Note: I am not sure that my regex's are correct since my textbook at school does not explain/teach regex's of this form but only of the math form such as for DFA's/NFA
I would appreciate any suggestions or hints
Question:
(a) find all occurrences of three letter words in text that begin with `a' and end with 'e';
(b) find all occurrences of words in text that begin with `m' and end with 'r';
My Approach:
a) ^[a][a-zA-Z][e]$ (how to distinguish between 3 letter words and all words?)
b) ^[m][a-zA-Z][r]$
Also I want to use these regex's in linux so would the following command work?:
grep '^[a][a-zA-Z][e]$' 'usr/dir/.../text.txt'
or should I use egrep in this way:
find . -text "*.txt" -print0 | xargs -0 egrep '^[a][a-zA-Z][e]$'
You can use grep -w with an alternation of regex for both the matches:
grep -w 'a[a-zA-Z]e\|m[a-zA-Z]*r' file.txt
You can use the word boundary \b to match the start and the end of a word:
a) find all occurrences of three letter words in text that begin with `a' and end with 'e';
grep -o '\ba[a-zA-Z]e\b'
The pattern matches a word boundary, then a following a, a single character and a following e and a word boundary.
b) find all occurrences of words in text that begin with `m' and end with 'r';
grep -o '\bm[a-zA-Z]*r\b'
The pattern matches a word boundary, an m zero ore more characters (thorugh the * quantifier), an r and a word boundary again.
Further I'm using the options -o which outputs every match on its own line rather than outputting the whole line of input which contains a match.
Btw, thanks to the option -w - matching only whole words - you can even simplify the above patterns to:
a)
grep -wo 'a[a-zA-Z]e'
and b)
grep -wo 'm[a-zA-Z]*r'
Thanks to #anubhava!
You asked for egrep. egrep can't help to simplify or optimize the patterns. grep is absolutely fine.
In your examples, you're only going to match full lines with three characters, matching the letters you expect.
The '^' indicates the beginning of the line
The '$' indicates the end of the line
In order to pull out only three letter words you're going to have to match on some whitespace. For instance
grep ' a[a-Z]e ' 'usr/dir/.../text.txt'
however this will miss all instances of three letter words at the beginning or end of your line
here is an issue using egrep and grep to match whitespace/start of line
First of all, egrep is extended grep and is the same as calling grep with option -E. Secondly, you don't need to use find and xargs in many cases as there is -r option that will search recursively in files within specified path.
Your regular expression fits basic (not extended) regular expression language supported by grep, therefore egrep is not needed.
I would simplify this to
grep -r '^a[a-zA-Z]e$' /usr/share/dict/
and this
grep -r '^m[a-zA-Z]*r$' /usr/share/dict/
Related
I have a large binary file. I want to extract certain strings from it and copy them to a new text file.
For example, in:
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7cacscKLrrok9bwC3Z64NTnZM-^G
I want to take the number '7' (after the #^#^#E) and every character after it stopping at the Z ('ignoring the M-^G).
I want to copy this 7cacscKLrrok9bwC3Z64NTnZ to a new file.
There will be multiple such strings in one file. The end will always be denoted by the M- (which I don't want copied). The start will always be denoted by a 7 (which I do want copied).
Unfortunately, my knowledge of grep, sed, etc, does not extend to this level. Can someone please suggest a viable way to achieve this?
cat -v filename | grep [7][A-Z,a-z] will show all strings with a '7' followed by a letter but that's not much.
Thank you.
I've noticed that my requirements are rather more complicated.
(I've performed the correct - I hope - formatting this time). Thanks to 'tshiono' for his (?) answer to the earlier submission.
I want to check the ending of a string and, if it ends in M-, grep another string that follows it (with junk in between). If the string does not end in M-, then I don't want it copied (let alone any other strings).
So what I would like is:
grep -a -Po "7[[:alnum:]]+(?=M-)" file_name and if the ending is M- then grep -a -Po "5x[[:alnum:]]+(?=\^)" file_name to copy the string that starts with 5x and ends with a ^.
In this example:
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7cacscKLrrok9bwC3Z64NTnZM-^GwM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk^89038432nowefe
The outcome would be:
7cacscKLrrok9bwC3Z64NTnZ
5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk
However, if the ending is not M- (more precisely, if the ending is ^S), then do not try the second grep and do not record anything at all.
