Bash: Extract Range with Regular Expressioin (maybe sed?) - regex

I have a file that is similar to this:
<many lines of stuff>
SUMMARY:
<some lines of stuff>
END OF SUMMARY
I want to extract just the stuff between SUMMARY and END OF SUMMARY. I suspect I can do this with sed but I am not sure how. I know I can modify the stuff in between with this:
sed "/SUMMARY/,/END OF SUMMARY/ s/replace/with/" fileName
(But not sure how to just extract that stuff).
I am Bash on Solaris.

sed -n "/SUMMARY/,/END OF SUMMARY/p" fileName

If Perl is fine you can use:
perl -e 'print $1 if(`cat FILE_NAME`=~/SUMMARY:\n(.*?)END OF SUMMARY/s);'

If you don't want to print the marker lines:
sed '1,/SUMMARY/d;/END OF SUMMARY/,$d' filename

This should work using (FreeBSD) sed as well:
sed -E -n -e '/^SUMMARY:/,/^END OF SUMMARY/{ /^SUMMARY:/d; /^END OF SUMMARY/d; p;}' file.txt

You can do this with awk:
$ echo 'many
lines
of
stuff
SUMMARY:
this is the summary
over two lines
END OF SUMMARY' | awk '
BEGIN {e=0}
/^END OF SUMMARY$/ {e=0}
{if (e==1) {print}}
/^SUMMARY:$/ {e=1}'
which outputs:
this is the summary
over two lines
Not all implementations of awk will require the BEGIN clause but I always like to include explicit initialisation.
It works by using an echo flag (e) to decide whether you're in the summary section or not.

On Solaris , use nawk
#!/bin/bash
nawk '
/SUMMARY/{
gsub(".*SUMMARY:","");
f=1
}
/END OF SUMMARY/{f=0;
gsub("END OF SUMMARY.*","")
}f' file
output
$ cat file
1 2 3 <many lines of stuff>
4 5 6 SUMMARY: 7 8 9
<some lines of stuff>
END OF SUMMARY blah
blah
$ ./shell.sh
7 8 9
<some lines of stuff>

Here's yet another sed version just doing a multi-line print & quit (which may be suitable for extracting a range of lines from a large file):
sed -E -n -e '/^SUMMARY:$/{n;h;};/^END OF SUMMARY$/!H;/^END OF SUMMARY$/{g;p;q;}' fileName | sed 1d
For a multi-line sed script pretty well explained see:
http://ilfilosofo.com/blog/2008/04/26/sed-multi-line-search-and-replace/

Related

Inserting a "," in a particular position of a text

(I put a exact text and command I executed so would be looking a bit messy.)
I have a .TXT file looking like
11111111111111111111111111111111111111111111111111111111111111111111111
11111111111111111111111111111111111111111111111111111111111111111111111
And outcome I am looking for would be like
11111111111111,1111111,11,1,111,1111111111111,1,11111111,1111111111111111,111,111
11111111111111,1111111,11,1,111,1111111111111,1,11111111,1111111111111111,111,111
Command I have tried is
sed -i 's/\(.\{14\}\)\(.\{7\}\)\(.\{2\}\)\(.\{1\}\)\(.\{3\}\)\(.\{13\}\)\(.\{1\}\)\(.\{8\}\)\(.\{16\}\)\(.\{3\}\)/\1,\2,\3,\4,\5,\6,\7,\8,\9,\10,/' SOME.TXT
And outcome I have got was
11111111111111,1111111,11,1,111,1111111111111,1,11111111,1111111111111111,1111111111111110,111
11111111111111,1111111,11,1,111,1111111111111,1,11111111,1111111111111111,1111111111111110,111
I have literally no idea why these 0s suddenly popped out and ' , ' doesn't appear in the position where I command even though it worked half way.
Is this a bug or something in sed command?
It is printing 0 in output because sed capture groups and their back-references can be up to 9 only and \10 is interpreted as \1 followed by literal 0.
You can solve it easily using FIELDWIDTHS feature of gnu-awk:
awk -v OFS=, 'BEGIN { FIELDWIDTHS = "14 7 2 1 3 13 1 8 16 3 *" } {$1 = $1} 1' file
11111111111111,1111111,11,1,111,1111111111111,1,11111111,1111111111111111,111,111
11111111111111,1111111,11,1,111,1111111111111,1,11111111,1111111111111111,111,111
Just for academic exercise, here is a working sed to solve this using 2 substitutions:
sed -E 's/(.{14})(.{7})(.{2})(.)(.{3})(.{13})(.)(.{8})(.+)/\1,\2,\3,\4,\5,\6,\7,\8,\9/; s/(.+,.{16})(.{3})(.*)/\1,\2,\3/' file
sed can't reference capture groups > 9, Perl can:
perl -i -pe 's/(.{14})(.{7})(.{2})(.)(.{3})(.{13})(.)(.{8})(.{16})(.{3})/$1,$2,$3,$4,$5,$6,$7,$8,$9,$10,/' SOME.TXT
If you insist to use sed, you can do something like:
sed 's/./&,/68;s/./&,/65;s/./&,/49;s/./&,/41;s/./&,/40;s/./&,/27;s/./&,/24;s/./&,/23;s/./&,/21;s/./&,/14' test.txt
11111111111111,1111111,11,1,111,1111111111111,1,11111111,1111111111111111,111,111
11111111111111,1111111,11,1,111,1111111111111,1,11111111,1111111111111111,111,111

