So I have a bunch of data that all looks like this:
janitor#1/2 of dorm#1/1
president#4/1 of class#2/2
hunting#1/1 hat#1/2
side#1/2 of hotel#1/1
side#1/2 of hotel#1/1
king#1/2 of hotel#1/1
address#2/2 of girl#1/1
one#2/1 in family#2/2
dance#3/1 floor#1/2
movie#1/2 stars#5/1
movie#1/2 stars#5/1
insurance#1/1 office#1/2
side#1/1 of floor#1/2
middle#4/1 of December#1/2
movie#1/2 stars#5/1
one#2/1 of tables#2/2
people#1/2 at table#2/1
Some lines have prepositions, others don't so I thought I could use regular expressions to clean it up. What I need is each noun, the # sign and the following number on its own line. So for example, the first lines of output should look like this in the final file:
janitor#1
dorm#1
president#4
etc...
The list is stored in a file called NPs. My code to do this is:
cat NPs | grep -E '\b(\w*[#][1-9]).' >> test
When I open test, however, it's the exact same as the input file. Any input as to what I'm missing? It doesn't seem like it should be a hard operation, so maybe I'm missing something about syntax? I'm using this command from a shell script that is called in bash.
Thanks in advance!
This should do what you need.
The -o option will show only the part of a matching line that matches the PATTERN.
grep -Eo '[a-z#]+[1-9]' NPs > test
or even the -P option, which Interprets the PATTERN as a Perl regular expression
grep -Po '[\w#]*(?=/)' NPs > test
Using grep:
$ grep -o "\w*[#]\w*" inputfile
janitor#1
dorm#1
president#4
class#2
hunting#1
hat#1
side#1
hotel#1
side#1
hotel#1
king#1
hotel#1
address#2
girl#1
one#2
family#2
dance#3
floor#1
movie#1
stars#5
movie#1
stars#5
insurance#1
office#1
side#1
floor#1
middle#4
ecember#1
movie#1
stars#5
one#2
tables#2
people#1
table#2
grep variations extracting entire lines from text, if they match pattern. If you need to modify lines, you should use sed, like
cat NPs | sed 's/^\(\b\w*[#][1-9]\).*$/\1/g'
You need sed, not grep. (Or awk, or perl.) It looks like this would do what you want:
cat NPs | sed 's?/.*??'
or simply
sed 's?/.*??' NPs
s means "substitute". The next character is the delimiter between regular expressions. Usually it's "/", but since you need to search for "/", I used "?" instead. "." refers to any character, and "*" says "zero or more of what preceded me". Whatever is between the last two delimiters is the replacement string. In this case it's empty, so you're replacing "/" followed by zero or more of any character, with the empty string.
EDIT: Oh, I see now that you wanted to extract the last item on the line, too. Well, I'm sure that others' suggested regexps would work. If it were my problem, I'd probably filter the file in two steps, perhaps piping the results from one step to the next, or using multiple substitutions with sed: First delete the "of"s and middle spaces, and add newlines, and then run sed as above. It's not as cool as doing it all in one regexp, but each step is easier to understand. For even more simplicity and uncoolness, use three steps, replacing " of " with space in the first step. Since others have provided complete solutions, I won't work out the details.
Grep by default just searches for the text, so in your case it is printing the lines that match. I think you want to investigate sed instead to perform the replacement. (And you don't need to cat the file, just grep PATTERN filename)
To get your output on separate lines, this worked for me:
sed 's|/.||g' NPs | sed 's/ .. /=/' | tr "=" "\n"
This uses two seds in a row to do different substitutions, and tr to insert line feeds.
The -o option in grep, which causes it to print out only the matching text, as described in another answer, is probably even simpler!
An awk version:
awk '/#/ {print $NF}' RS="/" NPs
janitor#1
dorm#1
president#4
class#2
hunting#1
hat#1
side#1
hotel#1
side#1
hotel#1
king#1
hotel#1
address#2
girl#1
one#2
family#2
dance#3
floor#1
movie#1
stars#5
movie#1
stars#5
insurance#1
office#1
side#1
floor#1
middle#4
December#1
movie#1
stars#5
one#2
tables#2
people#1
table#2
Related
new to regex and have a problem. I want to replace hyphens with underscores in certain places in a file. To simplify things, let's say I want to replace the first hyphen. Here's an example "file":
dont-touch-these-hyphens
leaf replace-these-hyphens
I want to replace hyphens in all lines found by
grep -P "leaf \w+-" file
I tried
sed -i 's/leaf \(\w+\)-/leaf \1_/g' file
but nothing happens (wrong replacement would have been better than nothing). I've tried a few tweaks but still nothing. Again, I'm new to this so I figure the above "should basically work". What's wrong with it, and how do I get what I want? Thanks.
