I need to change Finnish-Czech dictionary into the Czech-Finnish dictionary.
I tried this command:
sed -ne 's/\([^a-z A-Z]*\) \(.*\)$/\2 \1/ p' finnish-czech.txt
But the first back-reference doesn't work. I realized the end of the back-reference is false and instead of taking only first column it takes everything.
The seperator is <TAB>:
sed -r 's/^([^\t]*)\t([^\t]*)$/\2\t\1/p' finnish-czech.txt
Finnish field match( ^([^\t]*)) then TAB(\t) then czech filed match (([^\t]*)$
This is a simple job for awk:
awk '{ print $2 "\t" $1; }' <finnish-czech.txt
For each line, this prints the second field, then a tab, then the first field.
One possible complication is that your file seems to have carriage-returns preceding the newlines - you will probably want to remove them with tr -d '\r' or similar.
Related
I have a text file in the below format . I need to remove the text between the first and second semicolon (delimiter ), but retain the second semicolon
$cat test.txt
abc;def;ghi;jkl
mno;pqr;stu,xxx
My expected output
abc;ghi;jkl
mno;stu,xxx
I tried using sed 's/^([^;][^;]*);.*$/\1/', but it removes everything after the first semicolon. I also tried with cut -d ';' -f2, this only give the 2nd field as output.
Using cut
cut -d";" -f2 --complement file
-d is for delimeter, i.e ";" in your case
-f is for field, i.e keep the fields listed
--complement is to reverse the selection, i.e remove the fields listed
So:
$ cat test.txt
abc;def;ghi;jkl
mno;pqr;stu;xxx
$ cut -d";" -f2 --complement test.txt
abc;ghi;jkl
mno;stu;xxx
You may use this sed:
sed 's/;[^;]*//' file
abc;ghi;jkl
mno;stu,xxx
You can do it directly by simply removing the 2nd occurrence of the characters in question, e.g.
sed 's/[^;]*;//2' test.txt
Example Use/Output
$ sed 's/[^;]*;//2' test.txt
abc;ghi;jkl
mno;stu,xxx
A thanks to #EdMorton for improvements here as well.
If you did want to use awk, you could simply replace the 2nd field with nothing as well, e.g.
awk -F';' '{sub(/;[^;]*/,"")}1' test.txt
(same output)
With a thanks to #EdMorton for the improvement to the original.
Or as Cyrus suggest with cut, deleting field 2, e.g.
cut -d';' -f-1,3- test.txt
(same output)
Trying to fix OP's attempts here, with sed you could try following code. Simple explanation would be, create 1st back reference which has value till 1st occurrence of ; then from 1st ; to 2nd ; don't keep it in backreference and keep rest of the value in 2nd back reference. Finally while substituting substitute it with 1st and 2nd back reference values.
sed -E 's/^([^;]*);[^;]*;(.*)/\1;\2/' Input_file
OR as per Ed's comment please try following;
sed -E 's/^([^;]*);[^;]*/\1/' Input_file
super lazy awk solution
gawk/mawk/mawk2 'sub(/;[^;]+/,"")'
a more verbose solution but makes it clearer what it's doing
g/mawk 'BEGIN {FS=";+"; OFS=";"} ($2="")||($0=$0)&&($1=$1)'
clean out 2nd field, but since null string is assigned in, it returns 0 (false), thus requiring logical or || to continue.
$0=$0 plus $1=$1 to clean up extra ;, which will also print it.
Suppose i have | delimeted file,
Line1: 1|2|3|4
Line2: 5|6|7|8
Line3: 9|9|1|0
Now i need to read 3 field at second line which is 7 in above example how i can do that using Cut or Sed Command. I'm new to unix please help
A job for awk:
awk -F '|' 'NR==2{print $3}' file
or
awk -F '|' -v row=2 -v col=3 'NR==row{print $col}' file
Output:
7
This should work:
sed -n '2p' file |awk -F '|' '{print $3}'
This might work for you (GNU sed):
sed -rn '2s/^(([^|]*)\|?){3}.*/\2/p' file
Turn off automatic printing by setting the -n option, turn on easier regexp declaration by -r option. Use pattern matching and back references to replace the whole of the second line by the third field of the same line and print the result.