Find all text between $...$ delimiters using bash script

I have a text file, and I'm trying to get an array of strings containing between $..$ delimiters (LaTeX formulas) using bash script. My current code doesn't work, result is empty:
#!/bin/bash
array=($(grep -o '\$([^\$]*)\$' test.txt))
echo ${array[#]}
I tested this regex here, it finds the matches. I use the following test string:
b5f1e7$bfc2439c621353$d1ce0$629f$b8b5
Expected result is
bfc2439c621353 629f
But echo returns empty. Although if I use '[0-9]\+' it works:
5 1 7 2439 621353 1 0 629 8 5
What do I do wrong?
How about:
grep -o '\$[^$]*\$' test.txt | tr -d '$'
This is basically performing your original grep (but without the brackets, which were causing it to not match), then removing the first/last characters from each match.
You may use awk with input field separator as $:
s='b5f1e7$bfc2439c621353$d1ce0$629f$b8b5'
awk -F '$' '{for (i=2; i<=NF; i+=2) print $i}' <<< "$s"
Note that this awk command doesn't validate input. If you want awk to allow for only valid inputs then you may use this gnu awk command with FPAT:
awk -v FPAT='\\$[^$]*\\$' '{for (i=1; i<=NF; i++) {gsub(/\$/, "", $i); print $i}}' <<< "$s"
bfc2439c621353
629f
What about this?
grep -Eo '\$[^$]+\$' a.txt | sed 's/\$//g'
I'm using sed to replace the $.
Try escaping your braces:
tst> grep -o '\$\([^\$]*\)\$' test.txt
$bfc2439c621353$
$629f$
of course, you then have to strip out the $ signs (-o prints the entire match). You can try sed instead:
tst> sed 's/[^\$]*\$\([^\$]*\)\$[^\$]*/\1\n/g' test.txt
bfc2439c621353
629f
Why is your expected output given b5f1e7$bfc2439c621353$d1ce0$629f$b8b5 the two elements bfc2439c621353 629f rather than the three elements bfc2439c621353 d1ce0 629f?
Here's a single grep command to extract those:
$ grep -Po '\$\K[^\$]*(?=\$)' <<<'b5f1e7$bfc2439c621353$d1ce0$629f$b8b5'
bfc2439c621353
d1ce0
629f
(This requires GNU grep as compiled with libpcre for -P)
This uses \$\K (equivalent to (?<=\$)to look behind at the first $ and (?=\$) to look ahead to the next $. Since these are lookarounds, they are not absorbed by grep in the process and therefore d1ce0 is available to be found.
Here's a single POSIX sed command to extract those:
$ sed 's/^[^$]*\$//; s/\$[^$]*$//; s/\$/\n/g' \
<<<'b5f1e7$bfc2439c621353$d1ce0$629f$b8b5'
bfc2439c621353
d1ce0
629f
This does not use any GNU notation and should work on any POSIX-compatible system (such as OS X). It removes the leading and trailing portions that aren't wanted, then replaces each $ with a newline.
Using bash regex:
var="b5f1e7\$bfc2439c621353\$d1ce0\$629f\$b8b5" # string to var
while [[ $var =~ ([^$]*\$)([^$]*)\$(.*) ]] # matching
do
echo -n "${BASH_REMATCH[2]} " # 2nd element has the match
var="${BASH_REMATCH[3]}" # 3rd is the rest of the string
done
echo # trailing newline
bfc2439c621353 629f

