I am relatively new to linux I want to search a pattern in a file which starts with "Leonard is" and ends on "champion"
Also this pattern might be placed in multiple lines
the input file(input.txt) may look like:
1 rabbit eats carrot Leonard is a champion
2 loin is the king of
3 jungle Leonard is a
4 Champion
5 Leonard is An exemplary
6 Champion
i would want to have all the occurrences of my pattern ignoring all the other characters other than the pattern in the output file:
1 Leonard is a champion
3 Leonard is a
4 Champion
5 Leonard is An exemplary
6 Champion
i have been very close with the following command:
cat input.txt | grep -ioE "Leonard.*Champion$"
as this command only returns
1 Leonard is a champion
ignoring all the patterns occurring in multiple line
if any other approach of searching other than grep is useful kindly let me know Thanks!!
Perl to the rescue:
perl -l -0777 -e 'print for <> =~ /(.*Leonard(?s:.*?)[Cc]hampion.*)/g' -- input.txt
-l adds newlines to prints
-0777 reads the whole file instead of processing it line by line
the diamond operator <> reads the input
.*? is like .*, i.e. it matches anything, but the ? means the shortest possible match is enough. That prevents the regex from matching everything between the first Leonard and last Champion.
. in a regex doesn't match a newline normally, but it does with the s modifier. (?s:.*?) localizes the changed behaviour, so other dots still don't match newlines.
You're looking for \s which stands for whitespace. + stands for one or more
Pattern: Leonard is a\s+Champion
See: https://regex101.com/r/qiNXhf/1
I use this tool with 0 knowledge of regex in my mind, and it helps me a lot. See the notes on the right bottom, where all these signs are explained.
The "." is referenced as "any character except new line", therefore, what you're trying to achieve with . is not possible, I suggest using \s with an addition of * or + as well (as suggested above), but need to find out how to implement it with the "grep" reg expression. There are also nice tools for regex testing - https://regexr.com/ for example.
Related
I am trying to replace string between two strings in a file with the command below. There could be any number of such patterns in the file. This is just an example.
sed 's/word1.*word2/word1/' 1.txt
There are two instances where 'word1' followed by 'word2' occurs in the sample source file I'm testing. Content of the 1.txt file
word1---sjdkkdkjdk---word2 I want this text----word1---jhfnkfnsjkdnf----word2 I need this also
Result is as below.
word1 I need this also
Expected Output :
word1 I want this text----word1 I need this also
Can anybody help me with this please?
I looked at other stack-overflow questionnaire but they discuss about replacing only one instance of the pattern.
Regular expressions are greedy - they match the longest possible string, so everything from the first 'word1' to the last 'word2'. Not sure if any version of sed supports non-greedy regexps... you could just use perl, though, which does:
perl -pe 's/word1.*?word2/word1/g' 1.txt
should do the trick. That ? changes the meaning of the prior * from 'match as many times as possible as long as the rest of the pattern matches' to 'match as few times as possible as long as the rest of the pattern matches'.
$ sed 's/#/#A/g; s/{/#B/g; s/}/#C/g; s/word1/{/g; s/word2/}/g; s/{[^{}]*}/word1/g; s/}/word2/g; s/{/word1/g; s/#C/}/g; s/#B/{/g; s/#A/#/g' file
word1 I want this text----word1 I need this also
It's lengthy and looks complicated but it's a technique that is used fairly often and is really just a series of simple steps to robustly convert word1 to { and word2 to } so you're dealing with characters instead of strings in the actual substitution s/{[^{}]*}/word1/g and so can use a negated bracket expression to avoid the greedy regexp taking up too much of the line.
See https://stackoverflow.com/a/35708616/1745001 for more info on the general approach used here to be able to turn strings into characters that cannot be present in the input by the time the real work takes place and then restore them again afterwards.
If you only have two instances of the word1-word2 pattern on a line, this should work:
sed 's/\(word1\).*word2\(.*\)\(word1\).*word2\(.*\)/\1\2\3\4/' 1.txt
I grab the parts we want to keep inside escaped brackets \( and \) then I can refer to those parts as \1 \2 and so on.
I created a file via Linux terminal and named it people, the contents of this file are as follows..
