I'm cleaning up a LaTeX file, and I'm in a situation where I need to distinguish absolute value |x| from the set "such that" symbol i.e. {x | x < 0}.
The first step for me is to find all lines containing an odd number of | characters (i.e. the pipe symbol).
In principle, I know how to do this, but I've tried the following regex command with no luck.
egrep '^[^\|]*\|([^\|]*\|[^\|]*\|)*[^\|]*$'
The idea is that a matching line contains, in order:
The line start
0 or more non-pipe characters
Exactly one pipe character
0 or more copies of text containing exactly 2 pipes
The line end
However, for some reason this isn't working.
I run the command on the following file:
\[
S = \{ x | x < 0}
y = |x|
\]
and none of the lines match.
I suspect I'm making a silly mistake somewhere, possibly to do with escaping the pipe characters,
but I'm stumped as to what's wrong.
Can anybody tell me either how to fix this, or provide an alternate expression which matches lines containing an odd number of pipe characters?
Inside the [], | is not a special character so should not be escaped by \. Try:
egrep '^[^|]*\|([^|]*\|[^|]*\|)*[^|]*$'
Better to use awk for this purpose:
awk -F '|' '!(NF%2)'
TESTING:
echo "a|bc|d|erg" | awk -F '|' '!(NF%2)'
OUTPUT:
a|bc|d|erg
echo "abc|d|ergxy" | awk -F '|' '!(NF%2)'
OUTPUT:
how about:
awk -F'|' 'NF&&(NF-1)%2' file
example:
kent$ cat file
|foo|bar
| | | | |
||||||
|||||||
kent$ awk -F'|' 'NF&&(NF-1)%2' file
| | | | |
|||||||
Perl, which is cross platform (Windows too) and generally installed everywhere these days, is my axe of choice:
perl -ne 'print if (s/\|/\|/g) %2 == 1' file
script.sed
#!/bin/sed -nf
# Save to hold
h
# Delete all non | chars
s#[^|]##g
# Odd match
/^\(||\)*|$/ {
# Fetch hold
g
s#^#odd\t:#
}
# Even match
/^\(||\)\+$/ {
# Fetch hold
g
s#^#even\t:#
}
# No match
/^$/ {
# Fetch hold
g
s#^#none\t:#
}
# Print
p
data.txt
do|odd
do|odd|match|me
|even match|me
do|even match|me
do|even match|also|me|please
no-match
shell
sed -nf script.sed data.txt
stdout
odd :do|odd
odd :do|odd|match|me
even :|even match|me
even :do|even match|me
even :do|even match|also|me|please
none :
none :no-match
Related
I want to Filter all content after match with the content and bring the first value after the "."
I have an output something like this:
Output:
product: 13.6.0.35_0
More specifically, I need only the first two digits and the first digit after the dot, remembering that we should not cling to the values in the issue, but rather on the method of filtering the content.
Expected:
13.6
I tried something like:
echo "product: 13.6.0.35_0" | grep -ow '\w*13\w*'
If you need to use grep with the current logic, you can use
echo "product: 13.6.0.35_0" | grep -ow '13\.[0-9]*' | head -1
where 13\.[0-9]* matches 13, . and zero or more digits (as whole word due to w option) and head -1 gets the first match.
You may also use sed or awk:
sed -En 's/.* ([0-9]+\.[0-9]+).*/\1/p' <<< "product: 13.6.0.35_0"
awk -F'[[:space:].]' '{print $2"."$3}' <<< "product: 13.6.0.35_0"
See the online demo.
The sed command matches any text up to space, then matches the space and captures the two subsequent dot-separated numbers into Group 1 (\1) and then the rest of the line is matched and replaced with Group 1 value that is printed (as the default line output is suppressed with -n).
In the awk command, the field separator is set to whitespace and . with -F'[[:space:].]' and the {print $2"."$3} part prints the second and third field values joined with a ..
