grep with indentation-based context - grep

Is there a grep tool that shows context based on indentation? For each match, I would like to see all the most recent lines with a lower indent level, but skip all the lines at the same indent level in between. Example:
$ contextgrep rstrip pep8.py
class Checker(object):
def check_logical(self):
if self.verbose >= 2:
print(self.logical_line[:80].rstrip())

Here's a blog post about a literal Haskell implementation of "outline grep".

Related

How to type AND in regex word matching

I'm trying to do a word search with regex and wonder how to type AND for multiple criteria.
For example, how to type the following:
(Start with a) AND (Contains p) AND (Ends with e), such as the word apple?
Input
apple
pineapple
avocado
Code
grep -E "regex expression here" input.txt
Desired output
apple
What should the regex expression be?
In general you can't implement and in a regexp (but you can implement then with .*) but you can in a multi-regexp condition using a tool that supports it.
To address the case of ands, you should have made your example starts with a and includes p and includes l and ends with e with input including alpine so it wasn't trivial to express in a regexp by just putting .*s in between characters but is trivial in a multi-regexp condition:
$ cat file
apple
pineapple
avocado
alpine
Using &&s will find both words regardless of the order of p and l as desired:
$ awk '/^a/ && /p/ && /l/ && /e$/' file
apple
alpine
but, as you can see, you can't just use .*s to implement and:
$ grep '^a.*p.*l.*e$' file
apple
If you had to use a single regexp then you'd have to do something like:
$ grep -E '^a.*(p.*l|l.*p).*e$' file
apple
alpine
two ways you can do it
all that "&&" is same as negating the totality of a bunch of OR's "||", so you can write the reverse of what you want.
at a single bit-level, AND is same as multiplication of the bits, which means, instead of doing all the && if u think it's overly verbose, you can directly "multiply" the patterns together :
awk '/^a/ * /p/ * /e$/'
so by multiplying them, you're doing the same as performing multiple logical ANDs all at once
(but only use the short hand if inputs aren't too gigantic, or when savings from early exit are known to be negligible.
don't think of them as merely regex patterns - it's easier for one to think of anything not inside an action block, what's typically referred to as pattern, as
any combination and collection of items that could be evaluated for a boolean outcome of TRUE or FALSE in the end
e.g. POSIX-compliant expressions that work in the space include
sprintf()
field assignments, etc
(even decrementing NR - if there's such a need)
but not
statements like next, print, printf(),
delete array etc, or any of the loop structures
surprisingly though, getline is directly doable
in the pattern space area (with some wrapper workaround)

Grep Filenames from ls for specific part of them

I want to extract a specific part out of the filenames to work with them.
Example:
ls -1
REZ-Name1,Surname1-02-04-2012.png
REZ-Name2,Surname2-07-08-2013.png
....
So I want to get only the part with the name.
How can this be achieved ?
There are several ways to do this. Here's a loop:
for file in REZ-*-??-??-????.png
do
name=${file#*-}
name=${name%-??-??-????.png}
echo "($name)"
done
Given a variety of filenames with all sorts of edge cases from spacing, additional hyphens and line feeds:
REZ-Anna-Maria,de-la-Cruz-12-32-2015.png
REZ-Bjørn,Dæhlie-01-01-2015.png
REZ-First,Last-12-32-2015.png
REZ-John Quincy,Adams-11-12-2014.png
REZ-Ridiculous example # this is one filename
is ridiculous,but fun-22-11-2000.png # spanning two lines
it outputs:
(Anna-Maria,de-la-Cruz)
(Bjørn,Dæhlie)
(First,Last)
(John Quincy,Adams)
(Ridiculous example
is ridiculous,but fun)
If you're less concerned with correctness, you can simplify it further:
$ ls | grep -o '[^-]*,[^-]*'
Maria,de
Bjørn,Dæhlie
First,Last
John Quincy,Adams
is ridiculous,but fun
In this case, cut makes more sense than grep:
ls -l | cut -f2 -d-
cut the second field from the input, using '-' as the field delimiter. That other guy's answer will correctly handle some cases mine will not, but for one off uses, I generally find the semantics of cut to be much easier to remember.

Pattern matching using grep

Assuming we have one input string like
Nice
And we have the pattern
D*A*C*N*a*g*.h*ca*e
then "Nice" will match the pattern. (* means 0 or more occurrence, . means one char)
I think using grep is better than java in this case(maybe). How can I do it in grep?
Use the same regular expression:
grep 'D*A*C*N*a*g*.h*ca*e' <<EOF
Nice
EOF
If the input is "Nicely" it still prints it! How does it work?
The current regex looks for the pattern anywhere on the line. If it must match exactly (the whole line), then add anchors to start (^) and end ($) of line:
grep '^D*A*C*N*a*g*.h*ca*e$' <<EOF
Nice
Nicely
Darce
Darcy
Darcey
EOF

Grep for beginning and end of line?

