I have a csh script and no matter what I do it never echoes the right tab space.
This is what i want:
header 1 header 2 header 3 header 4
Case 1:
echo header 1 header 2 header 3 header 4 | tabify
Output: It will replace all spaces to tab
Case 2:
echo "header 1\\theader 2\\theader 3\\theader 4"
or
echo header 1\\theader 2\\theader 3\\theader 4
Output: header 1\\theader 2\\theader 3\\theader 4
Case 3:
echo -e "header 1\\theader 2\\theader 3\\theader 4"
Output: -e header 1\\theader 2\\theader 3\\theader 4
Help? T.T
CShell sucks, but printf will save the day once more.
$ printf 'header 1\theader 2\theader 3\n'
header 1 header 2 header 3
(Please ignore Stack Overflow's inability to show tabs)
Related
Hello Sed/Regexp experts, Need some help,
I have a file with below contents, need to replace tabs as space inside double quotes.
Note \t is tab.
1 \t 2 \t 3 \t "4 \t 5 \t 6" \t 7
Expected output:
1 \t 2 \t 3 \t "4 5 6" \t 7
Matching quotes and tired replacing the tabs to space but it replaces the content inside the quotes.
sed '/\s/s/".*"/" "/' 1.txt
Thanks
Here is a sed solution using label:
sed -E -e :a -e 's/("[^\t"]*)\t([^"]*")/\1 \2/; ta' file
1 2 3 "4 5 6" 7
However, it is easier to do this using awk by using " as field delimiter and change every even numbered field (which will be inside the quote):
awk '
BEGIN {FS=OFS="\""}
{
for (i=2; i<=NF; i+=2)
gsub(/\t/, " ", $i)
} 1' file
1 2 3 "4 5 6" 7
With your shown samples Only, please try following awk code. Written and tested in GNU awk using RT variable of awk to deal with values between "....".
awk -v RS='"[^*]*"' 'RT{gsub(/\t/,OFS,RT);ORS=RT;print};END{ORS="";print}' Input_file
with python using indexes and regex - re.sub
st = r'1 2 3 "4 5 6" 7'
l_ind = st.index('"')
r_ind = st.rindex('"')
new_st = st[:l_ind] + re.sub(r'\s+', r' ', st[l_ind:r_ind]) + st[r_ind:]
1 2 3 "4 5 6" 7
another version using re.sub and re.findall
re.sub(r'".*?"',re.sub(r'\s+', r' ', re.findall(r'".*?"', st)[0]), st)
1 2 3 "4 5 6" 7
re.findall(r'".*?"', st)[0] - find the string in double quotes
re.sub(r'\s+', r' ', - compress the multiple space to one inside the double quoted string
re.sub(r'".*?"', - substitute the original double quoted string with the new one.
This might work for you (GNU sed):
sed -E ':a;s/^([^"]*("[^"\t]*"[^"]*)*"[^"\t]*)\t/\1 /;ta' file
Replace the first tab within matched double quotes with a space and repeat until failure.
N.B. This solution caters for lines with multiple matching double quotes.
i am trying to delete everything between bracket <>, i can do it if one line only has one <>, but if a line has more than one, it seems to delete everything inside the outer <>.
echo "hi, <how> are you" | sed 's/<.*>//'
result: hi, are you
echo "hi, <how> are <you>? " | sed 's/<.*>//'
result: hi, ?
the first echo is working fine, but if one sentense has more than one <>, it can not classify.
expected input: 1 <a> 2 <b> 3 <c> 4 <d> ...... 1000 <n>
expected out come: 1 2 3 4 .... 1000
thanks
Using awk:
# using gsub - recommended
$ echo "1 <a> 2 <b> 3 <c> 4 <d> ...... 1000 <n>" | awk 'gsub(/<[^>]*>/,"")'
1 2 3 4 ...... 1000
# OR using FS and OFS
$ echo "1 <a> 2 <b> 3 <c> 4 <d> ...... 1000 <n>" | awk -F'<[^>]*>' -v OFS='' '$1=$1'
1 2 3 4 ...... 1000
Following awk will be helpful to you.
echo "hi, <how> are <you>? " | awk '{for(i=1;i<=NF;i++){if($i~/<.*>/){$i=""}}} 1'
OR
echo "1 <a> 2 <b> 3 <c> 4 <d> ...... 1000 <n>" | awk '{for(i=1;i<=NF;i++){if($i~/<.*>/){$i=""}}} 1'
Explanation: Simply going through all the fields of the line(by starting a for loop which starts from i=1 to till the value of NF(number of fields)), there I am checking if a field's value is satisfying regex <.*>(means it has ) then I am nullifying it.
* matches zero or more times with greedy. use the negation character class <[^>]*>
echo "hi, <how> are <you>? " | sed 's/<[^>]*>//g'
I would like to get multi-line text in between horizontal delimiter and ignore anything else before and after the delimiter.