A pure shell solution using the builtin read , Parameter Expansion and curly braces for command groupings.
echo "product: 13.6.0.35_0" | { read -r _ value; echo "${value%.*.*}" ; }
You can also use cut:
echo 'product: 13.6.0.35_0' | cut -d ' ' -f2 | cut -d '.' -f1-2
13.6
I reached the expected output, it's simple but it works:
var=$(echo "product: 13.6.0.35_0" | grep -Eo '[[:digit:]]+' | sed -n 1,2p)
echo ${var} | sed 's/ /./g'
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.
I would like to convert strings like:
abc=123.24|127.9|2891;xyz;hgy
to:
abc=123.24,127.9,2891;xyz;hgy
This is close:
echo "abc=123.24|127.9|2891;xyz;hgy" | sed -r 's/(=)([0-9.]+)\|/\1\2,/g'
but returns:
abc=123.24,127.9|2891;xyz;hgy
How can I do the rest of the numbers in a similar fashion if the number of bar-separated numbers is variable?
Clarification:
I hate it when people do not give me the whole picture in questions, but my original description above did just that. The small example is embedded in a much larger line that includes "|" separated text. I want to replace only the "|" with "," between numbers that follow an equal sign. Here is an entire line as an example:
chr1 69511 rs75062661 A G . QSS_ref ASP;BaseCounts=375,3,118,4;CAF=[0.348,0.652];COMMON=1;EFF=NON_SYNONYMOUS_CODING(MODERATE|MISSENSE|Aca/Gca|T141A|305|OR4F5|protein_coding|CODING|ENST00000335137|1|1);GNO;HRun=0;HaplotypeScore=0.0000;KGPROD;KGPhase1;LowMQ=0.0280,0.0580,500;MQ=49.32;MQ0=14;MSigDb=ACEVEDO_METHYLATED_IN_LIVER_CANCER_DN,KEGG_OLFACTORY_TRANSDUCTION,REACTOME_GPCR_DOWNSTREAM_SIGNALING,REACTOME_OLFACTORY_SIGNALING_PATHWAY,REACTOME_SIGNALING_BY_GPCR,chr1p36;NORMALT=86;NORMREF=228;NSM;NT=het;OTHERKG;QSS=8;QSS_NT=6;REF;RS=75062661;RSPOS=69511;S3D;SAO=0;SGT=AG->AG;SOMATIC;SSR=0;TQSS=1;TQSS_NT=2;TUMALT=15;TUMREF=227;TUMVAF=0.06198347107438017;TUMVARFRACTION=0.1485148514851485;VC=SNV;VLD;VP=0x050200000a05140116000100;WGT=1;dbNSFP_1000Gp1_AC=1424;dbNSFP_1000Gp1_AF=0.652014652014652;dbNSFP_1000Gp1_AFR_AC=162;dbNSFP_1000Gp1_AFR_AF=0.32926829268292684;dbNSFP_1000Gp1_AMR_AC=235;dbNSFP_1000Gp1_AMR_AF=0.649171270718232;dbNSFP_1000Gp1_ASN_AC=500;dbNSFP_1000Gp1_ASN_AF=0.8741258741258742;dbNSFP_1000Gp1_EUR_AC=527;dbNSFP_1000Gp1_EUR_AF=0.6952506596306068;dbNSFP_29way_logOdds=4.1978;dbNSFP_29way_pi=0.1516:0.0:0.6258:0.2226;dbNSFP_ESP6500_AA_AF=0.544101;dbNSFP_ESP6500_EA_AF=0.887429;dbNSFP_Ensembl_geneid=ENSG00000186092;dbNSFP_Ensembl_transcriptid=ENST00000534990|ENST00000335137;dbNSFP_FATHMM_score=0.51;dbNSFP_GERP++_NR=2.31;dbNSFP_GERP++_RS=1.15;dbNSFP_Interpro_domain=GPCR|_rhodopsin-like_superfamily_(1)|;dbNSFP_LRT_Omega=0.000000;dbNSFP_LRT_pred=N;dbNSFP_LRT_score=0.000427;dbNSFP_MutationAssessor_pred=neutral;dbNSFP_MutationAssessor_score=-1.295;dbNSFP_MutationTaster_pred=N;dbNSFP_MutationTaster_score=0.000162;dbNSFP_Polyphen2_HDIV_pred=B;dbNSFP_Polyphen2_HVAR_pred=B;dbNSFP_SIFT_score=0.950000;dbNSFP_Uniprot_aapos=141;dbNSFP_Uniprot_acc=Q8NH21;dbNSFP_Uniprot_id=OR4F5_HUMAN;dbNSFP_aaalt=A;dbNSFP_aapos=189|141;dbNSFP_aaref=T;dbNSFP_cds_strand=+;dbNSFP_codonpos=1;dbNSFP_fold-degenerate=0;dbNSFP_phyloP=0.267000;dbNSFP_refcodon=ACA;dbSNPBuildID=131 AU:CU:DP:FDP:GU:SDP:SUBDP:TU 228,232:3,3:322:4:86,109:0:0:1,2 227,228:0,0:244:1:15,16:0:0:1,2