In this example:
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7cacscKLrrok9bwC3Z64NTnZ^SGwM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk^89038432nowefe
The outcome would be null (nothing copied) as the 7cacs... string ends in ^S.
Is grep the correct tool? Grep a file and if the condition in the grep command is 'yes' then issue a different grep command but if the condition is 'no' then do nothing.
Thanks again.
I have noticed one addition modification.
Can one add an OR command to the second part? Grep if the second string starts with 5x OR 6x?
In the example below, grep -aPo "7[[:alnum:]]+M-.*?5x[[:alnum:]]+\^" filename | grep -aPo "7[[:alnum:]]+(?=M-)|5x[[:alnum:]]+(?=\^)" will extract the strings starting with 7 and the strings starting with 5x.
How can one change the 5x to 5x or 6x?
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7cacscKLrrok9bwC3Z64NTnZM-^GwM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk^89038432nowefe
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7AAAAAscKLrrok9bwC3Z64NTnZM-^GwM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM6x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk^89038432nowefe
In this example, the desired outcome would be:
7cacscKLrrok9bwC3Z64NTnZ
5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk
7AAAAAscKLrrok9bwC3Z64NTnZ
6x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk
UPDATE MARCH 09:
I need to create a series of complex grep (or perl) commands to extract strings from a series of binary files.
I need two strings from the binary file.
The first string will always start with a 1.
The first string will end with a letter or number. The next letter will always be a lower case k. I do not want this k character.
The difficulty is that the ending k will not always be the first k in the string. It might be the first k but it might not.
After the k, there is a second string. The second string will always start with an A or a B.
The ending of the second string will be in one of two forms:
a) it will end with a space then display the first three characters from the first string in lower case followed by a )
b) it will end with a ^K then display the first three characters from the first string in lower case.
For example:
1pppsx9YPar8Rvs75tJYWZq3eo8PgwbckB4m4zT7Yg042KIDYUE82e893hY ppp)
Should be:
1pppsx9YPar8Rvs75tJYWZq3eo8Pgwbc and B4m4zT7Yg042KIDYUE82e893hY - delete the k and the space then ppp.
For example:
1zzzsx9YPkr8Rvs75tJYWZq3eo8PgwbckA2m4zT7Yg042KIDYUE82e893hY^Kzzz
Should be:
1zzzsx9YPkar8Rvs75tJYWZq3eo8Pgwbc and A4m4zT7Yg042KIDYUE82e893hY - delete the second k and the ^Kzzz.
In the second example, we see that the first k is part of the first string. It is the k before the A that breaks up the first and second strings.
I hope there is a super grep expert who can help! Many thanks!
If your grep supports -P option, would you please try:
grep -a -Po "7[[:alnum:]]+(?=M-)" file
The -a option forces grep to read the input as a text file.
The -P option enables the perl-compatible regex.
The -o option tells grep to print only the matched substring(s).
The pattern (?=M-) is a zero-width lookahead assertion (introduced in
Perl) without including it in the result.
Alternatively you can also say with sed:
sed 's/M-/\n/g' file | sed -n 's/.*\(7[[:alnum:]]\+\).*/\1/p'
The first sed command splits the input file into miltiple lines by
replacing the substring M- with a newline.
It has two benefits: it breaks the lines to allow multiple matches with
sed and excludes the unnecessary portion M- from the input.
The next sed command extracts the desired pattern from the input.
It assumes your sed accepts \n in the replacement, which is
a GNU extension (not POSIX compliant). Otherwise please try (in case you are working on bash):
sed 's/M-/\'$'\n''/g' file | sed -n 's/.*\(7[[:alnum:]]\+\).*/\1/p'
[UPDATE]
(The requirement has been updated by the OP and the followings are solutions according to it.)
Let me assume the string which starts with 7 and ends with M- is always followed
by another (no more and no less than one) string which starts with 5x and ends
with ^ (ascii caret character) with junks in between.
Then would you please try the following:
grep -aPo "7[[:alnum:]]+M-.*?5x[[:alnum:]]+\^" file | grep -aPo "7[[:alnum:]]+(?=M-)|5x[[:alnum:]]+(?=\^)"
It executes the task in two steps (two cascaded greps).