You can simplify things by using two distinct regex's ; one for matching the lines that need processing, and one for matching what must be modified.
You can try something like this:
$ sed '/^leaf/ s/-/_/' file
dont-touch-these-hyphens
leaf replace_these-hyphens
Just use awk:
$ awk '$1=="leaf"{ sub(/-/,"_",$2) } 1' file
dont-touch-these-hyphens
leaf replace_these-hyphens
It gives you much more precise control over what you're matching (e.g. the above is doing a string instead of regexp comparison on "leaf" and so would work even if that string contained regexp metacharacters like . or *) and what you're replacing (e.g. the above only does the replacement in the text AFTER leaf and so would continue to work even if leaf itself contained -s):
$ cat file
dont-touch-these-hyphens
leaf-foo.*bar replace-these-hyphens
leaf-foobar dont-replace-these-hyphens
Correct output:
$ awk '$1=="leaf-foo.*bar"{ sub(/-/,"_",$2) } 1' file
dont-touch-these-hyphens
leaf-foo.*bar replace_these-hyphens
leaf-foobar dont-replace-these-hyphens
Wrong output:
$ sed '/^leaf-foo.*bar/ s/-/_/' file
dont-touch-these-hyphens
leaf_foo.*bar replace-these-hyphens
leaf_foobar dont-replace-these-hyphens
(note the "-" in leaf-foo being replaced by "_" in each of the last 2 lines, including the one that does not start with the string "leaf-foo.*bar").
That awk script will work as-is using any awk on any UNIX box.
I have a string in text file where i want to replace the version number. Quotation marks can vary from ' to ". Also spaces around = can be there and can be not as well:
$data['MODULEXXX_VERSION'] = "1.0.0";
For testing i use
echo "_VERSION'] = \"1.1.1\"" | sed "s/\(_VERSION.*\)[1-9]\.[1-9]\.[1-9]/\11.1.2/"
which works perfectly.
When i change it to search in the file (the file has the same string):
sed "s/\(_VERSION.*\)[1-9]\.[1-9]\.[1-9]/\11.1.2/" -i test.php
, it does not find anything.
After after playing with the search part of regex, i found one more odd thing:
sed "s/\(_VERSION.*\)[1-9]\./\1***/" -i test.php
works and changes the string to $data['MODULEXXX_VERSION'] = "***0.0";, but
sed "s/\(_VERSION.*\)[1-9]\.[1-9]/\1***/" -i test.php
does not find anything anymore. Why?
I am using Ubuntu 17.04 desktop.
Anyone can explain what am I doing wrong? What would be the best command for replacing version numbers in the file for the string $data['MODULEXXX_VERSION'] = "***0.0";?
The main problem is that [1-9] doesn't match the 0s in the version number. You need to use [0-9].
Besides that, you may use the following sed command:
sed -r 's/(.*_VERSION['\''"]]\s*=\s*).*/\1"1.0.1";/' conf.php
This doesn't look at the current value, it simply replaces everything after the =.
I've used -r which enables extended posix regular expressions which makes it a bit simpler to formulate the pattern.
Another, probably cleaner attempt is to store the conf.php as a template like conf.php.tpl and then use a template engine to render the file. Or if you really want to use sed, the file may look like:
$data['FOO_VERSION'] = "FOO_VERSION_TPL";
Then just use:
sed 's/FOO_VERSION_TPL/1.0.1/' conf.php.tpl > conf.php
If there are multiple values to replace:
sed \
-e 's/FOO/BAR/' \
-e 's/HELLO/WORLD/' \
conf.php.tpl > conf.php
But I recommend a template engine instead of sed. That becomes more important when the content of the variables to replace may contain characters special to regular expressions.
I want to run ack or grep on HTML files that often have very long lines. I don't want to see very long lines that wrap repeatedly. But I do want to see just that portion of a long line that surrounds a string that matches the regular expression. How can I get this using any combination of Unix tools?
You could use the grep options -oE, possibly in combination with changing your pattern to ".{0,10}<original pattern>.{0,10}" in order to see some context around it:
-o, --only-matching
Show only the part of a matching line that matches PATTERN.
-E, --extended-regexp
Interpret pattern as an extended regular expression (i.e., force grep to behave as egrep).
For example (from #Renaud's comment):
grep -oE ".{0,10}mysearchstring.{0,10}" myfile.txt
Alternatively, you could try -c:
-c, --count
Suppress normal output; instead print a count of matching lines
for each input file. With the -v, --invert-match option (see
below), count non-matching lines.
Pipe your results thru cut. I'm also considering adding a --cut switch so you could say --cut=80 and only get 80 columns.
You could use less as a pager for ack and chop long lines: ack --pager="less -S" This retains the long line but leaves it on one line instead of wrapping. To see more of the line, scroll left/right in less with the arrow keys.