The address of the substitution command is limited to only the second line.
The regexp groups the non-delimited characters followed by a delimiter a specific number of times. The second group, only retains the non-delimited characters for the specific number. Each grouping is replaced by the next and so the last grouping is reported, the .* consumes the remainder of the line and so only the third field (contents of second group) is printed.
N.B. the delimiter would be present following the final column and is therefore optional \|?
I have a file that, occasionally, has split lines. The split is signaled by the fact that the line starts with a space, empty line or a nonnumeric character. E.g.
40403813|7|Failed|No such file or directory|1
40403816|7|Hi,
The Conversion System could not be reached.|No such file or directory||1
40403818|7|Failed|No such file or directory|1
...
I'd like join the split line back with the previous line (as mentioned below):
40403813|7|Failed|No such file or directory|1
40403816|7|Hi, The Conversion System could not be reached.|No such file or directory||1
40403818|7|Failed|No such file or directory|1
...
using a Unix command like sed/awk. I'm not clear how to join a line with the preceeding one.
Any suggestion?
awk to the rescue!
awk -v ORS='' 'NR>1 && /^[0-9]/{print "\n"} NF' file
only print newline when the current line starts with a digit, otherwise append rows (perhaps you may want to add a space to ORS if the line break didn't preserve the space).
Don't do anything based on the values of the strings in your fields as that could go wrong. You COULD get a wrapping line that starts with a digit, for example. Instead just print after every complete record of 5 fields:
$ awk -F'|' '{rec=rec $0; nf+=NF} nf>=5{print rec; nf=0; rec=""}' file
40403813|7|Failed|No such file or directory|1
40403816|7|Hi, The Conversion System could not be reached.|No such file or directory||1
40403818|7|Failed|No such file or directory|1
Try:
awk 'NF{printf("%s",$0 ~ /^[0-9]/ && NR>1?RS $0:$0)} END{print ""}' Input_file
OR
awk 'NF{printf("%s",/^[0-9]/ && NR>1?RS $0:$0)} END{print ""}' Input_file
It will check if each line starts from a digit or not if yes and greater than line number 1 than it will insert a new line with-it else it will simply print it, also it will print a new line after reading the whole file, if we not mention it, it is not going to insert that at end of the file reading.
If you only ever have the line split into two, you can use this sed command:
sed 'N;s/\n\([^[:digit:]]\)/\1/;P;D' infile
This appends the next line to the pattern space, checks if the linebreak is followed by something other than a digit, and if so, removes the linebreak, prints the pattern space up to the first linebreak, then deletes the printed part.
If a single line can be broken across more than two lines, we have to loop over the substitution:
sed ':a;N;s/\n\([^[:digit:]]\)/\1/;ta;P;D' infile
This branches from ta to :a if a substitution took place.
To use with Mac OS sed, the label and branching command must be separate from the rest of the command:
sed -e ':a' -e 'N;s/\n\([^[:digit:]]\)/\1/;ta' -e 'P;D' infile
If the continuation lines always begin with a single space:
perl -0000 -lape 's/\n / /g' input
If the continuation lines can begin with an arbitrary amount of whitespace:
perl -0000 -lape 's/\n(\s+)/$1/g' input
It is probably more idiomatic to write:
perl -0777 -ape 's/\n / /g' input
You can use sed when you have a file without \r :
tr "\n" "\r" < inputfile | sed 's/\r\([^0-9]\)/\1/g' | tr '\r' '\n'
How to use "grep" shell command to show specific word from a line starting with a specific word.
