"sed \'s/\t/ /g\' "
Thank you! It is part of code like this
string cmd = "sed \'s/\t/ /g\' " + string(filename) + " > sample.clean";
system(cmd.c_str());
ifstream infile("sample.clean");
if (!infile){
cerr << "error loading file after clean-up!\n";
return 0;
}
It replaces tabs with spaces, but it does so poorly. Within sed itself, 'y/\t/ /' is better, but sed is the wrong tool and it is even better written tr '\t' ' '
It replaces tab characters \t with spaces.
it seems like it replaces tabulations (\t) by spaces. The g means that the replacement should not stop at the first match on a line.
Related
I want to use sed to replace some strings containing '^' in a text file
The original string (" are part of the string)
"POINT(5.94462 45.569924)^^geo:wktLiteral"
the expected substituted string
"POINT(5.94462 45.569924)"^^geo:wktLiteral
I've tried
sed -i 's/\^\^geo:wktLiteral\"/\"\^\^geo:wktLiteral/' myFile
or
sed -i 's/\x5e\x5egeo:wktLiteral\"/\"\x5e\x5egeo:wktLiteral/' myFile
these two solutions do not work.
Any help is welcome.
If you switch to the WSL Linux terminal, you can use a simpler
sed -i 's/\(\^\^geo:wktLiteral\)"/"\1/' file
Details
\(\^\^geo:wktLiteral\) - Group 1: ^^geo:wktLiteral string
" - a " char
"\1 - replacement: a " char and Group 1 value.
I have a tab-delimited file of genetic variants with an INFO column of many semicolon-delimited tags:
Chr Start End Ref Alt ExAC_ALL ExAC_AFR ExAC_AMR ExAC_EAS ExAC_FIN ExAC_NFE ExAC_OTH ExAC_SAS Otherinfo QUAL DP Chr Start Ref Alt QUAL FILTER INFO
1 15847952 15847952 G C . . . . . . . . . 241.9 76196 1 15847952 . G C 241.9 PASS AC=2;AF=0;AN=18332;BaseQRankSum=0.731;ClippingRankSum=-0.731;DP=76196;ExcessHet=3.1;FS=0;InbreedingCoeff=-0.0456;MLEAC=2;MLEAF=0;MQ=38.93;MQRankSum=0.515;NEGATIVE_TRAIN_SITE;QD=10.52;ReadPosRankSum=0.89;SOR=0.481;VQSLOD=-1.406 culprit=MQ
1 15847963 15847963 A C . . . . . . . . . 1607.1 126156 1 15847963 . A C 1607.1 PASS AC=2;AF=0;AN=22004;BaseQRankSum=0.851;ClippingRankSum=-0.419;DP=126156;ExcessHet=3.4904;FS=0;InbreedingCoeff=0.0299;MLEAC=2;MLEAF=0;MQ=59.29;MQRankSum=0.18;QD=1.55;ReadPosRankSum=0.067;SOR=0.651;VQSLOD=0.995 culprit=QD
1 15847964 15847966 GCC - . . . . . . . . . 1607.1 126156 1 15847963 . AGCC A 1607.1 PASS AC=63;AF=0.003;AN=22004;BaseQRankSum=0.851;ClippingRankSum=-0.419;DP=126156;ExcessHet=3.4904;FS=0;InbreedingCoeff=0.0299;MLEAC=55;MLEAF=0.002;MQ=59.29;MQRankSum=0.18;QD=1.55;ReadPosRankSum=0.067;SOR=0.651;VQSLOD=0.995 culprit=QD
1 15847978 15847978 C T . . . . . . . . . 648.41 234344 1 15847978 . C T 648.41 PASS AC=9;AF=0;AN=25894;BaseQRankSum=-0.572;ClippingRankSum=-0.404;DP=234344;ExcessHet=3.348;FS=2.639;InbreedingCoeff=-0.0098;MLEAC=6;MLEAF=0;MQ=58.71;MQRankSum=-0.456;NEGATIVE_TRAIN_SITE;QD=4.13;ReadPosRankSum=-0.456;SOR=0.452;VQSLOD=-1.238 culprit=QD
I want to split the first 3 semicolon-delimited terms in the INFO column:
AC=2;AF=0;AN=18332
So that they become:
AC=2 AF=0 AN=18332 BaseQRankSum=0.731;ClippingRankSum=-0.731;DP=76196;ExcessHet=3.1;FS=0;InbreedingCoeff=-0.0456;MLEAC=2;MLEAF=0;MQ=38.93;MQRankSum=0.515;NEGATIVE_TRAIN_SITE;QD=10.52;ReadPosRankSum=0.89;SOR=0.481;VQSLOD=-1.406 culprit=M
So far I've tried the following expression with sed:
sed -i .bk 's/\(A.=.*\);/\1 /g' allChr_ExAC38.hg38_multianno.txt
But this yields no changes.