The replacement in this line is of the string:
dbNSFP_aapos=189|141
with:
dbNSFP_aapos=189,141
why not:
sed 's/|/,/g'
kent$ echo "abc=123.24|127.9|2891;xyz;hgy"|sed 's/|/,/g'
abc=123.24,127.9,2891;xyz;hgy
Without using perl you can use
input="abc=123;def=hello123test;dbNSFP_aapos=189|141|142;dbNSFP_aaref=T;another=test|hello;"
sed -r 's/(.*=)([0-9.]+\|)+(.*)/\1'$(sed -r 's/(.*=)([0-9.]+\|)+(.*)/\2/' <<< $input | tr '|' ,)'\3/' <<< $input
Output:
abc=123;def=hello123test;dbNSFP_aapos=189,141,142;dbNSFP_aaref=T;another=test|hello;
Replace <<< $input with your file or whatever you actually have as input :)
Explanation:
We have three capturing groups in the regex (I restructured the groups from the OP), the second will contain only the string where the replacement of the | is to happen, while the first and third contain everything before and after the second group.
See the demo # regex101.
Within the second command ($(...)) we grab the second capturing group with sed and replace every | inside with a comma. This substitution is then used to be put in the place of the second group within the other sed-call.
As alternative, you can try with perl and its evalutation flag:
echo "..." | perl -pe 's{=([\d.|]+)}{"=" . (join ",", split /\|/, $1)}eg'
It searches for a string after an equal sign, splits it with | and join it with commas.
Using tr
echo "abc=123.24|127.9|2891;xyz;hgy" | tr \| ,
abc=123.24,127.9,2891;xyz;hgy
Assuming semicolon is your field separator, how about something like
tr `;\n' '\n;' | sed '/=[0-9.|]*$/s/|/,/g' | tr '\n;' ';\n'
This has a serious flaw; it fails in weird ways for the first and last fields on a line. If you can't live with that, maybe try this:
awk -F ';' '{ for(i=1; i<=NF; ++i) if ($i ~ /=([0-9.]+\|)+[0-9.]+$/) gsub(/\|/,",",$i); print }'
I'm working in verilog and need to edit a specific line within a unique block, but am unsure of how to proceed
file.v
...
block1 block1(
.port1(port1),
.port2(port2),
);
block2 block2(
.(port2)(port2),
.(port3)(port3)
);
....
I need to somehow remove the " , " for port2 in block1. without modifying block2. There are also multiple blocks else where that contains port2.
block1 block1(
.port1(port1),
.port2(port2)
);
I've been trying ranges of awk and sed lines, but not getting the results to modify the file successfully. Any suggestions or solutions is much appreciated
This will remove any comma that occurs just before the end of a block (whitespace then );):
perl -0777 -pe 's/,(?=\s*\);)//g'
Notes:
-0777 causes perl to slurp all the input in as a single string. This is required because
we know there's newlines in between so we don't want to read line-by-line
there might be empty lines between the comma and the parentheses so reading by "paragraph" won't work either.
