Each file's name starts with "input". One example of the files look like:
0.0005
lii_bk_new
traj_new.xyz
0
73001
146300
I want to delete the lines which only includes '0' and the expected output is:
0.0005
lii_bk_new
traj_new.xyz
73001
146300
I have tried with
sed -i 's/^0\n//g' input_*
and
grep -RiIl '^0\n' input_* | xargs sed -i 's/^0\n//g'
but neither works.
Please give some suggestions.
Could you please try changing your attempted code to following, run it on a single Input_file once.
sed 's/^0$//' Input_file
OR as per OP's comment to delete null lines:
sed 's/^0$//;/^$/d' Input_file
I have intentionally not put -i option here first test this in a single file of output looks good then only run with -i option on multiple files.
Also problem in your attempt was, you are putting \n in regex of sed which is default separator of line, we need to put $ in it to tell sed delete those lines which starts and ends with 0.
In case you want to take backup of files(considering that you have enough space available in your file system) you could use -i.bak option of sed too which will take backup of each file before editing(this isn't necessary but for safer side you have this option too).
$ sed '/^0$/d' file
0.0005
lii_bk_new
traj_new.xyz
73001
146300
In your regexp you were confusing \n (the literal LineFeed character which will not be present in the string sed is analyzing since sed reads one \n-separated line at a time) with $ (the end-of-string regexp metacharacter which represents end-of-line when the string being parsed is a line as is done with sed by default).
The other mistake in your script was replacing 0 with null in the matching line instead of just deleting the matching line.
Please give some suggestions.
I would use GNU awk -i inplace for that following way:
awk -i inplace '!/^0$/' input_*
This simply will preserve all lines which do not match ^0$ i.e. (start of line)0(end of line). If you want to know more about -i inplace I suggest reading this tutorial.
Related
Let's say I have a line looking like this:
/Users/random/354765478/Tests/StoreTests/Base64Tests.swift
In this example, I would like to get the result:
Tests/StoreTests/Base64Tests.swift
How can I do if I want to get everything before the first pattern match (either Sources or Tests) using sed or awk?
I am using sed 's/^.*\(Tests.*\).*$/\1/' right now but it's falling:
echo '/Users/random/354765478/Tests/StoreTests/Base64Tests.swift' | sed 's/^.*\(Tests\)/\1/'
Tests.swift
Here's another example using Sources (which seems to work):
echo '/Users/random/741672469/Sources/Store/StoreDataSource.swift' | sed 's/^.*\(Sources\)/\1/'
Sources/Store/StoreDataSource.swift
I would like to get everything before the first, and not the last Sources or Tests pattern match.
Any help would be appreciated!
How can I do if I want to get everything before the first pattern match (either Sources or Tests).
Easier to use a grep -o here:
grep -Eo '(Sources|Tests)/.*' file
Tests/StoreTests/Base64Tests.swift
Sources/Store/StoreDataSource.swift
# where input file is
cat file
/Users/random/354765478/Tests/StoreTests/Base64Tests.swift
/Users/random/741672469/Sources/Store/StoreDataSource.swift
Breakdown:
Regex pattern (Sources|Tests)/.* match any text that starts with Sources/ or Tests/ until end of the line.
-E: enables extended regex mode
-o: prints only matched text instead of full line
Alternatively you may use this awk as well:
awk 'match($0, /(Sources|Tests)\/.*/) {
print substr($0, RSTART)
}' file
Tests/StoreTests/Base64Tests.swift
Sources/Store/StoreDataSource.swift
Or this sed:
sed -E 's~.*/((Sources|Tests)/.*)~\1~' file
Tests/StoreTests/Base64Tests.swift
Sources/Store/StoreDataSource.swift
With your shown samples please try following GNU grep. This will look for very first match of /Sources OR /Tests and then print values from these strings to till end of the value.
grep -oP '^.*?\/\K(Sources|Tests)\/.*' Input_file
Using sed
$ sed -E 's~([^/]*/)+((Tests|Sources).*)~\2~' input_file
Tests/StoreTests/Base64Tests.swift
would like to get everything before the first, and not the last
Sources or Tests pattern match.