Ex:
I want to print a string "myFTPpath/folderName/" from the line starting with searchStr in the below mentioned line.
searchStr:somestring:myFTPpath/folderName/:somestring
Something like this with awk:
awk -F: '/^searchStr/{print $3}' File
From all the lines starting with searchStr, print the 3rd field (field seperator set as :)
Sample:
AMD$ cat File
someStr:somestring:myFTPpath/folderName/:somestring
someStr:somestring:myFTPpath/folderName/:somestring
searchStr:somestring:myFTPpath/folderName/:somestring
someStr:somestring:myFTPpath/folderName/:somestring
AMD$ awk -F: '/^searchStr/{print $3}' File
myFTPpath/folderName/
Remember that grep isn't the only tool that can usefully do searches.
In this particular case, where the lines are naturally broken into fields, awk is probably the best solution, as #A.M.D's answer suggests.
For more general case edits, however, remember sed's -n option, which suppresses printing out a line after edits:
sed -n 's/searchStr:[^:]*:\([^:]*\):.*/\1/p' input-file
The -n suppresses automatic printing of the line, and the trailing /p flag explicitly prints out lines on which there is a substitution.
This matching pattern is fiddly – use awk in this fielded case – but don't forget sed -n.
You could get the desired output with grep itself but you need to enable -P and -o parameters.
$ echo 'searchStr:somestring:myFTPpath/folderName/:somestring' | grep -oP '^searchStr:[^:]*:\K[^:]*'
myFTPpath/folderName/
\K discards the characters which are matched previously from printing at the final leaving only the characters which are matched by the pattern exists next to \K. Here we used \K instead of a variable length positive lookbehind assertion.
I have a text file ("INPUT.txt") of the format:
A<LF>
B<LF>
C<LF>
D<LF>
X<LF>
Y<LF>
Z<LF>
<EOF>
which I need to reformat to:
A:B:C:D:X:Y:Z<LF>
<EOF>
I know you can do this with 'sed'. There's a billion google hits for doing this with 'sed'. But I'm trying to emphasis readability, simplicity, and using the correct tool for the correct job. 'sed' is a line editor that consumes and hides newlines. Probably not the right tool for this job!
I think the correct tool for this job would be 'tr'. I can replace all the newlines with colons with the command:
cat INPUT.txt | tr '\n' ':'
There's 99% of my work done. I have a problem, now, though. By replacing all the newlines with colons, I not only get an extraneous colon at the end of the sequence, but I also lose the carriage return at the end of the input. It looks like this:
A:B:C:D:X:Y:Z:<EOF>
Now, I need to remove the colon from the end of the input. However, if I attempt to pass this processed input through 'sed' to remove the final colon (which would now, I think, be a proper use of 'sed'), I find myself with a second problem. The input is no longer terminated by a newline at all! 'sed' fails outright, for all commands, because it never finds the end of the first line of input!
It seems like appending a newline to the end of some input is a very, very common task, and considering I myself was just sorely tempted to write a program to do it in C (which would take about eight lines of code), I can't imagine there's not already a very simple way to do this with the tools already available to you in the Linux kernel.
This should do the job (cat and echo are unnecessary):
tr '\n' ':' < INPUT.TXT | sed 's/:$/\n/'
Using only sed:
sed -n ':a; $ ! {N;ba}; s/\n/:/g;p' INPUT.TXT
Bash without any externals:
string=($(<INPUT.TXT))
string=${string[#]/%/:}
string=${string//: /:}
string=${string%*:}
Using a loop in sh:
colon=''
while read -r line
do
string=$string$colon$line
colon=':'
done < INPUT.TXT
Using AWK:
awk '{a=a colon $0; colon=":"} END {print a}' INPUT.TXT
Or:
awk '{printf colon $0; colon=":"} END {printf "\n" }' INPUT.TXT
Edit:
Here's another way in pure Bash:
string=($(<INPUT.TXT))
saveIFS=$IFS
IFS=':'
newstring="${string[*]}"
IFS=$saveIFS
Edit 2:
Here's yet another way which does use echo:
echo "$(tr '\n' ':' < INPUT.TXT | head -c -1)"
Old question, but
paste -sd: INPUT.txt
Here's yet another solution: (assumes a character set where ':' is
octal 72, eg ascii)
perl -l72 -pe '$\="\n" if eof' INPUT.TXT