sed & regex expression

I'm trying to add a 'chr' string in the lines where is not there. This operation is necessary only in the lines that have not '##'.
At first I use grep + sed commands, as following, but I want to run the command overwriting the original file.
grep -v "^#" 5b110660bf55f80059c0ef52.vcf | grep -v 'chr' | sed 's/^/chr/g'
So, to run the command in file I write this:
sed -i -E '/^#.*$|^chr.*$/ s/^/chr/' 5b110660bf55f80059c0ef52.vcf
This is the content of the vcf file.
##FORMAT=<ID=DP4,Number=4,Type=Integer,Description="#ref plus strand,#ref minus strand, #alt plus strand, #alt minus strand">
#CHROM POS ID REF ALT QUAL FILTER INFO FORMAT 24430-0009S21_GM17-12140
1 955597 95692 G T 1382 PASS VARTYPE=1;BGN=0.00134309;ARL=150;DER=53;DEA=55;QR=40;QA=39;PBP=1091;PBM=300;TYPE=SNP;DBXREF=dbSNP:rs115173026,g1000:0.2825,esp5400:0.2755,ExAC:0.2290,clinvar:rs115173026,CLNSIG:2,CLNREVSTAT:mult,CLNSIGLAB:Benign;SGVEP=AGRN|+|NM_198576|1|c.45G>T|p.:(p.Pro15Pro)|synonymous GT:DP:AD:DP4 0/1:125:64,61:50,14,48,13
chr1 957898 82729935 G T 1214 off_target VARTYPE=1;BGN=0.00113362;ARL=149;DER=50;DEA=55;QR=38;QA=40;PBP=245;PBM=978;NVF=0.53;TYPE=SNP;DBXREF=dbSNP:rs2799064,g1000:0.3285;SGVEP=AGRN|+|NM_198576|2|c.463+56G>T|.|intronic GT:DP:AD:DP4 0/1:98:47,51:9,38,10,41
If I understand what is your expected result, try:
sed -ri '/^(#|chr)/! s/^/chr/' file
Your question isn't clear and you didn't provide the expected output so we can't test a potential solution but if all you want is to add chr to the start of lines where it's not already present and which don't start with # then that's just:
awk '!/^(#|chr)/{$0="chr" $0} 1' file
To overwrite the original file using GNU awk would be:
awk -i inplace '!/^(#|chr)/{$0="chr" $0} 1' file
and with any awk:
awk '!/^(#|chr)/{$0="chr" $0} 1' file > tmp && mv tmp file
This can be done with a single sed invocation. The script itself is something like the following.
If you have an input of format
$ echo -e '#\n#\n123chr456\n789chr123\nabc'
#
#
123chr456
789chr123
abc
then to prepend chr to non-commented chrless lines is done as
$ echo -e '#\n#\n123chr456\n789chr123\nabc' | sed '/^#/ {p
d
}
/chr/ {p
d
}
s/^/chr/'
which prints
#
#
123chr456
789chr123
chrabc
(Note the multiline sed script.)
Now you only need to run this script on a file in-place (-i in modern sed versions.)

How to display words as per given number of letters?

I have created this basic script:
#!/bin/bash
file="/usr/share/dict/words"
var=2
sed -n "/^$var$/p" /usr/share/dict/words
However, it's not working as required to be (or still need some more logic to put in it).
Here, it should print only 2 letter words but with this it is giving different output
Can anyone suggest ideas on how to achieve this with sed or with awk?
it should print only 2 letter words
Your sed command is just searching for lines with 2 in text.
You can use awk for this:
awk 'length() == 2' file
Or using a shell variable:
awk -v n=$var 'length() == n' file
What you are executing is:
sed -n "/^2$/p" /usr/share/dict/words
This means: all lines consisting in exactly the number 2, nothing else. Of course this does not return anything, since /usr/share/dict/words has words and not numbers (as far as I know).
If you want to print those lines consisting in two characters, you need to use something like .. (since . matches any character):
sed -n "/^..$/p" /usr/share/dict/words
To make the number of characters variable, use a quantifier {} like (note the usage of \ to have sed's BRE understand properly):
sed -n "/^.\{2\}$/p" /usr/share/dict/words
Or, with a variable:
sed -n '/^.\{'"$var"'\}$/p' /usr/share/dict/words
Note that we are putting the variable outside the quotes for safety (thanks Ed Morton in comments for the reminder).
Pure bash... :)
file="/usr/share/dict/words"
var=2
#building a regex
str=$(printf "%${var}s")
re="^${str// /.}$"
while read -r word
do
[[ "$word" =~ $re ]] && echo "$word"
done < "$file"
It builds a regex in a form ^..$ (the number of dots is variable). So doing it in 2 steps:
create a string of the desired length e.g: %2s. without args the printf prints only the filler spaces for the desired length e.g.: 2
but we have a variable var, therefore %${var}s
replace all spaces in the string with .
but don't use this solution. It is too slow, and here are better utilities for this, best is imho grep.
file="/usr/share/dict/words"
var=5
grep -P "^\w{$var}$" "$file"
Try awk-
awk -v var=2 '{if (length($0) == var) print $0}' /usr/share/dict/words
This can be shortened to
awk -v var=2 'length($0) == var' /usr/share/dict/words
which has the same effect.
To output only lines matching 2 alphabetic characters with grep:
grep '^[[:alpha:]]\{2\}$' /usr/share/dict/words
GNU awk and mawk at least (due to empty FS):
$ awk -F '' 'NF==2' /usr/share/dict/words #| head -5
aa
Ab
ad
ae
Ah
Empty FS separates each character on its own field so NF tells the record length.