First thing is to understand reason of that, you are using
sed 's/^.*\(Tests.*\).*$/\1/'
observe that * is greedy, i.e. it will match as much as possible, therefore it will always pick last Tests, if it would be non-greedy it would find first Tests but sed does not support this, if you are using linux there is good chance that you have perl command which does support that, let file.txt content be
/Users/random/354765478/Tests/StoreTests/Base64Tests.swift
then
perl -p -e 's/^.*?(Tests.*)$/\1/' file.txt
gives output
Tests/StoreTests/Base64Tests.swift
Explanation: -p -e means engage sed-like mode, alterations in regular expression made: brackets no longer require escapes, first .* (greedy) changed to .*? (non-greedy), last .* deleted as superfluous (observe that capturing group will always extended to end of line)
(tested in perl 5, version 30, subversion 0)
I have got the following URL:
https://xcg5847#git.rz.bankenit.de/scm/smat/sma-mes-test.git
I need to pull out smat-mes-test and smat:
git config --local remote.origin.url|sed -n 's#.*/\([^.]*\)\.git#\1#p'
sma-mes-test
This works. But I also need the project name, which is smat
I am not really familiar to complex regex and sed, I was able to find the other command in another post here. Does anyone know how I am able to extract the smat value here?
With your shown samples please try following awk code. Simple explanation would be, setting field separator(s) as / and .git for all the lines and in main program printing 3rd last and 3nd last elements from the line.
your_git_command | awk -F'/|\\.git' '{print $(NF-2),$(NF-1)}'
Your sed is pretty close. You can just extend it to capture 2 values and print them:
git config --local remote.origin.url |
sed -E 's~.*/([^/]+)/([^.]+)\.git$~\1 \2~'
smat sma-mes-test
If you want to populate shell variable using these 2 values then use this read command in bash:
read v1 v2 < <(git config --local remote.origin.url |
sed -E 's~.*/([^/]+)/([^.]+)\.git$~\1 \2~')
# check variable values
declare -p v1 v2
declare -- v1="smat"
declare -- v2="sma-mes-test"
Using sed
$ sed -E 's#.*/([^/]*)/#\1 #' input_file
smat sma-mes-test.git
I would harness GNU AWK for this task following way, let file.txt content be
https://xcg5847#git.rz.bankenit.de/scm/smat/sma-mes-test.git
then
awk 'BEGIN{FS="/"}{sub(/\.git$/,"",$NF);print $(NF-1),$NF}' file.txt
gives output
smat sma-mes-test
Explanation: I instruct GNU AWK that field separator is slash character, then I replace .git (observe that . is escaped to mean literal dot) adjacent to end ($) in last field ($NF), then I print 2nd from end field ($(NF-1)) and last field ($NF), which are sheared by space, which is default output field separator, if you wish to use other character for that purpose set OFS (output field separator) in BEGIN. If you want to know more about NF then read 8 Powerful Awk Built-in Variables – FS, OFS, RS, ORS, NR, NF, FILENAME, FNR
(tested in gawk 4.2.1)
Why not sed 's!.*/\(.*/.*\)!\1!'?
string=$(config --local remote.origin.url | tail -c -21)
var1=$(echo "${string}" | cut -d'/' -f1)
var2=$(echo "${string}" | cut -d'/' -f2 | sed s'#\.git##')
If you have multiple urls with variable lengths, this will not work, but if you only have the one, it will.
var1=smat
var2=sma-mes-test.git
If I did have something variable, personally I would replace all of the forward slashes with carriage returns, throw them into a file, and then export the last and second last lines with ed, which would give me the two last segments of the url.
Regular expressions literally give me a migraine headache, but as long as I can get everything on its' own line, I can quite easily bypass the need for them entirely.
