I would like to replace the last column values with another value using vim editor or sed command.I tried the below command but it replaces the data which is already present in my 2nd column sometimes.
:%s/,3/,2/
My sample data:
410339,166,1430,3
410340,112,1840,3
410341,109,1315,3
410342,123,1435,3
410343,230,3200,3
410344,857,36975,3
410345,125,4440,3
410346,105,1460,3
410348,122,1150,3
410349,314,2380,3
410350,132,4650,3
410351,136,7465,3
410352,103,1775,3
410353,101,1095,3
410354,101,1360,3
You'll need to indicate that you want the end-of-line:
:%s/,3$/2/
To complete your attempt in Vim, just anchor the matched expression to the end of the line with $.
:%s/,3$/,2/
Produces:
410339,166,1430,2
410340,112,1840,2
410341,109,1315,2
410342,123,1435,2
410343,230,3200,2
410344,857,36975,2
410345,125,4440,2
410346,105,1460,2
410348,122,1150,2
410349,314,2380,2
410350,132,4650,2
410351,136,7465,2
410352,103,1775,2
410353,101,1095,2
410354,101,1360,2
I would use awk:
awk '{$NF=2}1' FS=, OFS=,
This unconditionally makes the last column have the value 2. If the desired replacement is a string, you will need to use quotes or the more flexible:
awk '{$NF=r}1' FS=, OFS=, r="the string to put in the last column"
You can restrict the replacement to those columns that have the value 3 with:
awk '$NF==3{$NF=r}1' FS=, OFS=, r="the string to put in the last column"
And, to do this in vim, just do:
:%! awk ...
With GNU sed:
If the last column is only ever '3':
sed -i 's/3$/2/' file
If the last column could be something like '13':
sed -i 's/,3$/,2/' file
The -i flag:
-i[SUFFIX], --in-place[=SUFFIX]
edit files in place (makes backup if extension supplied)
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.
This is the string I have:
my_file1.txt-myfile2.txt_my_file3.txt
I want to remove all the characters after the first "_" that follows the first ".txt".
From the above example, I want the output to be my_file1.txt-myfile2.txt. I have to search for first occurrence of ".txt" and continue parsing until I find the underscore character, and remove everything from there on.
Is it possible to do it in sed/awk/cut etc commands?
You can't do this job with cut but you can with sed and awk:
$ sed 's/\.txt/\n/g; s/\([^\n]*\n[^_]*\)_.*/\1/; s/\n/.txt/g' file
my_file1.txt-myfile2.txt
$ awk 'match($0,/\.txt[^_]*_/){print substr($0,1,RSTART+RLENGTH-2)}' file
my_file1.txt-myfile2.txt
Could you please try following, written based on your shown samples.
awk '{sub(/\.txt_.*/,".txt")} 1' Input_file
Simply substituting everything from .txt_ to till last of line with .txt and printing the line here
I have the file teste.txt with the following content:
02183101399205000 GBTD9VBYMBQ 04455927964
02183101409310000 XBQMPL1C93B 27699484827
54183101003651000 1WFG3SNVDG9 71530894204
I execute the command
sed -e 's/^\(.\{18\}\)[0-9]/\1#/g' teste.txt
The result is:
02183101399205000 GBTD9VBYMBQ 04455927964
02183101409310000 XBQMPL1C93B 27699484827
54183101003651000 #WFG3SNVDG9 71530894204
Only the 19th position in line 3 is changed from 1 to #.
I would like to know how can I change all numeric characters from the 19th to the 30th position.
The expected result is:
02183101399205000 GBTD#VBYMBQ 04455927964
02183101409310000 XBQMPL#C##B 27699484827
54183101003651000 #WFG#SNVDG# 71530894204
An awk command to accomplish your goal:
awk '{ gsub(/[0-9]/,"#",$2); print }' teste.txt
This might work for you (GNU sed):
sed -r 's/./&\n/30;s//\n&/19;h;s/[0-9]/#/g;H;x;s/\n.*\n(.*)\n.*\n(.*)\n.*/\2\1/' file
Surround the string, which is from the 19th to the 30th character, by newlines and make a copy. Replace all digits by #'s. Append this string to the original and use pattern matching to rearrange the strings to make a new string with the unchanged parts either side of the changed part, at the same time discarding the introduced newlines.
An alternative method, utilising the fact the the fields are space separated:
sed -r ':a;s/( \S*)[0-9](\S* )/\1#\2/;ta' file
In fact the two methods can be combined:
sed -r 's/./&\n/30;s//\n&/19;:a;s/(\n.*)[0-9](.*\n)/\1#\2/;ta;s/\n//g' file
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 \|?
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.