I would like to extract 1, 10, and 100 from:
1 one -args 123
10 ten -args 123
100 one hundred -args 123
However this regex returns 100:
echo -e " 1 one\n 10 ten\n100 one hundred" | grep -Po '^(?=[ ]*)\d+(?=.*)'
100
Not ignoring the preceding spaces returns the numbers (but of course with undesired spaces):
echo -e " 1 one\n 10 ten\n100 one hundred" | grep -Po '^[ ]*\d+(?=.*)'
1
10
100
Have I misunderstood non capturing regex groups in grep / Perl (grep version 2.2, Perl as the -P flag should use its regex) or is this a bug? I notice the release notes for 2.6 says "This release fixes an unexpectedly large number of flaws, from outright bugs (surprisingly many, considering this is "grep")".
If someone with 2.6 could try these examples that would be valuable to determine if this is a bug (in 2.2) or intended behaviour.
The issue is what is considered a 'match' by grep. In the absence of telling grep part of the total match is not what you want, it prints everything up to the end of the match regardless of matching groups.
Given:
$ echo "$txt"
1 one -args 123
10 ten -args 123
100 one hundred -args 123
You can get just the first column of digits without leading spaces several ways.
With GNU grep:
$ echo "$txt" | grep -Po '^[ ]*\K\d+'
1
10
100
Here \K is equivalent to a look behind assertion that resets the match text of the match to be what comes after. The left hand, before the \K, is required to match, but is not included in match text printed by grep.
Demo
awk:
$ echo "$txt" | awk '/^[ ]*[0-9]+/{print $1}'
sed:
$ echo "$txt" | sed 's/^[ ]*\([0-9]*\).*/\1/'
Perl:
$ echo "$txt" | perl -lne 'print $1 if /^[ ]*\K(\d+)/'
And then if you want the matches on a single line, run through xargs:
$ echo "$txt" | grep -Po '^[ ]*\K(\d+)' | xargs
1 10 100
Or, if you are using awk or Perl, just change the way it is printed to not include a carriage return.
You can delete the unwanted spaces this way :
echo -e " 1 one\n 10 ten\n100 one hundred" | grep -Po '^[ ]*(\d+)' | tr -d ' '
As for your question of why it is not working, it is not a bug, it is working as intended, you just misinterpreted how it should work.
If we focus on this ^(?=[ ]*)\d+:
The (?=[ ]*) part is a lookahead assertion. So it means that the regex engine tries to check if the ^ is followed by zero or more spaces. But the assertion itself is not part of the match, so in reality this code means :
- Match a ^ that is followed by 0 or more spaces
- After this ^, match one or more digits
So your code will only match when a digit is the first character of the line. The lookahead won't help you on your use case.
I think the anchor messes with the lookahead, which could be a lookbehind, but they can't be ambiguous (I always run into that one). So the following would work:
echo -e " 1 one\n 10 ten\n100 one hundred" | grep -Po '(?=[ ]*)\d+(?=.*)'
As for a better tool, I would use awk as it is suited to any column driven data. So if you were running it off of ps you could do something like:
ps | awk '/stuff you want to look for here/{print $1}'
awk will take care of all the white space by default
Related
So I have tried to write a Bash script to extract the 10 most common double-vowels words from a file, like good, teeth, etc.
Here is what I have so far:
grep -E -o '[aeiou]{2}' $1|tr 'A-Z' 'a-z' |sort|uniq -c|sort -n | tail -10
I tried to use grep with flag E, then find the pattern match, such as 'aa', 'ee', 'ii' , etc, but it is not working at all,
enter image description here, what I got back, just 'ai', 'ea', something like this. Can anyone help me figure how to do pattern match in bash script?
You can simply match any amount of letters before or after a repeated vowel with this POSIX ERE regex with a GNU grep:
grep -oE '[[:alpha:]]*([aeiou])\1[[:alpha:]]*' words.txt
FreeBSD (non-GNU) grep does not support a backreference in the pattern, so you will have to list all possible vowel sequences:
grep -oE '[[:alpha:]]*(aa|ee|ii|oo|uu)[[:alpha:]]*' words.txt
See the online demo:
#!/bin/bash
s='Some good feed
Soot and weed'
grep -oE '[[:alpha:]]*([aeiou])\1[[:alpha:]]*' <<< "$s"
Details:
[[:alpha:]]* - zero or more letters
(aa|ee|ii|oo|uu) - one of the char sequences, aa, ee, ii, oo or uu (| is an alternation operator in a POSIX ERE regex)
([aeiou]) - Group 1: a vowel
\1 - the same vowel as in Group 1
[[:alpha:]]* - zero or more letters
See the diagram:
Simple way to change your regex: replace [aeiou]{2} with aa|ee|ii|oo|uu. (This does not fix the issue of only finding the match rather than the full word.)