Ideally I was looking for a way to tell sed to replace the first 3 occurences of a semicolon ; for a tab, but 's/;/ /g3' doesn't seem to mean that.
Use Perl instead of sed:
perl -i.bk -pe '$c = 0; s/;/\t/ while $c++ < 3' -- file.txt
You can try this awk
awk '{for(i=1;i<4;i++)sub(";","\t")}1' infile
The .* in your regex is greedy, and will match as much text as possible on the line, up to just before the last semicolon (but not beyond, because then the entire regex won't match at all).
You cannot mix /3 and /g; the latter means, replace all occurrences on every line, so it is directly at odds with the /3 which says to replace only a maximum of three occurrences on a line.
"No changes" seems wrong, though; if your regex matched at all, the last semicolon on matching lines will have been replaced.
Some regex engines support non-greedy matching, but sed isn't one of them. As long as there is a single delimiter character you can use to limit the greediness, using that is a much better solution anyway. In your case, simply replace . with [^;] to say "any character except (newline or) semicolon" instead of "any character (except newline)."
sed 's/\(A.=[^;]*\);/\1 /3' allChr_ExAC38.hg38_multianno.txt
(This will print to standard output for verification; put back the -i .bk once you see the result is correct.)
Based on your example data, perhaps consider replacing the remaining . in the expression with [A-Z] and [^;] with [^;=] or even [0-9]. The more specific you can make your regex, the better.
Could you please try following and let me know if this helps you.
awk '
FNR==1{
print;
next}
{
num=split($(NF-1),array,";");
for(i=4;i<=num;i++){
val=val?val ";"array[i]:array[i]};
$(NF-1)=array[1] OFS array[2] OFS array[3] OFS val;
val="";
$1=$1
}
1
' OFS="\t" Input_file
This might work for you (GNU sed):
sed -i.bak 's/;/\n/3;h;y/;/\t/;G;s/\n.*\n/\t/' file
Replace the third ; with a newline, make a copy of the line, replace all ;'s with \t's, append the copy and replace the end of the first line to the middle of the second line with a \t.
Since by definition a line is demarcated by a newline, lines cannot contain a newline unless they are introduced by a programmer.
If the number of occurrences is reasonable you can pipe sed multiple times i.e.
sed -E -e 's/[0-9]{4}/****/'| sed -E -e 's/[0-9]{4}/****/'| sed -E -e 's/[0-9]{4}/****/'
will mask first 3 4-digit groups of credit card number like so
Input:
1234 5678 9101 1234
Output:
**** **** **** 1234
What is a sed script that will remove the "\n" character but only if it is inside "" characters (delimited string), not the \n that is actually at the end of the (virtual) line?