-p causes perl to print the input after modifications.
the regex is the trickiest part
it finds a comma and then looks ahead to match zero or more whitespace characters (includes spaces, tabs, newlines, etc) followed by a close parenthesis and a semicolon.
the lookahead text is not part of the matched text (lookaheads are known as "zero width assertions") -- the matched text will be just the comma
if there's a match, replace the comma with an empty string.
the g flag says do this globally in the string.
This might do the job for you
sed '/block1 block1/,/);/{s/\((port2)\),/\1/}' file.v
how about:
awk -v RS="" '/block1/{sub("port2),","port2)")}7' file
I guess you want to remove commas located after a closing paren ()) followed by a newline and a closing paren and a semicolon ();)?
In this case this might work for you:
sed -r ':a;N;s/\),\n\s*\);/)\n);/;P;D;ba'
| | | |---------| |---| | | |
| | | | | | | -- branch to label "a"
| | | | | | -- delete up to first newline of pattern space
| | | | | -- print up to first newline of pattern space
| | | | -- replace pattern
| | | -- search pattern
| | -- substitute
| -- read next line into pattern space (append)
-- branch label "a"
I am trying to loop through each line in a file and find and extract letters that start with ${ and end with }. So as the final output I am expecting only SOLDIR and TEMP(from inputfile.sh).
I have tried using the following script but it seems it matches and extracts only the second occurrence of the pattern TEMP. I also tried adding g at the end but it doesn't help. Could anybody please let me know how to match and extract both/multiple occurrences on the same line ?
inputfile.sh:
.
.
SOLPORT=\`grep -A 4 '\[LocalDB\]' \${SOLDIR}/solidhac.ini | grep \${TEMP} | awk '{print $2}'\`
.
.
script.sh:
infile='inputfile.sh'
while read line ; do
echo $line | sed 's%.*${\([^}]*\)}.*%\1%g'
done < "$infile"
May I propose a grep solution?
grep -oP '(?<=\${).*?(?=})'
It uses Perl-style lookaround assertions and lazily matches anything between '${' and '}'.
Feeding your line to it, I get
$ echo "SOLPORT=\`grep -A 4 '[LocalDB]' \${SOLDIR}/solidhac.ini | grep \${TEMP} | awk '{print $2}'\`" | grep -oP '(?<=\${).*?(?=})'
SOLDIR
TEMP
This might work for you (but maybe only for your specific input line):
sed 's/[^$]*\(${[^}]\+}\)[^$]*/\1\t/g;s/$[^{$]\+//g'
Extracting multiple matches from a single line using sed isn't as bad as I thought it'd be, but it's still fairly esoteric and difficult to read:
$ echo 'Hello ${var1}, how is your ${var2}' | sed -En '
# Replace ${PREFIX}${TARGET}${SUFFIX} with ${PREFIX}\a${TARGET}\n${SUFFIX}
s#\$\{([^}]+)\}#\a\1\n#
# Continue to next line if no matches.
/\n/!b
# Remove the prefix.
s#.*\a##
# Print up to the first newline.
P
# Delete up to the first newline and reprocess what's left of the line.
D
'
var1
var2
And all on one line:
sed -En 's#\$\{([^}]+)\}#\a\1\n#;/\n/!b;s#.*\a##;P;D'
Since POSIX extended regexes don't support non-greedy quantifiers or putting a newline escape in a bracket expression I've used a BEL character (\a) as a sentinel at the end of the prefix instead of a newline. A newline could be used, but then the second substitution would have to be the questionable s#.*\n(.*\n.*)##, which might involve a pathological amount of backtracking by the regex engine.