Building on Andrew's answer (re: matching double vowels):
$ cat words.txt
good food;foul make chicken,eek too brave
eye you yuu something:three food too tu too
$ grep -E -o '\<[[:alnum:]]*(aa|ee|ii|oo|uu)[[:alnum:]]*\>' words.txt
good
food
eek
too
yuu
three
food
too
too
The grep finds only words (\< and \> represent word boundaries) with letters and/or digits and containing a dual vowel, printing each word on a separate line.
Applying the rest of OP's counting/sorting logic:
$ grep -E -o '\<[[:alnum:]]*(aa|ee|ii|oo|uu)[[:alnum:]]*\>' words.txt | sort | uniq -c | sort -n
1 eek
1 good
1 three
1 yuu
2 food
3 too
Problem
I want to get any text that consists of 1 to three digits followed by a % but without the % using sed.
What I tried
So i guess the following regex should match the right pattern : [0-9]{1,3}%.
Then i can use this sed command to catch the three digits and only print them :
sed -nE 's/.*([0-9]{1,3})%.*/\1/p'
Example
However when i run it, it shows :
$ echo "100%" | sed -nE 's/.*([0-9]{1,3})%.*/\1/p'
0
instead of
100
Obviously, there's something wrong with my sed command and i think the problem comes from here :
[0-9]{1,3}
which apparently doesn't do what i want it to do.
edit:
Solution
The .* at the start of sed -nE 's/.*([0-9]{1,3})%.*/\1/p' "ate" the two first digits.
The right way to write it, according to Wicktor's answer, is :
sed -nE 's/(.*[^0-9])?([0-9]{1,3})%.*/\2/p'
The .* grabs all digits leaving just the last of the three digits in 100%.
Use
sed -nE 's/(.*[^0-9])?([0-9]{1,3})%.*/\2/p'
Details
(.*[^0-9])? - (Group 1) an optional sequence of any 0 or more chars up to the non-digit char including it
([0-9]{1,3}) - (Group 2) one to three digits
% - a % char
.* - the rest of the string.
The match is replaced with Group 2 contents, and that is the only value printed since n suppresses the default line output.
It will be easier to use a cut + grep option:
echo "abc 100%" | cut -d% -f1 | grep -oE '[0-9]{1,3}'
100
echo "100%" | cut -d% -f1 | grep -oE '[0-9]{1,3}'
100
Or else you may use this awk:
echo "100%" | awk 'match($0, /[0-9]{1,3}%/){print substr($0, RSTART, RLENGTH-1)}'
100
Or else if you have gnu grep then use -P (PCRE) option:
echo "abc 100%" | ggrep -oP '[0-9]{1,3}(?=%)'
100
This might work for you (GNU sed):
sed -En 's/.*\<([0-9]{1,3})%.*/\1/p' file
This is a filtering exercise, so use the -n option.
Use a back reference to capture 1 to 3 digits, followed by % and print the result if successful.
N.B. The \< ensures the digits start on a word boundary, \b could also be used. The -E option is employed to reduce the number of back slashes which would normally be necessary to quote (,),{ and } metacharacters.
Question
Let's say I have one line of text with a number placed somewhere (it could be at the beginning, in the middle or at the end of the line).
How to match and keep the first number found in a line using sed?
Minimal example
Here is my attempt (following this page of a tutorial on regular expressions) and the output for different positions of the number:
$echo "SomeText 123SomeText" | sed 's:.*\([0-9][0-9]*\).*:\1:'
3
$echo "123SomeText" | sed 's:.*\([0-9][0-9]*\).*:\1:'
3
$echo "SomeText 123" | sed 's:.*\([0-9][0-9]*\).*:\1:'
3
As you can only the last digit is kept in the process whereas the desired output should be 123...
Using sed:
echo "SomeText 123SomeText 456" | sed -r 's/^[^0-9]*([0-9]+).*$/\1/'
123
You can also do this in gnu awk:
echo "SomeText 123SomeText 456" | awk '{print gensub(/^[^0-9]*([0-9]+).*$/, "\\1", $0)}'
123
To complement the sed solutions, here's an awk alternative (assuming that the goal is to extract the 1st number on each line, if any (i.e., ignore lines without any numbers)):
awk -F'[^0-9]*' '/[0-9]/ { print ($1 != "" ? $1 : $2) }'
-F'[^0-9]*' defines any sequence of non-digit chars. (including the empty string) as the field separator; awk automatically breaks each input line into fields based on that separator, with $1 representing the first field, $2 the second, and so on.