I am trying to modify the name of a large number of files, all of them with the following structure:
4.A.1 Introduction to foo.txt
2.C.3 Lectures on bar.pdf
3.D.6 Processes on baz.mp4
5.A.8 History of foo.txt
And I want to add a leading zero to the last digit:
4.A.01 Introduction to foo.txt
2.C.03 Lectures on bar.pdf
3.D.06 Processes on baz.mp4
5.A.08 History of foo.txt
At first I am trying to get the new names with sed (FreeBSD implementation):
ls | sed 's/\.[0-9]/0&/'
But I get the zero before the .
Note: replacing the second dot would also work. I am also open to use awk.
While it may have worked for you here, in general slicing and dicing ls output is fragile, whether using sed or awk or anything else. Fortunately one can accomplish this robustly in plain old POSIX sh using globbing and fancy-pants parameter expansions:
for f in [[:digit:]].[[:alpha:]].[[:digit:]]\ ?*; do
# $f = "[[:digit:]].[[:alpha:]].[[:digit:]] ?*" if no files match.
if [ "$f" != '[[:digit:]].[[:alpha:]].[[:digit:]] ?*' ]; then
tail=${f#*.*.} # filename sans "1.A." prefix
head=${f%"$tail"} # the "1.A." prefix
mv "$f" "${head}0${tail}"
fi
done
(EDIT: Filter out filenames that don't match desired format.)
This pipeline should work for you:
ls | sed 's/\.\([0-9]\)/.0\1/'
The sed command here will capture the digit and replace it with a preceding 0.
Here, \1 references the first (and in this case only) capture group - the parenthesized expression.
I am also open to use awk.
Let file.txt content be:
4.A.1 Introduction to foo.txt
2.C.3 Lectures on bar.pdf
3.D.6 Processes on baz.mp4
5.A.8 History of foo.txt
then
awk 'BEGIN{FS=OFS="."}{$3="0" $3;print}' file.txt
outputs
4.A.01 Introduction to foo.txt
2.C.03 Lectures on bar.pdf
3.D.06 Processes on baz.mp4
5.A.08 History of foo.txt
Explanation: I set dot (.) as both field seperator and output field seperator, then for every line I add leading 0 to third column ($3) by concatenating 0 and said column. Finally I print such altered line.
(tested in GNU Awk 5.0.1)
This might work for you (GNU sed):
sed 's/^\S*\./&0/' file
This appends 0 after the last . in the first string of non-empty characters in each line.
In case it helps somebody else, as an alternative to #costaparas answer:
ls | sed -E -e 's/^([0-9][.][A-Z][.])/\10/' > files
To then create the script the files:
cat files | awk '{printf "mv \"%s\" \"%s\"\n", $0, $0}' | sed 's/\.0/\./' > movefiles.sh
I have a file which has the following content:
10 tiny toes
tree
this is that tree
5 funny 0
There are spaces at the end of the file. I want to get the line number of the last row of a file (that has characters). How do I do that in SED?
This is easily done with awk,
awk 'NF{c=FNR}END{print c}' file
With sed it is more tricky. You can use the = operator but this will print the line-number to standard out and not in the pattern space. So you cannot manipulate it. If you want to use sed, you'll have to pipe it to another or use tail:
sed -n '/^[[:blank:]]*$/!=' file | tail -1
You can use following pseudo-code:
Replace all spaces by empty string
Remove all <beginning_of_line><end_of_line> (the lines, only containing spaces, will be removed like this)
Count the number of remaining lines in your file
It's tough to count line numbers in sed. Some versions of sed give you the = operator, but it's not standard. You could use an external tool to generate line numbers and do something like:
nl -s ' ' -n ln -ba input | sed -n 's/^\(......\)...*/\1/p' | sed -n '$p'
but if you're going to do that you might as well just use awk.
This might work for you (GNU sed):
sed -n '/\S/=' file | sed -n '$p'
For all lines that contain a non white space character, print a line number. Pipe this output to second invocation of sed and print only the last line.