The 1st grep narrows down the input data into the candidate substring
which will include the desired two sequences and junks in between.
The regex .*? in between matches any (ascii or binary) characters
except for a newline character.
The trailing ? enables the shortest match
which avoids the overrun due to the greedy nature of regex. The regex is intended to match junks in between.
The 2nd grep includes two regex's merged with a pipe | meaning logical OR.
Then it extracts two desired sequences.
A potential problem of grep solution is that grep is a line oriented command
and cannot include the newline character in the matched string.
If a newline character is included in the junks in between (I'm not sure about the possibility), the above solution will fail.
As a workaround, perl will provide flexible manipulations with binary data.
perl -0777 -ne '
while (/(7[[:alnum:]]+)M-.*?(5x[[:alnum:]]+)\^/sg) {
printf("%s\n%s\n", $1, $2);
}
' file
The regex is mostly same as that of grep because the -P option of grep means
perl-compatible.
It can capture multiple patterns at once in variables $1 and $2 hence just one regex is enough.
The -0777 option to the perl command tells perl to slurp all data
at once.
The s option at the end the regex makes a dot match a newline character.
The g option enables the global (multiple) match.
[UPDATE2]
In order to make the regex match either 5x or 6x, replace 5x with (5|6)x.
Namely:
grep -aPo "7[[:alnum:]]+M-.*?(5|6)x[[:alnum:]]+\^" file | grep -aPo "7[[:alnum:]]+(?=M-)|(5|6)x[[:alnum:]]+(?=\^)"
As mentioned before, the pipe | means OR. The OR operator has the lowest priority in the evaluation, hence you need to enclose them with parens in this case.
If there is a possibility any other number than 5 or 6 may appear, it will be safer to put [[:digit:]] instead, which matches any one digit betweeen 0 and 9:
grep -aPo "7[[:alnum:]]+M-.*?[[:digit:]]x[[:alnum:]]+\^" file | grep -aPo "7[[:alnum:]]+(?=M-)|[[:digit:]]x[[:alnum:]]+(?=\^)"
[UPDATE3]
(Answering the OP's requirement on March 9th)
Let me start with a perl code which regex will be relatively easier
to explain.
perl -0777 -ne 'while (/(1(.{3}).+)k([AB].*)[\013 ]\2/g){print "$1 $3\n"}' file
Output:
1pppsx9YPar8Rvs75tJYWZq3eo8Pgwbc B4m4zT7Yg042KIDYUE82e893hY
1zzzsx9YPkr8Rvs75tJYWZq3eo8Pgwbc A2m4zT7Yg042KIDYUE82e893hY
[Explanation of regex]
(1(.{3}).+)k([AB].*)[\013 ]\2
( start of the 1st capture group referred by $1 later
1 literal "1"
( start of the 2nd capture group referred by \2 later
.{3} a sequence of the identical three characters such as ppp or zzz
) end of the 2nd capture group
.+ followed by any characters with "greedy" match which may include the 1st "k"
) end of the 1st capture group
k literal "k"
( start of the 3rd capture group referred by $3 later
[AB].* the character "A" or "B" followed by any characters
) end of the 3rd capture group
[\013 ] followed by ^K or a whitespace
\2 followed by the capture group 2 previously assigned
When implementing it with grep, we will encounter a limitation of grep.
Although we want to extract multiple patterns from the input file,
the -e option (which can specify multiple search patterns) does not
work with -P option. Then we need to split the regex into two patterns
such as:
grep -Po "(1(.{3}).+)(?=k([AB].*)[\013 ]\2)" file
grep -Po "(1(.{3}).+)k\K([AB].*)(?=[\013 ]\2)" file
And the result will be:
1pppsx9YPar8Rvs75tJYWZq3eo8Pgwbc
1zzzsx9YPkr8Rvs75tJYWZq3eo8Pgwbc
B4m4zT7Yg042KIDYUE82e893hY
A2m4zT7Yg042KIDYUE82e893hY
Please be noted the order of output is not same as the order of appearance in the original file.
Another option will be to introduce ripgrep or rg which is a fast
and versatile version of grep. You may need to install ripgrep with
sudo apt install ripgrep or using other package handling tool.