For example, I want to turn this file
"lalala","lalalslalsa"
"lalalala","lkjasjdf
asdfasfd"
"lalala","dasdf"
(line 2 has an embedded \n ) into this one
"lalala","lalalslalsa"
"lalalala","lkjasjdf \\n asdfasfd"
"lalala","dasdf"
(Line 2 and 3 are now joined, and the real line feed was replaced with the character string \\n (or any other easy to spot character string, I'm not picky))
I don't just want to remove every other newline as a previous question asked, nor do I want to remove ALL newlines, just those that are inside quotes. I'm not wedded to sed, if awk would work, that's fine too.
The file being operated on is too large to fit in memory all at once.
sed is an excellent tool for simple substitutions on a single line but for anything else you should use awk., e.g:
$ cat tst.awk
{
if (/"$/) {
print prev $0
prev = ""
}
else {
prev = prev $0 " \\\\n "
}
}
$ awk -f tst.awk file
"lalala","lalalslalsa"
"lalalala","lkjasjdf \\n asdfasfd"
"lalala","dasdf"
Below was my original answer but after seeing #NeronLeVelu's approach of just testing for a quote at the end of the line I realized I was doing this in a much too complicated way. You could just replace gsub(/"/,"&") % 2 below with /"$/ and it'd work the same but the above code is a simpler implementation of the same functionality and will now handle embedded escaped double quotes as long as they aren't at the end of a line.
$ cat tst.awk
{ $0 = saved $0; saved="" }
gsub(/"/,"&") % 2 { saved = $0 " \\\\n "; next }
{ print }
$ awk -f tst.awk file
"lalala","lalalslalsa"
"lalalala","lkjasjdf \\n asdfasfd"
"lalala","dasdf"
The above only stores 1 output line in memory at a time. It just keeps building up an output line from input lines while the number of double quotes in that output line is an odd number, then prints the output line when it eventually contains an even number of double quotes.
It will fail if you can have double quotes inside your quoted strings escaped as \", not "", but you don't show that in your posted sample input so hopefully you don't have that situation. If you have that situation you need to write/use a real CSV parser.
sed -n ':load
/"$/ !{N
b load
}
:cycle
s/^\(\([^"]*"[^"]*"\)*\)\([^"]*"[^"]*\)\n/\1\3 \\\\n /
t cycle
p' YourFile
load the lines in working buffer until a close line (ending with ") is found or end reach
replace any \n that is after any couple of open/close " followed by a single " with any other caracter that " between from the start of file by the escapped version of new line (in fact replace starting string + \n by starting string and escaped new line)
if any substitution occur, retry another one (:cycle and t cycle)
print the result
continue until end of file
thanks to #Ed Morton for remark about escaped new line
I have a special file with this kind of format :
title1
_1 texthere
title2
_2 texthere
I would like all newlines starting with "_" to be placed as a second column to the line before
I tried to do that using sed with this command :
sed 's/_\n/ /g' filename
but it is not giving me what I want to do (doing nothing basically)
Can anyone point me to the right way of doing it ?
Thanks
Try following solution:
In sed the loop is done creating a label (:a), and while not match last line ($!) append next one (N) and return to label a:
:a
$! {
N
b a
}
After this we have the whole file into memory, so do a global substitution for each _ preceded by a newline:
s/\n_/ _/g
p
All together is:
sed -ne ':a ; $! { N ; ba }; s/\n_/ _/g ; p' infile
That yields:
title1 _1 texthere
title2 _2 texthere
If your whole file is like your sample (pairs of lines), then the simplest answer is
paste - - < file
Otherwise
awk '
NR > 1 && /^_/ {printf "%s", OFS}
NR > 1 && !/^_/ {print ""}
{printf "%s", $0}
END {print ""}
' file
This might work for you (GNU sed):
sed ':a;N;s/\n_/ /;ta;P;D' file
This avoids slurping the file into memory.
or:
sed -e ':a' -e 'N' -e 's/\n_/ /' -e 'ta' -e 'P' -e 'D' file
A Perl approach:
perl -00pe 's/\n_/ /g' file
Here, the -00 causes perl to read the file in paragraph mode where a "line" is defined by two consecutive newlines. In your example, it will read the entire file into memory and therefore, a simple global substitution of \n_ with a space will work.