Alternative:
grep -n '\S' file | sed -n '$s/:.*//p'
In the following file I want to replace all the ; by , with the exception that, when there is a string (delimited with two "), it should not replace the ; inside it.
Example:
Input
A;B;C;D
5cc0714b9b69581f14f6427f;5cc0714b9b69581f14f6428e;1;"5cc0714b9b69581f14f6427f;16a4fba8d13";xpto;
5cc0723b9b69581f14f64285;5cc0723b9b69581f14f64294;2;"5cc0723b9b69581f14f64285;16a4fbe3855";xpto;
5cc072579b69581f14f6428a;5cc072579b69581f14f64299;3;"5cc072579b69581f14f6428a;16a4fbea632";xpto;
output
A,B,C,D
5cc0714b9b69581f14f6427f,5cc0714b9b69581f14f6428e,1,"5cc0714b9b69581f14f6427f;16a4fba8d13",xpto,
5cc0723b9b69581f14f64285,5cc0723b9b69581f14f64294,2,"5cc0723b9b69581f14f64285;16a4fbe3855",xpto,
5cc072579b69581f14f6428a,5cc072579b69581f14f64299,3,"5cc072579b69581f14f6428a;16a4fbea632",xpto,
For sed I have: sed 's/;/,/g' input.txt > output.txt but this would replace everything.
The regex for the " delimited string: \".*;.*\" .
(A regex for hexadecimal would be better -- something like: [0-9a-fA-F]+)
My problem is combining it all to make a grep -o / sed that replaces everything except for that pattern.
The file size is in the order of two digit Gb (max 99Gb), so performance is important. Relevant.
Any ideas are appreciated.
sed is for doing simple s/old/new on individual strings. grep is for doing g/re/p. You're not trying to do either of those tasks so you shouldn't be considering either of those tools. That leaves the other standard UNIX tool for manipulating text - awk.
You have a ;-separated CSV that you want to make ,-separated. That's simply:
$ awk -v FPAT='[^;]*|"[^"]+"' -v OFS=',' '{$1=$1}1' file
A,B,C,D
5cc0714b9b69581f14f6427f,5cc0714b9b69581f14f6428e,1,"5cc0714b9b69581f14f6427f;16a4fba8d13",xpto,
5cc0723b9b69581f14f64285,5cc0723b9b69581f14f64294,2,"5cc0723b9b69581f14f64285;16a4fbe3855",xpto,
5cc072579b69581f14f6428a,5cc072579b69581f14f64299,3,"5cc072579b69581f14f6428a;16a4fbea632",xpto,
The above uses GNU awk for FPAT. See What's the most robust way to efficiently parse CSV using awk? for more details on parsing CSVs with awk.
If I get correctly your requirements, one option would be to make a three pass thing.
From your comment about hex, I'll consider nothing like # will come in the input so you can do (using GNU sed) :
sed -E 's/("[^"]+);([^"]+")/\1#\2/g' original > transformed
sed -i 's/;/,/g' transformed
sed -i 's/#/;/g' transformed
The idea being to replace the ; when within quotes by something else and write it to a new file, then replace all ; by , and then set back the ; in place within the same file (-i flag of sed).
The three pass can be combined in a single command with
sed -E 's/("[^"]+);([^"]+")/\1#\2/g;s/;/,/g;s/#/;/g' original > transformed
That said, there's probably a bunch of csv parser witch already handle quoted fields that you can probably use in the final use case as I bet this is just an intermediary step for something else later in the chain.
From Ed Morton's comment: if you do it in one pass, you can use \n as replacement separator as there can't be a newline in the text considered line by line.
This might work for you (GNU sed):
sed -E ':a;s/^([^"]*("[^"]*"[^"]*)*"[^";]*);/\1\n/;ta;y/;/,/;y/\n/;/' file
Replace ;'s inside double quotes with newlines, transpose ;'s to ,'s and then transpose newlines to ;'s.