how to extract the last word of line in notepad++ - regex

I want to extract the last word of each line in notepad++. For this purpose i used the regular expression (\w+)$ in find what text box, it showed all the last word highlighted after the search. but i don't know how i will extract these words. Below is a look of my file
gene expression
gene
B activation
B
surface receptor
cell activation
proliferation
T lymphocyte
oxygen intermediate
activation
B
complex
expression
signaling cascade
tyrosine kinase activity
tyrosine kinase
A2
metabolite
formation
B activation
B

You can
Search: (.* )(\w+)$
Replace with: $2
to delete all others except the last word in each line.

To extract the last word of each line, you can use the command line cc.gc ^-1w of the plugin ConyEdit.

Related

RegEx in Notepad++ to find lines with less or more than n pipes

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.

Regexextract over multiple lines within one cell

In Google Sheets, I have this in one cell:
Random stuff blah blah 123456789
<Surname, Name><123456><A><100><B><200>
<Surname2, Name2><456789><A><300><B><400>
Some more random stuff
And would like to match the strings within <> brackets. With = REGEXEXTRACT(A4, "<(.*)>") I got thus far:
Surname, Name><123456><A><100><B><200
which is nice, but it is only the first line. The desired output would be this (maybe including the <> at the beginning/end, it doesn't really matter):
Surname, Name><123456><A><100><B><200>
<Surname2, Name2><456789><A><300><B><400
or simply:
Surname, Name><123456><A><100><B><200><Surname2, Name2><456789><A><300><B><400
How to get there?
Please try:
=SUBSTITUTE(regexextract(substitute(A4,char(10)," "),"<(.*)>"),"> <",">"&char(10)&"<")
Starting in the middle, the substitute replaces line breaks (char(10)) with spaces. This enables the regexextract the complete (ie multi-line) string to work on, with the same pattern as already familiar to OP. SUBSTITUTE then reinstates the relevant space (identified as being immediately surrounded by > and <) with a line break.
Google sheets uses RE2 syntax. You can set the multi-line and s flags in order to match multiple lines. The following will match all characters over multiple lines in cell A2.
=REGEXEXTRACT(A2, "(?ms)^(.*)$")
REGEXEXTRACT(A1,"text1(?ms)(.*)text2")
So, in this case:
REGEXEXTRACT(A1,"<(?ms)(.*)>")

Grep for Pattern in File in R

In a document, I'm trying to look for occurences of a 12-digit string which contains alpha and numerals. A sample string is: "PXB111X2206"
I'm trying to get the line numbers that contain this string in R using the below:
FileInput = readLines("File.txt")
prot_pattern="([A-Z0-9]{12})";
prot_string<-grep(prot_pattern,FileInput)
prot_string
This worked fine until it hit a document containing all upper-case titles and returned a line containing the word "CONCENTRATIO"
The string I am trying to look for is: "PXB111X2206". I am expecting the grep to return the line numbers containing the string : "PXB111X2206". It however is returning the line number containing the word: "CONCENTRATIO"
What is wrong with my expression above? Any idea what I am doing wrong here?
Here is some sample input:
Each design objective described herein is significantly important, yet it is just one aspect of what it takes to achieve a successful project.
A successful project is one where project goals are identified early on and where the >interdependencies of all building systems are coordinated concurrently from the planning and programming phase.
CONCENTRATION:
The areas of concentration for design objectives: accessible, aesthetics, cost effective, >functional/operational, historic preservation, productive, secure/safe, and sustainable and >their interrelationships must be understood, evaluated, and appropriately applied.
Each of these design objectives is presented in the design objectives document number. >PXB111X2206.
>
Thanks & Regards,
Simak
You are using a very powerful tool for a very simple task, the expression
[A-Z0-9]{12}
will match any alphanumeric 12 sized uppercased string, for example the word "CONCENTRATIO", however, your "PXB111X2206" is not even 12 symbols long, so it is not possible that is being matched. If you only want to match "PXB111X2206" you only have to use it as a regular expression itself, for example, if you file contents are:
foo
CONCENTRATIO.
bazz
foo bar bazz PXB111X2206 foo bar bazz
foo
bar
bazz
and you use:
grep('PXB111X2206',readLines("File.txt"))
then R will only match line 4 as you would wish.
EDIT
If you are looking for that specific pattern try:
grep('[A-Z]{3}[0-9]{3}[A-Z]{1}[0-9]{4}',readLines("File.txt"))
That expression will match strings like 'AAADDDADDDD' where A is an capital letter, and D a digit, the regular expression contains a group (symbols inside square brackets) and a quantifier (the number inside the brackets) that tells how many of the previous symbol will the expression accept, if no quantifier is present it assumes it is 1.
Let's take a look at what your regular expression means. [A-Z0-9] means any capitalized letter or number and {12} means the previous expression must occur exactly 12 times. The string CONCENTRATIO is 12 capitaized letters, so it's no surprise that grep picks it up. If you want to take out the matches that match to just letters or just numbers you could try something like
allleters <- grep("[A-Z]{12}",strings)
allnumbers <-grep("[0-9]{12}",strings)
both <- grep("[A-Z0-9]{12}",strings)
the matches you wanted would then be something like
both <- both[!both %in% union(allletters,allnumbers)]
Someone with better regexfu might have a more elegant solution, but this will work too.

What would be the best approach to this substitution in Vim?

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

Find lines matching regex and select a different part of the line

I have two lines like below:
/pace =builtin\administrators Type=0x0 Flags=0x13 AccessMask=0x1f01ff
/pace =domain\user Type=0x0 Flags=0x13 AccessMask=0x1f01ff
Need to create a regular expression where it only select 0x1f01ff where the line have domain\user.
This is what I have created but it select /pace =domain\user Type=0x0 Flags=0x13 AccessMask=:
^(.+domain(.*)accessmask=)
try this:
^.+domain\\user.+AccessMask=([^\s]+)
It matches any line that has domain\user and then get the value of accessmask (any character that is not a whitespace)