I have the following alias setup for ack to do this:
alias ick='ack -i --pager="less -R -S"'
grep -oE ".\{0,10\}error.\{0,10\}" mylogfile.txt
In the unusual situation where you cannot use -E, use lowercase -e instead.
Explanation:
cut -c 1-100
gets characters from 1 to 100.
The Silver Searcher (ag) supports its natively via the --width NUM option. It will replace the rest of longer lines by [...].
Example (truncate after 120 characters):
$ ag --width 120 '#patternfly'
...
1:{"version":3,"file":"react-icons.js","sources":["../../node_modules/#patternfly/ [...]
In ack3, a similar feature is planned but currently not implemented.
Taken from: http://www.topbug.net/blog/2016/08/18/truncate-long-matching-lines-of-grep-a-solution-that-preserves-color/
The suggested approach ".{0,10}<original pattern>.{0,10}" is perfectly good except for that the highlighting color is often messed up. I've created a script with a similar output but the color is also preserved:
#!/bin/bash
# Usage:
# grepl PATTERN [FILE]
# how many characters around the searching keyword should be shown?
context_length=10
# What is the length of the control character for the color before and after the
# matching string?
# This is mostly determined by the environmental variable GREP_COLORS.
control_length_before=$(($(echo a | grep --color=always a | cut -d a -f '1' | wc -c)-1))
control_length_after=$(($(echo a | grep --color=always a | cut -d a -f '2' | wc -c)-1))
grep -E --color=always "$1" $2 |
grep --color=none -oE \
".{0,$(($control_length_before + $context_length))}$1.{0,$(($control_length_after + $context_length))}"
Assuming the script is saved as grepl, then grepl pattern file_with_long_lines should display the matching lines but with only 10 characters around the matching string.
I put the following into my .bashrc:
grepl() {
$(which grep) --color=always $# | less -RS
}
You can then use grepl on the command line with any arguments that are available for grep. Use the arrow keys to see the tail of longer lines. Use q to quit.
Explanation:
grepl() {: Define a new function that will be available in every (new) bash console.
$(which grep): Get the full path of grep. (Ubuntu defines an alias for grep that is equivalent to grep --color=auto. We don't want that alias but the original grep.)
--color=always: Colorize the output. (--color=auto from the alias won't work since grep detects that the output is put into a pipe and won't color it then.)
$#: Put all arguments given to the grepl function here.
less: Display the lines using less
-R: Show colors
S: Don't break long lines
Here's what I do:
function grep () {
tput rmam;
command grep "$#";
tput smam;
}
In my .bash_profile, I override grep so that it automatically runs tput rmam before and tput smam after, which disabled wrapping and then re-enables it.
ag can also take the regex trick, if you prefer it:
ag --column -o ".{0,20}error.{0,20}"
I have a ";" delimited file:
aa;;;;aa
rgg;;;;fdg
aff;sfg;;;fasg
sfaf;sdfas;;;
ASFGF;;;;fasg
QFA;DSGS;;DSFAG;fagf
I'd like to process it replacing the missing value with a \N .
The result should be:
aa;\N;\N;\N;aa
rgg;\N;\N;\N;fdg
aff;sfg;\N;\N;fasg
sfaf;sdfas;\N;\N;\N
ASFGF;\N;\N;\N;fasg
QFA;DSGS;\N;DSFAG;fagf
I'm trying to do it with a sed script:
sed "s/;\(;\)/;\\N\1/g" file1.txt >file2.txt
But what I get is
aa;\N;;\N;aa
rgg;\N;;\N;fdg
aff;sfg;\N;;fasg
sfaf;sdfas;\N;;
ASFGF;\N;;\N;fasg
QFA;DSGS;\N;DSFAG;fagf
You don't need to enclose the second semicolon in parentheses just to use it as \1 in the replacement string. You can use ; in the replacement string:
sed 's/;;/;\\N;/g'
As you noticed, when it finds a pair of semicolons it replaces it with the desired string then skips over it, not reading the second semicolon again and this makes it insert \N after every two semicolons.
A solution is to use positive lookaheads; the regex is /;(?=;)/ but sed doesn't support them.
But it's possible to solve the problem using sed in a simple manner: duplicate the search command; the first command replaces the odd appearances of ;; with ;\N, the second one takes care of the even appearances. The final result is the one you need.
The command is as simple as:
sed 's/;;/;\\N;/g;s/;;/;\\N;/g'
It duplicates the previous command and uses the ; between g and s to separe them. Alternatively you can use the -e command line option once for each search expression:
sed -e 's/;;/;\\N;/g' -e 's/;;/;\\N;/g'
Update:
The OP asks in a comment "What if my file have 100 columns?"