That is not very efficient for very large files though. If your data is too large to fit in memory, use this:
perl -ne 'chomp;
s/^_// ? print "$l " : print "$l\n" if $. > 1;
$l=$_;
END{print "$l\n"}' file
Here, the file is read line by line (-n) and the trailing newline removed from all lines (chomp). At the end of each iteration, the current line is saved as $l ($l=$_). At each line, if the substitution is successful and a _ was removed from the beginning of the line (s/^_//), then the previous line is printed with a space in place of a newline print "$l ". If the substitution failed, the previous line is printed with a newline. The END{} block just prints the final line of the file.
I would be happy if anyone can suggest me command (sed or AWK one line command) to divide each line of file in equal number of part. For example divide each line in 4 part.
Input:
ATGCATHLMNPHLNTPLML
Output:
ATGCA THLMN PHLNT PLML
This should work using GNU sed:
sed -r 's/(.{4})/\1 /g'
-r is needed to use extended regular expressions
.{4} captures every four characters
\1 refers to the captured group which is surrounded by the parenthesis ( ) and adds a space behind this group
g makes sure that the replacement is done as many times as possible on each line
A test; this is the input and output in my terminal:
$ echo "ATGCATHLMNPHLNTPLML" | sed -r 's/(.{4})/\1 /g'
ATGC ATHL MNPH LNTP LML
I suspect awk is not the best tool for this, but:
gawk --posix '{ l = sprintf( "%d", 1 + (length()-1)/4);
gsub( ".{"l"}", "& " ) } 1' input-file
If you have a posix compliant awk you can omit the --posix, but --posix is necessary for gnu awk and since that seems to be the most commonly used implementation I've given the solution in terms of gawk.
This might work for you (GNU sed):
sed 'h;s/./X/g;s/^\(.*\)\1\1\1/\1 \1 \1 \1/;G;s/\n/&&/;:a;/^\n/bb;/^ /s/ \(.*\n.*\)\n\(.\)/\1 \n\2/;ta;s/^.\(.*\n.*\)\n\(.\)/\1\2\n/;ta;:b;s/\n//g' file
Explanation:
h copy the pattern space (PS) to the hold space (HS)
s/./X/g replace every character in the HS with the same non-space character (in this case X)
s/^\(.*\)\1\1\1/\1 \1 \1 \1/ split the line into 4 parts (space separated)
G append a newline followed by the contents of the HS to the PS
s/\n/&&/ double the newline (to be later used as markers)
:a introduce a loop namespace
/^\n/bb if we reach a newline we are done and branch to the b namespace
/^ /s/ \(.*\n.*\)\n\(.\)/\1 \n\2/;ta; if the first character is a space add a space to the real line at this point and repeat
s/^.\(.*\n.*\)\n\(.\)/\1\2\n/;ta any other character just bump along and repeat
:b;s/\n//g all done just remove the markers and print out the result
This work for any length of line, however is the line is not exactly divisible by 4 the last portion will contain the remainder as well.
perl
perl might be a better choice here:
export cols=4
perl -ne 'chomp; $fw = 1 + int length()/$ENV{cols}; while(/(.{1,$fw})/gm) { print $1 . " " } print "\n"'
This re-calculates field-width for every line.
coreutils
A GNU coreutils alternative, field-width is chosen based on the first line of infile:
cols=4
len=$(( $(head -n1 infile | wc -c) - 1 ))
fw=$(echo "scale=0; 1 + $len / 4" | bc)
cut_arg=$(paste -d- <(seq 1 $fw 19) <(seq $fw $fw $len) | head -c-1 | tr '\n' ',')
Value of cut_arg is in the above case:
1-5,6-10,11-15,16-
Now cut the line into appropriate chunks:
cut --output-delimiter=' ' -c $cut_arg infile