An example would be:-
Some text here before any delimiter
----------
Line 1
Line 2
Line 3
Line 4
----------
Line 1
Line 2
Line 3
Line 4
----------
Some text here after last delimiter
And I would like to get
Line 1
Line 2
Line 3
Line 4
Line 1
Line 2
Line 3
Line 4
How do I do this with awk / sed with regex? Thanks.
You can try this.
file: a.awk:
BEGIN { RS = "-+" }
{
if ( NR > 1 && RT != "" )
{
print $0
}
}
run: awk -f a.awk data_file
If you can comfortably fit the entire file into memory, and if Perl is acceptable instead of awk or sed,
perl -0777 -pe 's/\A.*?\n-{10}\n//s;
s/(.*\n)-{10}\n.*?\Z/\1/s;
s/\n-{10}\n/\n\n\n/g' file >newfile
The main FAQs here are the -0777 option (slurp mode) and the /s (dot matches newlines) regex flag.
This might work for you:
sed '1,/^--*$/d;:a;$!{/\(^\|\n\)--*$/!N;//!ba;s///p};d' file
Most command-line programs just operate on one line at a time.
Can I use a common command-line utility (echo, sed, awk, etc) to concatenate every set of two lines, or would I need to write a script/program from scratch to do this?
$ cat myFile
line 1
line 2
line 3
line 4
$ cat myFile | __somecommand__
line 1line 2
line 3line 4
sed 'N;s/\n/ /;'
Grab next line, and substitute newline character with space.
seq 1 6 | sed 'N;s/\n/ /;'
1 2
3 4
5 6
$ awk 'ORS=(NR%2)?" ":"\n"' file
line 1 line 2
line 3 line 4
$ paste - - < file
line 1 line 2
line 3 line 4
Not a particular command, but this snippet of shell should do the trick:
cat myFile | while read line; do echo -n $line; [ "${i}" ] && echo && i= || i=1 ; done
You can also use Perl as:
$ perl -pe 'chomp;$i++;unless($i%2){$_.="\n"};' < file
line 1line 2
line 3line 4
Here's a shell script version that doesn't need to toggle a flag:
while read line1; do read line2; echo $line1$line2; done < inputfile
So my dear SOers, Let me be direct to the point:
specification: filter a text file using pairs of patterns.
Example: if we have a file:
line 1 blabla
line 2 more blabla
line 3 **PAT1a** blabla
line 4 blabla
line 5 **PAT1b** blabla
line 6 blabla
line 7 **PAT2a** blabla
line 8 blabla
line 9 **PAT2b** blabla
line 10 **PAT3a** blabla
line 11 blabla
line 12 **PAT3b** blabla
more and more blabla
should give:
line 3 **PAT1a** blabla
line 4 blabla
line 5 **PAT1b** blabla
line 7 **PAT2a** blabla
line 8 blabla
line 9 **PAT2b** blabla
line 10 **PAT3a** blabla
line 11 blabla
line 12 **PAT3b** blabla
I know how to filer only one part of it using 'sed':
sed -n -e '/PAT1a/,/PAT1b/{p}'
But how to filter all the snippets, do i need to write those pairs of patterns in a configuration file, read a pair from it, use the sed cmd above, go to next pair...?
Note: Suppose PAT1, PAT2 and PAT3, etc share no common prefix(like 'PAT' in this case)
One thing more: how to make a newline in quota text in this post without leaving a whole blank line?
I assumed the pattern pairs are given as a separate file. Then, when they appear in order in the input, you could use this awk script:
awk 'NR == FNR { a[NR] = $1; b[NR] = $2; next }
!s && $0 ~ a[i+1] { s = 1 }
s
s && $0 ~ b[i+1] { s = 0; i++ }' patterns.txt input.txt
And a more complicated version when the patterns can appear out of order:
awk 'NR == FNR { a[++n] = $1; b[n] = $2; next }
{ for (i = 1; !s && i <= n; i++) if ($0 ~ a[i]) s = i; }
s
s && $0 ~ b[s] { s = 0 }' patterns.txt input.txt
Awk.
$ awk '/[0-9]a/{o=$0;getline;$0=o"\n"$0;print;next}/[0-9]b/' file
line 3 PAT1a blabla
line 4 blabla
line 5 PAT1b blabla
line 7 PAT2a blabla
line 8 blabla
line 9 PAT2b blabla
line 10 PAT3a blabla
line 11 blabla
line 12 PAT3b blabla
Note: Since you said "share no common prefix", then I use the number and [ab] pattern for regex
Use the b command to skip all lines between the patterns and the d command to delete all other lines:
sed -e '/PAT1a/,/PAT1b/b' -e '/PAT2a/,/PAT2b/b' -e '/PAT3a/,/PAT3b/b' -e d