An advantage of ripgrep is it supports -r (replace) option in which
you can make use of the backreferences:
rg -N -Po "(1(.{3}).+)k([AB].*)[\013 ]\2" -r '$1 $3' file
The -r '$1 $3' option prints the 1st and the 3rd capture groups and the result will be the same as perl.
In the general case, you can use the strings utility to pluck out ASCII from binary files; then of course you can try to grep that output for patterns that you find interesting.
Many traditional Unix utilities like grep have internal special markers which might get messed up by binary input. For example, the character \xFF was used for internal purposes by some versions of GNU grep so you can't grep for that character even if you can figure out a way to represent it in the shell (Bash supports $'\xff' for example).
A traditional approach would be to run hexdump or a similar utility, and then grep that for patterns. However, more modern scripting languages like Perl and Python make it easy to manipulate arbitrary binary data.
perl -ne 'print if m/\xff\xff/' </dev/urandom
This might work for you (GNU sed):
sed -En '/\n/!{s/M-\^G/\n/;s/7[^\n]*\n/\n&/};/^7[^\n]*/P;D' file
Split each line into zero or more lines that begin with 7 and end just before M-^G and only print such lines.
I want to grep the shortest match and the pattern should be something like:
<car ... model=BMW ...>
...
...
...
</car>
... means any character and the input is multiple lines.
You're looking for a non-greedy (or lazy) match. To get a non-greedy match in regular expressions you need to use the modifier ? after the quantifier. For example you can change .* to .*?.
By default grep doesn't support non-greedy modifiers, but you can use grep -P to use the Perl syntax.
Actualy the .*? only works in perl. I am not sure what the equivalent grep extended regexp syntax would be. Fortunately you can use perl syntax with grep so grep -P would work but grep -E which is same as egrep would not work (it would be greedy).
See also: http://blog.vinceliu.com/2008/02/non-greedy-regular-expression-matching.html
grep
For non-greedy match in grep you could use a negated character class. In other words, try to avoid wildcards.
For example, to fetch all links to jpeg files from the page content, you'd use:
grep -o '"[^" ]\+.jpg"'
To deal with multiple line, pipe the input through xargs first. For performance, use ripgrep.
My grep that works after trying out stuff in this thread:
echo "hi how are you " | grep -shoP ".*? "
Just make sure you append a space to each one of your lines
(Mine was a line by line search to spit out words)
Sorry I am 9 years late, but this might work for the viewers in 2020.
So suppose you have a line like "Hello my name is Jello".
Now you want to find the words that start with 'H' and end with 'o', with any number of characters in between. And we don't want lines we just want words. So for that we can use the expression:
grep "H[^ ]*o" file
This will return all the words. The way this works is that: It will allow all the characters instead of space character in between, this way we can avoid multiple words in the same line.
Now you can replace the space character with any other character you want.
Suppose the initial line was "Hello-my-name-is-Jello", then you can get words using the expression:
grep "H[^-]*o" file
The short answer is using the next regular expression:
(?s)<car .*? model=BMW .*?>.*?</car>
(?s) - this makes a match across multiline
.*? - matches any character, a number of times in a lazy way (minimal
match)
A (little) more complicated answer is:
(?s)<([a-z\-_0-9]+?) .*? model=BMW .*?>.*?</\1>
This will makes possible to match car1 and car2 in the following text
<car1 ... model=BMW ...>
...
...
...
</car1>
<car2 ... model=BMW ...>
...
...
...
</car2>
(..) represents a capturing group
\1 in this context matches the sametext as most recently matched by
capturing group number 1
I know that its a bit of a dead post but I just noticed that this works. It removed both clean-up and cleanup from my output.
> grep -v -e 'clean\-\?up'
> grep --version grep (GNU grep) 2.20
I am trying to extract a single string out of a line having many segments in a key-value order, but I don't get it as it matches much more than I want to.
This is my example line:
|SEGA~1~MAGIC~DESCRIPTION~~~M~TEST~|SEGB~34~12.11.2011~3~M~O~|SEGC~HELLO~WORLD~|
This lines is a kind concatenation of many segments into one line. Now I want to extract the the string at index 2 in the segment starting with SEGA.