Mr. Smith
Mrs. Jenn Bewlite
Ms Carmichael
Dr Ivan James
Mrs Holly Alva Beswol
Mrs James Sheepwool
Mr. Hitchcock
How do I display lines that have the letter H followed later on the line by the letter o.
I have tried to use the following commands, but it didn't work.. maybe I have a typo.
$ egrep -w 'H|o' /home/liveuser/people
$ grep "H|o" people
You do not have a typo. What you are trying to do is called
"regular expressions".
How do I display lines that have the letter H followed later on the line by the letter o.
I assume you want to match H case insensitive, with any characters following until the character "o". Please post a comment to correct me if I've misunderstood.
grep -i "h.*o" /home/liveuser/people
This will do the trick in your case.
You can replace /home/liveuser/people with whatever path you want. I see you try to use just "people" which is fine if your current directory is /home/liveuser
Note that you can use "egrep" too.
Explanation
The -i flag makes grep case insensitive
As for the "h.*o" there are 4 things going on:
The "h" matches the character "h"
The "." matches any character. (Except for newline)
The "*" makes it so that the previous character or expression is matched multiple times. In this case it matches any amount of characters that isn't newline
The "o" makes sure that we have an "o" after the "h" with any characters in between.
A link to a page explaining regex in grep: Regular Expressions in grep
You have the wrong regexp syntax. Use H.*o, not H|o.
Like this?:
$ grep "H.*o" foo
Mrs Holly Alva Beswol
Mr. Hitchcock
One of my elder brother who is studying in Statistics. Now, he is writing his thesis paper in LaTeX. Almost all contents are written for the paper. And he took 5 number after point(e.g. 5.55534) for each value those are used for his calculation. But, at the last time his instructor said to change those to 3 number after point(e.g. 5.555) which falls my brother in trouble. Finding and correcting those manually is not easy. So, he told me to help.
I believe there is also a easy solution which is know to me. The snapshot of a portion of the thesis looks like-
&se($\hat\beta_1$)&0.35581&0.35573&0.35573\\
&mse($\hat\beta_1$)&.12945&.12947&.12947\\
\addlinespace
&$\hat\beta_2$&0.03329&0.03331&0.03331 \\
&se($\hat\beta_2$)&0.01593&0.01592&0.01591\\
&mse($\hat\beta_2$)&.000265&.000264&.000264 \\
\midrule
{n=100} & $\hat\beta_1$&-.52006&-.52001&-.51946\\
&se($\hat\beta_1$)&.22819&.22814&.22795\\
&mse($\hat\beta_1$)&.05247&.05244&.05234\\
\addlinespace
&$\hat\beta_2$&0.03134&0.03134&0.03133 \\
&se($\hat\beta_2$)&0.00979&0.00979&0.00979\\
&mse($\hat\beta_2$)&.000098&.000098&.000098
I want -
&se($\hat\beta_1$)&0.355&0.355&0.355\\
&mse($\hat\beta_1$)&.129&.129&.129\\
......................................................................
........................................................................
........................................................................
Note: Don't feel boring for the syntax(These are LaTeX syntax).
If anybody has solution or suggestion, please provide. Thank you.
In sed:
$ sed 's/\(\.[0-9]\{3\}\)[0-9]*/\1/g' file
&se($\hat\beta_1$)&0.355&0.355&0.355\\
&mse($\hat\beta_1$)&.129&.129&.129\\
ie. replace period starting numeric strings with at least 3 numbers with the leading period and three first numbers.
Here is the command in vim:
:%s/\.\d\{3}\zs\d\+//g
Explanation:
: entering command-mode
% is the range of all lines of the file
s substitution command
\.\d\{3}\zs\d\+ pattern you would like to change
\. literal point (.)
\d\{3} match 3 consecutive digits
\zs start substitution from here
\d\+ one or more digits
g Replace all occurrences in the line
Concerning grep and cat they have nothing to do with replacing text. These commands are only for searching and printing contents of files.
Instead, what you are looking is substitution there are lots of commands in Linux that can do that mainly sed, perl, awk, ex etc.
