameA
SectionName1,TitleName1,222,TheSubSectionNameB
SectionName1,TitleName1,3333,TheSubSectionNameC
SectionName2,TitleName2,444,TheSubSectionNameD
SectionName2,TitleName2,55555,TheSubSectionNameE
SectionName2,TitleName2,66,TheSubSectionNameF