Find regular expression in a file matching a given value

I have some basic knowledge on using regular expressions with grep (bash).
But I want to use regular expressions the other way around.
For example I have a file containing the following entries:
line_one=[0-3]
line_two=[4-6]
line_three=[7-9]
Now I want to use bash to figure out to which line a particular number matches.
For example:
grep 8 file
should return:
line_three=[7-9]
Note: I am aware that the example of "grep 8 file" doesn't make sense, but I hope it helps to understand what I am trying to achieve.
Thanks for you help,
Marcel
As others haven pointed out, awk is the right tool for this:
awk -F'=' '8~$2{print $0;}' file
... and if you want this tool to feel more like grep, a quick bash wrapper:
#!/bin/bash
awk -F'=' -v seek_value="$1" 'seek_value~$2{print $0;}' "$2"
Which would run like:
./not_exactly_grep.sh 8 file
line_three=[7-9]
My first impression is that this is not a task for grep, maybe for awk.
Trying to do things with grep I only see this:
for line in $(cat file); do echo 8 | grep "${line#*=}" && echo "${line%=*}" ; done
Using while for file reading (following comments):
while IFS= read -r line; do echo 8 | grep "${line#*=}" && echo "${line%=*}" ; done < file
This can be done in native bash using the syntax [[ $value =~ $regex ]] to test:
find_regex_matching() {
local value=$1
while IFS= read -r line; do # read from input line-by-line
[[ $line = *=* ]] || continue # skip lines not containing an =
regex=${line#*=} # prune everything before the = for the regex
if [[ $value =~ $regex ]]; then # test whether we match...
printf '%s\n' "$line" # ...and print if we do.
fi
done
}
...used as:
find_regex_matching 8 <file
...or, to test it with your sample input inline:
find_regex_matching 8 <<'EOF'
line_one=[0-3]
line_two=[4-6]
line_three=[7-9]
EOF
...which properly emits:
line_three=[7-9]
You could replace printf '%s\n' "$line" with printf '%s\n' "${line%%=*}" to print only the key (contents before the =), if so inclined. See the bash-hackers page on parameter expansion for a rundown on the syntax involved.
This is not built-in functionality of grep, but it's easy to do with awk, with a change in syntax:
/[0-3]/ { print "line one" }
/[4-6]/ { print "line two" }
/[7-9]/ { print "line three" }
If you really need to, you could programmatically change your input file to this syntax, if it doesn't contain any characters that need escaping (mainly / in the regex or " in the string):
sed -e 's#\(.*\)=\(.*\)#/\2/ { print "\1" }#'
As I understand it, you are looking for a range that includes some value.
You can do this in gawk:
$ cat /tmp/file
line_one=[0-3]
line_two=[4-6]
line_three=[7-9]
$ awk -v n=8 'match($0, /([0-9]+)-([0-9]+)/, a){ if (a[1]<n && a[2]>n) print $0 }' /tmp/file
line_three=[7-9]
Since the digits are being treated as numbers (vs a regex) it supports larger ranges:
$ cat /tmp/file
line_one=[0-3]
line_two=[4-6]
line_three=[75-95]
line_four=[55-105]
$ awk -v n=92 'match($0, /([0-9]+)-([0-9]+)/, a){ if (a[1]<n && a[2]>n) print $0 }' /tmp/file
line_three=[75-95]
line_four=[55-105]
If you are just looking to interpret the right hand side of the = as a regex, you can do:
$ awk -F= -v tgt=8 'tgt~$2' /tmp/file
You would like to do something like
grep -Ef <(cut -d= -f2 file) <(echo 8)
This wil grep what you want but will not display where.
With grep you can show some message:
echo "8" | sed -n '/[7-9]/ s/.*/Found it in line_three/p'
Now you would like to transfer your regexp file into such commands:
sed 's#\(.*\)=\(.*\)#/\2/ s/.*/Found at \1/p#' file
Store these commands in a virtual command file and you will have
echo "8" | sed -nf <(sed 's#\(.*\)=\(.*\)#/\2/ s/.*/Found at \1/p#' file)