/[0-9]/ is a pattern (condition) that ensures that output is only produced for lines that contain at least one digit, via its associated action (the {...} block) - in other words: lines containing NO number at all are ignored.
{ print ($1!="" ? $1 : $2) } prints the 1st field, if nonempty, otherwise the 2nd one; rationale: if the line starts with a number, the 1st field will contain the 1st number on the line (because the line starts with a field rather than a separator; otherwise, it is the 2nd field that contains the 1st number (because the line starts with a separator).
You can also use grep, which is ideally suited to this task. sed is a Stream EDitor, which is only going to indirectly give you what you want. With grep, you only have to specify the part of the line you want.
$ cat file.txt
SomeText 123SomeText
123SomeText
SomeText 123
$ grep -o '[0-9]\+' file.txt
123
123
123
grep -o prints only the matching parts of a line, each on a separate line. The pattern is simple: one or more digits.
If your version of grep is compatible with the -P switch, you can use Perl-style regular expressions and make the command even shorter:
$ grep -Po '\d+' file.txt
123
123
123
Again, this matches one or more digits.
Using grep is a lot simpler and has the advantage that if the line doesn't match, nothing is printed:
$ echo "no number" | grep -Po '\d+' # no output
$ echo "yes 123number" | grep -Po '\d+'
123
edit
As pointed out in the comments, one possible problem is that this won't only print the first matching number on the line. If the line contains more than one number, they will all be printed. As far as I'm aware, this can't be done using grep -o.
In that case, I'd go with perl:
perl -lne 'print $1 if /.*?(\d+).*/'
This uses lazy matching (the question mark) so only non-digit characters are consumed by the .* at the start of the pattern. The $1 is a back reference, like \1 in sed. If there are more than one number on the line, this only prints the first. If there aren't any at all, it doesn't print anything:
$ echo "no number" | perl -ne 'print "$1\n" if /.*?(\d+).*/'
$ echo "yes123number456" | perl -lne 'print $1 if /.*?(\d+).*/'
123
If for some reason you still really want to use sed, you can do this:
sed -n 's/^[^0-9]*\([0-9]\{1,\}\).*$/\1/p'
unlike the other answers, this is compatible with all version of sed and will only print lines that contain a match.
Try this sed command,
$echo "SomeText 123SomeText" | sed -r '/[^0-9]*([0-9][0-9]*)[^0-9]*/ s//\1 /g'
123
Another example,
$ echo "SomeText 123SomeText 456" | sed -r '/[^0-9]*([0-9][0-9]*)[^0-9]*/ s//\1 /g'
123 456
It prints all the numbers in a file and the captured numbers are separated by spaces while printing.
I need to extract a string with only single or double digit number in them. my file (test) looks like
test1correct
test12something
test123wrong
In the above example, i want to grep only for
test1correct and test12something
I tried this
grep "test[0-9]{1,2}" test but it gives me all the 3 lines.
use: grep "test[0-9]{1,2}[^0-9]"
Using lookaheads and lookbehinds you can specify "exactly one digit" or "exactly three digits" or whatever. This does exactly one digit:
echo 'WB123_4' | grep -Po '(?<![[:digit:]])([[:digit:]]{1})(?![[:digit:]])'
Result: 4
What it is doing is, find a digit that is not preceded by a digit, and also not followed by a digit. Also works for more than one digit. This does three digits, then at least one of anything else, then one digit:
echo 'WB123_4' | grep -Po '(?<![[:digit:]])([[:digit:]]{3})(?![[:digit:]]).+(?<![[:digit:]])([[:digit:]]{1})(?![[:digit:]])'
Result: 123_4
While I'm at it, this combination of grep and sed will find a string with three digits, then one or more of anything else, then one digit, and extract just those parts nicely. (There might have been another way to do that just in grep with groups.)
echo 'WB123_4' | grep -Po '(?<![[:digit:]])([[:digit:]]{3})(?![[:digit:]]).+(?<![[:digit:]])([[:digit:]]{1})(?![[:digit:]])' | sed -r -e 's/[^[:digit:]]+/ /'
Result: 123 4
Note: the -P flag to grep means to use Perl-style regular expressions, which lets you use lookaheads and lookbehinds.