I have a file where I want to grep for lines that start with either -rwx or drwx AND end in any number.
I've got this, but it isnt quite right. Any ideas?
grep [^.rwx]*[0-9] usrLog.txt
The tricky part is a regex that includes a dash as one of the valid characters in a character class. The dash has to come immediately after the start for a (normal) character class and immediately after the caret for a negated character class. If you need a close square bracket too, then you need the close square bracket followed by the dash. Mercifully, you only need dash, hence the notation chosen.
grep '^[-d]rwx.*[0-9]$' "$#"
See: Regular Expressions and grep for POSIX-standard details.
It looks like you were on the right track... The ^ character matches beginning-of-line, and $ matches end-of-line. Jonathan's pattern will work for you... just wanted to give you the explanation behind it
It should be noted that not only will the caret (^) behave differently within the brackets, it will have the opposite result of placing it outside of the brackets. Placing the caret where you have it will search for all strings NOT beginning with the content you placed within the brackets. You also would want to place a period before the asterisk in between your brackets as with grep, it also acts as a "wildcard".
grep ^[.rwx].*[0-9]$
This should work for you, I noticed that some posters used a character class in their expressions which is an effective method as well, but you were not using any in your original expression so I am trying to get one as close to yours as possible explaining every minor change along the way so that it is better understood. How can we learn otherwise?
You probably want egrep. Try:
egrep '^[d-]rwx.*[0-9]$' usrLog.txt
are you parsing output of ls -l?
If you are, and you just want to get the file name
find . -iname "*[0-9]"
If you have no choice because usrLog.txt is created by something/someone else and you absolutely must use this file, other options include
awk '/^[-d].*[0-9]$/' file
Ruby(1.9+)
ruby -ne 'print if /^[-d].*[0-9]$/' file
Bash
while read -r line ; do case $line in [-d]*[0-9] ) echo $line; esac; done < file
Many answers provided for this question. Just wanted to add one more which uses bashism-
#! /bin/bash
while read -r || [[ -n "$REPLY" ]]; do
[[ "$REPLY" =~ ^(-rwx|drwx).*[[:digit:]]+$ ]] && echo "Got one -> $REPLY"
done <"$1"
#kurumi answer for bash, which uses case is also correct but it will not read last line of file if there is no newline sequence at the end(Just save the file without pressing 'Enter/Return' at the last line).

Easiest way to remove Latex tag (but not its content)?

I am using TeXnicCenter to edit a LaTeX document.
I now want to remove a certain tag (say, emph{blabla}} which occurs multiple times in my document , but not tag's content (so in this example, I want to remove all emphasization).
What is the easiest way to do so?
May also be using another program easily available on Windows 7.
Edit: In response to regex suggestions, it is important that it can deal with nested tags.
Edit 2: I really want to remove the tag from the text file, not just disable it.
Using a regular expression do something like s/\\emph\{([^\}]*)\}/\1/g. If you are not familiar with regular expressions this says:
s -- replace
/ -- begin match section
\\emph\{ -- match \emph{
( -- begin capture
[^\}]* -- match any characters except (meaning up until) a close brace because:
[] a group of characters
^ means not or "everything except"
\} -- the close brace
and * means 0 or more times
) -- end capture, because this is the first (in this case only) capture, it is number 1
\} -- match end brace
/ -- begin replace section
\1 -- replace with captured section number 1
/ -- end regular expression, begin extra flags
g -- global flag, meaning do this every time the match is found not just the first time
This is with Perl syntax, as that is what I am familiar with. The following perl "one-liners" will accomplish two tasks
perl -pe 's/\\emph\{([^\}]*)\}/\1/g' filename will "test" printing the file to the command line
perl -pi -e 's/\\emph\{([^\}]*)\}/\1/g' filename will change the file in place.
Similar commands may be available in your editor, but if not this will (should) work.
Crowley should have added this as an answer, but I will do that for him, if you replace all \emph{ with { you should be able to do this without disturbing the other content. It will still be in braces, but unless you have done some odd stuff it shouldn't matter.
The regex would be a simple s/\\emph\{/\{/g but the search and replace in your editor will do that one too.
Edit: Sorry, used the wrong brace in the regex, fixed now.
\renewcommand{\emph}[1]{#1}
any reasonably advanced editor should let you do a search/replace using regular expressions, replacing emph{bla} by bla etc.

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