I want to extract the first instance of a string per line in linux. I am currently trying grep but it yields all the instances per line. Below I want the strings (numbers and letters) after "tn="...but only the first set per line. The actual characters could be any combination of numbers or letters. And there is a space after them. There is also a space before the tn=
Given the following file:
hello my name is dog tn=12g3 fun 23k3 hello tn=1d3i9 cheese 234kd dks2 tn=6k4k ksk
1263 chairs are good tn=k38493kd cars run vroom it95958 tn=k22djd fair gold tn=293838 tounge
Desired output:
12g3
k38493
Here's one way you can do it if you have GNU grep, which (mostly) supports Perl Compatible Regular Expressions with -P. Also, the non-standard switch -o is used to only print the part matching the pattern, rather than the whole line:
grep -Po '^.*?tn=\K\S+' file
The pattern matches the start of the line ^, followed by any characters .*?, where the ? makes the match non-greedy. After the first match of tn=, \K "kills" the previous part so you're only left with the bit you're interested in: one or more non-space characters \S+.
As in Ed's answer, you may wish to add a space before tn to avoid accidentally matching something like footn=.... You might also prefer to use something like \w to match "word" characters (equivalent to [[:alnum:]_]).
Just split the input in tn=-separators and pick the second one. Then, split again to get everything up to the first space:
$ awk -F"tn=" '{split($2,a, " "); print a[1]}' file
12g3
k38493kd
$ awk 'match($0,/ tn=[[:alnum:]]+/) {print substr($0,RSTART+4,RLENGTH-4)}' file
12g3
k38493kd
I'm trying to find the code that searches all palindromes in a dictionary file
this is what I got atm which is wrong :
sed -rn '/^([a-z])-([a-z])\2\1$/p' /usr/share/dict/words
Can somebody explain the code as well.
Found the right answer.
sed -n '/^\([a-z]\)\([a-z]\)\2\1$/p' /usr/share/dict/words
I have no idea why I used -
I also don't have an explenation for the \ ater each group
You can use the grep command as explained here
grep -w '^\(.\)\(.\).\2\1'
explanation The grep command searches for the first any three letters by using (.)(.). after that we are searching the same 2nd character and 1st character is occuring or not.
The above grep command will find out only 5 letters palindrome words.
extended version is proposed as well on that page; and works correctly for the first line but then crashes... there is surely some good to keep and maybe to adapt...
Guglielmo Bondioni proposed a single RE that finds all palindromes up to 19 characters long using 9 subexpressions and 9 back-references:
grep -E -e '^(.?)(.?)(.?)(.?)(.?)(.?)(.?)(.?)(.?).?\9\8\7\6\5\4\3\2\1' file
You can extend this further as much as you want :)
Perl to the rescue:
perl -lne 'print if $_ eq reverse' /usr/share/dict/words
Hate to say it, but while regex may be able to cook your breakfast, I don't think it can find a palindrome. According to the all-knowing Wikipedia:
In the automata theory, a set of all palindromes in a given alphabet is a typical example of a language that is context-free, but not regular. This means that it is impossible for a computer with a finite amount of memory to reliably test for palindromes. (For practical purposes with modern computers, this limitation would apply only to incredibly long letter-sequences.)
In addition, the set of palindromes may not be reliably tested by a deterministic pushdown automaton which also means that they are not LR(k)-parsable or LL(k)-parsable. When reading a palindrome from left-to-right, it is, in essence, impossible to locate the "middle" until the entire word has been read completely.
So a regular expression won't be able to solve the problem based on the problem's nature, but a computer program (or sed examples like #NeronLeVelu or #potong) will work.
explanation of your code
sed -rn '/^([a-z])-([a-z])\2\1$/p' /usr/share/dict/words
select and print line that correspond to :
A first (starting the line) small alphabetic character followed by - followed by another small alaphabetic character (could be the same as the first) followed by the last letter of the previous group followed by the first letter Letter1-Letter2Letter2Letter1 and the no other element (end of line)
sample:
a-bba
a is first letter
b second letter
b is \2
a is \1
But it's a bit strange for any work unless it came from a very specific dictionnary (limited to combination by example)
This might work for you (GNU sed):
sed -r 'h;s/[^[:alpha:]]//g;H;x;s/\n/&&/;ta;:a;s/\n(.*)\n(.)/\n\2\1\n/;ta;G;/\n(.*)\n\n\1$/IP;d' file
This copies the original string(s) to the hold space (HS), then removes everything but alpha characters from the string(s) and appends this to the HS. The second copy is then reversed and the current string(s) and the reversed copy compared. If the two strings are equal then the original string(s) is printed out otherwise the line is deleted.