Try this:
test[0-9]{1,2}[A-Za-z]+
cat tst--- tst file contains the following data
1
0
operator
4
5
5
cat tst | grep [0-9]--- while i grrp using using it return only 1
1
how to grep all the numbers in tst file?
Is there a way to tell sed to output only captured groups?
For example, given the input:
This is a sample 123 text and some 987 numbers
And pattern:
/([\d]+)/
Could I get only 123 and 987 output in the way formatted by back references?
The key to getting this to work is to tell sed to exclude what you don't want to be output as well as specifying what you do want. This technique depends on knowing how many matches you're looking for. The grep command below works for an unspecified number of matches.
string='This is a sample 123 text and some 987 numbers'
echo "$string" | sed -rn 's/[^[:digit:]]*([[:digit:]]+)[^[:digit:]]+([[:digit:]]+)[^[:digit:]]*/\1 \2/p'
This says:
don't default to printing each line (-n)
exclude zero or more non-digits
include one or more digits
exclude one or more non-digits
include one or more digits
exclude zero or more non-digits
print the substitution (p) (on one line)
In general, in sed you capture groups using parentheses and output what you capture using a back reference:
echo "foobarbaz" | sed 's/^foo\(.*\)baz$/\1/'
will output "bar". If you use -r (-E for OS X) for extended regex, you don't need to escape the parentheses:
echo "foobarbaz" | sed -r 's/^foo(.*)baz$/\1/'
There can be up to 9 capture groups and their back references. The back references are numbered in the order the groups appear, but they can be used in any order and can be repeated:
echo "foobarbaz" | sed -r 's/^foo(.*)b(.)z$/\2 \1 \2/'
outputs "a bar a".
If you have GNU grep:
echo "$string" | grep -Po '\d+'
It may also work in BSD, including OS X:
echo "$string" | grep -Eo '\d+'
These commands will match any number of digit sequences. The output will be on multiple lines.
or variations such as:
echo "$string" | grep -Po '(?<=\D )(\d+)'
The -P option enables Perl Compatible Regular Expressions. See man 3 pcrepattern or man 3 pcresyntax.
Sed has up to nine remembered patterns but you need to use escaped parentheses to remember portions of the regular expression.
See here for examples and more detail
you can use grep
grep -Eow "[0-9]+" file
run(s) of digits
This answer works with any count of digit groups. Example:
$ echo 'Num123that456are7899900contained0018166intext' \
| sed -En 's/[^0-9]*([0-9]{1,})[^0-9]*/\1 /gp'
123 456 7899900 0018166
Expanded answer.
Is there any way to tell sed to output only captured groups?
Yes. replace all text by the capture group:
$ echo 'Number 123 inside text' \
| sed 's/[^0-9]*\([0-9]\{1,\}\)[^0-9]*/\1/'
123
s/[^0-9]* # several non-digits
\([0-9]\{1,\}\) # followed by one or more digits
[^0-9]* # and followed by more non-digits.
/\1/ # gets replaced only by the digits.
Or with extended syntax (less backquotes and allow the use of +):
$ echo 'Number 123 in text' \
| sed -E 's/[^0-9]*([0-9]+)[^0-9]*/\1/'
123
To avoid printing the original text when there is no number, use:
$ echo 'Number xxx in text' \
| sed -En 's/[^0-9]*([0-9]+)[^0-9]*/\1/p'
(-n) Do not print the input by default.
(/p) print only if a replacement was done.
And to match several numbers (and also print them):
$ echo 'N 123 in 456 text' \
| sed -En 's/[^0-9]*([0-9]+)[^0-9]*/\1 /gp'
123 456
That works for any count of digit runs:
$ str='Test Num(s) 123 456 7899900 contained as0018166df in text'
$ echo "$str" \
| sed -En 's/[^0-9]*([0-9]{1,})[^0-9]*/\1 /gp'
123 456 7899900 0018166
Which is very similar to the grep command:
$ str='Test Num(s) 123 456 7899900 contained as0018166df in text'
$ echo "$str" | grep -Po '\d+'
123
456
7899900
0018166
About \d
and pattern: /([\d]+)/
Sed does not recognize the '\d' (shortcut) syntax. The ascii equivalent used above [0-9] is not exactly equivalent. The only alternative solution is to use a character class: '[[:digit:]]`.
The selected answer use such "character classes" to build a solution:
$ str='This is a sample 123 text and some 987 numbers'
$ echo "$str" | sed -rn 's/[^[:digit:]]*([[:digit:]]+)[^[:digit:]]+([[:digit:]]+)[^[:digit:]]*/\1 \2/p'
That solution only works for (exactly) two runs of digits.