Let's try and see if it works:
$ echo "0;1;;2;;;3;;;;4;;;;;5;;;;;;6;;;;;;;" | sed 's/;;/;\\N;/g;s/;;/;\\N;/g'
0;1;\N;2;\N;\N;3;\N;\N;\N;4;\N;\N;\N;\N;5;\N;\N;\N;\N;\N;6;\N;\N;\N;\N;\N;\N;
Look, ma! It works!
:-)
Update #2
I ignored the fact that the question doesn't ask to replace ;; with something else but to replace the empty/missing values in a file that uses ; to separate the columns. Accordingly, my expression doesn't fix the missing value when it occurs at the beginning or at the end of the line.
As the OP kindly added in a comment, the complete sed command is:
sed 's/;;/;\\N;/g;s/;;/;\\N;/g;s/^;/\\N;/g;s/;$/;\\N/g'
or (for readability):
sed -e 's/;;/;\\N;/g;' -e 's/;;/;\\N;/g;' -e 's/^;/\\N;/g' -e 's/;$/;\\N/g'
The two additional steps replace ';' when they found it at beginning or at the end of line.
You can use this sed command with 2 s (substitute) commands:
sed 's/;;/;\\N;/g; s/;;/;\\N;/g;' file
aa;\N;\N;\N;aa
rgg;\N;\N;\N;fdg
aff;sfg;\N;\N;fasg
sfaf;sdfas;\N;\N;
ASFGF;\N;\N;\N;fasg
QFA;DSGS;\N;DSFAG;fagf
Or using lookarounds regex in a perl command:
perl -pe 's/(?<=;)(?=;)/\\N/g' file
aa;\N;\N;\N;aa
rgg;\N;\N;\N;fdg
aff;sfg;\N;\N;fasg
sfaf;sdfas;\N;\N;
ASFGF;\N;\N;\N;fasg
QFA;DSGS;\N;DSFAG;fagf
The main problem is that you can't use several times the same characters for a single replacement:
s/;;/..../g: The second ; can't be reused for the next match in a string like ;;;
If you want to do it with sed without to use a Perl-like regex mode, you can use a loop with the conditional command t:
sed ':a;s/;;/;\\N;/g;ta;' file
:a defines a label "a", ta go to this label only if something has been replaced.
For the ; at the end of the line (and to deal with eventual trailing whitespaces):
sed ':a;s/;;/;\\N;/g;ta; s/;[ \t\r]*$/;\\N/1' file
this awk one-liner will give you what you want:
awk -F';' -v OFS=';' '{for(i=1;i<=NF;i++)if($i=="")$i="\\N"}7' file
if you really want the line: sfaf;sdfas;\N;\N;\N , this line works for you:
awk -F';' -v OFS=';' '{for(i=1;i<=NF;i++)if($i=="")$i="\\N";sub(/;$/,";\\N")}7' file
sed 's/;/;\\N/g;s/;\\N\([^;]\)/;\1/g;s/;[[:blank:]]*$/;\\N/' YourFile
non recursive, onliner, posix compliant
Concept:
change all ;
put back unmatched one
add the special case of last ; with eventually space before the end of line
This might work for you (GNU sed):
sed -r ':;s/^(;)|(;);|(;)$/\2\3\\N\1\2/g;t' file
There are 4 senarios in which an empty field may occur: at the start of a record, between 2 field delimiters, an empty field following an empty field and at the end of a record. Alternation can be employed to cater for senarios 1,2 and 4 and senario 3 can be catered for by a second pass using a loop (:;...;t). Multiple senarios can be replaced in both passes using the g flag.
I have a text file which has some below data:
AB-NJCFNJNVNE-802ac94f09314ee
AB-KJNCFVCNNJNWEJJ-e89ae688336716bb
AB-POJKKVCMMMMMJHHGG-9ae6b707a18eb1d03b83c3
AB-QWERTU-55c3375fb1ee8bcd8c491e24b2
I need to remove the data before the second hyphen (-) and produce another text file with the below output:
802ac94f09314ee
e89ae688336716bb
9ae6b707a18eb1d03b83c3
55c3375fb1ee8bcd8c491e24b2
I am pretty new to linux and trying sed command with unsuccessful attempts for the last couple of hours. How can I get the desired output with sed or any other useful command like awk?
You can use a simple cut call:
$ cat myfile.txt | cut -d"-" -f3- > myoutput.txt
Edit:
Some explanation, as requested in the comments:
cut breaks up a string of text to fields according to a given delimiter.
-d defines the delimiter, - in this case.
-f defines which fields to output. In this case, we want to eliminate everything before the second hyphen, or, in other words, return the third field and onwards (3-).
The rest of the command is just piping the output. cating the file into cut, and then saving the result to an output file.
Or, using sed:
cat myfile.txt | sed -e 's/^.\+-//'