So what I do is grep for this:
egrep -o 'SEGA(.*?)\~\|'
But it gives me the whole line, sometimes it gives me only the segment I am looking for. With the match I would split that segment by using the ~ character and take the third one.
Since I use .*? with the question mark I expected egrep to only match the content between SEGA and the very first occurrence of ~| which is right before SEGB and not the one at the end of SEGC or SEGB.
How can I tell grep to search for SEGA and give the whole content starting right after SEGA until THE VERY FIRST occurrence of ~|
You can use the -P(--perl-regexp) option in grep:
grep -oP '(?<=SEGA).*?(?=~\|)' file
If you want to include the trailing ~|, please remove the lookahead (?=...).
I think .*? (lazy) does not exit in egrep.
I'd suggest you break the line into lines on | and then grep from those:
$ echo "|SEGA~1~MAGIC~DESCRIPTION~~~M~TEST~|SEGB~34~12.11.2011~3~M~O~|SEGC~HELLO~WORLD~|" | sed -e 's/|/\n/g' | grep ^SEGA
SEGA~1~MAGIC~DESCRIPTION~~~M~TEST~
How do I search for x or more occurrences of a word using regular expressions and grep in a .txt file in a linux terminal, for example, find all lines with 4 or more "and"s in Sample.txt.
Try this:
egrep "and(.*?and){3}" data.txt
And to match "and" regardless of case ("And" or "AND", ...), but skip an "and" that is a part of another word (or name), try:
egrep -i "\band\b(.*?\band\b){3}" data.txt
The -i makes it ignore case, and the word boundaries, \b, will disregard occurrences like "Anand" and "Anderson".
If you need to match and but not bandit, use something like the following:
egrep '\band\b(.+?\band\b){3}' Sample.txt
I want to grep the shortest match and the pattern should be something like:
<car ... model=BMW ...>
...
...
...
</car>
... means any character and the input is multiple lines.
You're looking for a non-greedy (or lazy) match. To get a non-greedy match in regular expressions you need to use the modifier ? after the quantifier. For example you can change .* to .*?.
By default grep doesn't support non-greedy modifiers, but you can use grep -P to use the Perl syntax.
Actualy the .*? only works in perl. I am not sure what the equivalent grep extended regexp syntax would be. Fortunately you can use perl syntax with grep so grep -P would work but grep -E which is same as egrep would not work (it would be greedy).
See also: http://blog.vinceliu.com/2008/02/non-greedy-regular-expression-matching.html
grep
For non-greedy match in grep you could use a negated character class. In other words, try to avoid wildcards.
For example, to fetch all links to jpeg files from the page content, you'd use:
grep -o '"[^" ]\+.jpg"'
To deal with multiple line, pipe the input through xargs first. For performance, use ripgrep.
My grep that works after trying out stuff in this thread:
echo "hi how are you " | grep -shoP ".*? "
Just make sure you append a space to each one of your lines
(Mine was a line by line search to spit out words)
Sorry I am 9 years late, but this might work for the viewers in 2020.
So suppose you have a line like "Hello my name is Jello".
Now you want to find the words that start with 'H' and end with 'o', with any number of characters in between. And we don't want lines we just want words. So for that we can use the expression:
grep "H[^ ]*o" file
This will return all the words. The way this works is that: It will allow all the characters instead of space character in between, this way we can avoid multiple words in the same line.
Now you can replace the space character with any other character you want.
Suppose the initial line was "Hello-my-name-is-Jello", then you can get words using the expression:
grep "H[^-]*o" file
The short answer is using the next regular expression:
(?s)<car .*? model=BMW .*?>.*?</car>
(?s) - this makes a match across multiline
.*? - matches any character, a number of times in a lazy way (minimal
match)
A (little) more complicated answer is:
(?s)<([a-z\-_0-9]+?) .*? model=BMW .*?>.*?</\1>
This will makes possible to match car1 and car2 in the following text
<car1 ... model=BMW ...>
...
...
...
</car1>
<car2 ... model=BMW ...>
...
...
...
</car2>
(..) represents a capturing group
\1 in this context matches the sametext as most recently matched by
capturing group number 1
I know that its a bit of a dead post but I just noticed that this works. It removed both clean-up and cleanup from my output.
> grep -v -e 'clean\-\?up'
> grep --version grep (GNU grep) 2.20