Of course, as the answer is being executed inside the shell, we can define a couple of variables to make such answer shorter:
$ str='This is a sample 123 text and some 987 numbers'
$ d=[[:digit:]] D=[^[:digit:]]
$ echo "$str" | sed -rn "s/$D*($d+)$D+($d+)$D*/\1 \2/p"
But, as has been already explained, using a s/…/…/gp command is better:
$ str='This is 75577 a sam33ple 123 text and some 987 numbers'
$ d=[[:digit:]] D=[^[:digit:]]
$ echo "$str" | sed -rn "s/$D*($d+)$D*/\1 /gp"
75577 33 123 987
That will cover both repeated runs of digits and writing a short(er) command.
Give up and use Perl
Since sed does not cut it, let's just throw the towel and use Perl, at least it is LSB while grep GNU extensions are not :-)
Print the entire matching part, no matching groups or lookbehind needed:
cat <<EOS | perl -lane 'print m/\d+/g'
a1 b2
a34 b56
EOS
Output:
12
3456
Single match per line, often structured data fields:
cat <<EOS | perl -lape 's/.*?a(\d+).*/$1/g'
a1 b2
a34 b56
EOS
Output:
1
34
With lookbehind:
cat <<EOS | perl -lane 'print m/(?<=a)(\d+)/'
a1 b2
a34 b56
EOS
Multiple fields:
cat <<EOS | perl -lape 's/.*?a(\d+).*?b(\d+).*/$1 $2/g'
a1 c0 b2 c0
a34 c0 b56 c0
EOS
Output:
1 2
34 56
Multiple matches per line, often unstructured data:
cat <<EOS | perl -lape 's/.*?a(\d+)|.*/$1 /g'
a1 b2
a34 b56 a78 b90
EOS
Output:
1
34 78
With lookbehind:
cat EOS<< | perl -lane 'print m/(?<=a)(\d+)/g'
a1 b2
a34 b56 a78 b90
EOS
Output:
1
3478
I believe the pattern given in the question was by way of example only, and the goal was to match any pattern.
If you have a sed with the GNU extension allowing insertion of a newline in the pattern space, one suggestion is:
> set string = "This is a sample 123 text and some 987 numbers"
>
> set pattern = "[0-9][0-9]*"
> echo $string | sed "s/$pattern/\n&\n/g" | sed -n "/$pattern/p"
123
987
> set pattern = "[a-z][a-z]*"
> echo $string | sed "s/$pattern/\n&\n/g" | sed -n "/$pattern/p"
his
is
a
sample
text
and
some
numbers
These examples are with tcsh (yes, I know its the wrong shell) with CYGWIN. (Edit: For bash, remove set, and the spaces around =.)
Try
sed -n -e "/[0-9]/s/^[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\).*$/\1 \2 \3 \4 \5 \6 \7 \8 \9/p"
I got this under cygwin:
$ (echo "asdf"; \
echo "1234"; \
echo "asdf1234adsf1234asdf"; \
echo "1m2m3m4m5m6m7m8m9m0m1m2m3m4m5m6m7m8m9") | \
sed -n -e "/[0-9]/s/^[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\).*$/\1 \2 \3 \4 \5 \6 \7 \8 \9/p"
1234
1234 1234
1 2 3 4 5 6 7 8 9
$
You need include whole line to print group, which you're doing at the second command but you don't need to group the first wildcard. This will work as well:
echo "/home/me/myfile-99" | sed -r 's/.*myfile-(.*)$/\1/'
It's not what the OP asked for (capturing groups) but you can extract the numbers using:
S='This is a sample 123 text and some 987 numbers'
echo "$S" | sed 's/ /\n/g' | sed -r '/([0-9]+)/ !d'
Gives the following:
123
987
I want to give a simpler example on "output only captured groups with sed"
I have /home/me/myfile-99 and wish to output the serial number of the file: 99
My first try, which didn't work was:
echo "/home/me/myfile-99" | sed -r 's/myfile-(.*)$/\1/'
# output: /home/me/99
To make this work, we need to capture the unwanted portion in capture group as well:
echo "/home/me/myfile-99" | sed -r 's/^(.*)myfile-(.*)$/\2/'
# output: 99
*) Note that sed doesn't have \d
You can use ripgrep, which also seems to be a sed replacement for simple substitutions, like this
rg '(\d+)' -or '$1'
where ripgrep uses -o or --only matching and -r or --replace to output only the first capture group with $1 (quoted to be avoid intepretation as a variable by the shell) two times due to two matches.