I find many ways to do this, AWK, SED, UNIQ, but none of them are working on my file.
I want to delete duplicate lines. Here is an example of part of my file:
KTBX
KFSO
KCLK
KTBX
KFSO
KCLK
PAJZ
PAJZ
NOTE: I had to manually add line feeds when I cut and pasted from the file...for some reason it was putting all the variables on one line. Makes me think that my 44,000 line text file actually has only "1" line? Is there a way to modify it so I can delete dups?
You can see all non-printed characters with this command:
od -c oldfile
If all your records are on one line, you can use sed to replace a whitespace (space, tab, newline) with a linebreak:
sed -e 's/\s\+/\n/g' oldfile > oldfile.1
Once you have multiple lines, this awk one-liner:
awk '!x[$0]++' oldfile.1 > newfile
my outfile:
KTBX
KFSO
KCLK
PAJZ
Perl One-Liner:
perl -nle 'unless($hash{$_}++){print $_}' file
Related
I am cleaning up a dataset (csv dataset). I only want to consider registers in which all fields are complete and have the right type of values. This is what I tried:
sed -r '{
/regex_pattern/!d
more commands follow...
}' $1
The program works just fine and does what it is supposed to do. The problem is that it also removes the very first line (header line) since it does not match the specific regex_pattern. I know there is a way to specify the range in which the command should apply so for example:
sed '2,$ s/A/a/'
will do substitutions on data skipping the header line. Based on this logic I tried:
sed -r '{
2,$/regex_pattern/!d
more commands follow...
}' $1
so that the header line will be untouched however this code does not run at all.So what (and why) would be the right command to do what I am intending?
As an example, imagine my csv file is fruits.csv and that my regex_pattern is [0-9]+,[0-9]+
apples,oranges
20,5
7,3
,4
a,b
12,22
When I call the .sh script that contains the sed commands in should output:
apples,oranges
20,5
7,3
12,22
So, note that:
Header line was not deleted even though it does not match the regex_pattern.
Line number 4, i.e. ",4" was deleted as it does not match the regex_pattern.
Line number 5, i.e. "a,b" was deleted as it does not match the regex_pattern.
Any help is very much appreciated and I wish to thank you all in advance.
Kind regards.
You could write it like this, matching the whole line, starting at the second line:
sed -r '
2,${/^[0-9]+,[0-9]+$/!d}
' file
Output
apples,oranges
20,5
7,3
12,22
If you also want to allow single numbers or more than just 2 comma separated numbers:
sed -r '
2,${/^[0-9]+(,[0-9]+)*$/!d}
' file
Using sed
$ sed '2,${/[0-9]\+,[0-9]\+/!d}' input_file
apples,oranges
20,5
7,3
12,22
any one of these should work in gawk, mawk1/2, or macos nawk
mawk 'NF-_^(NF==NR)' FS='^[0-9]+,[0-9]+$'
nawk '(NF!=NR)!=NF' FS='^[0-9]+,[0-9]+$'
gawk 'NF-(NF!~NR)' FS='^[0-9]+,[0-9]+$'
'
apples,oranges
20,5
7,3
12,22
more concisely would be
mawk -F'[0-9]+,[0-9]+' '(NF<NR)-NF' # using FS
gawk '/[0-9]+,[0-9]+/^+(NF<NR)' # not using FS
nawk '(NF<NR)<=/([0-9]+,?){2}/' # same approach, rev. order
mawk '(NF~NR)-/[0-9]+,[0-9]+/' # truly fringe but
# concise syntax
nawk '(NF~NR)!=/([0-9]+,?){2}/' # same approach, to
# circumvent nawk peculiarities
sed is a bad choice for working with CSVs since it doesn't have any inbuilt functionality for working with fields, nor literal strings, nor variables, doesn't use EREs by default (all of the answers you have so far will only work with GNU sed), etc. To do what you specifically want with any awk in any shell on every Unix box is simply:
$ awk 'NR==1 || /[0-9]+,[0-9]+/' file
apples,oranges
20,5
7,3
12,22
which says "if the current line number (stored in NR) is 1 or the regexp matches the current line contents then print the line". Anything else you want to do with your CSV will also be easier with awk than with sed.
Meh, I would just preserve first line.
sed -r '
1{p;d}
/regex_pattern/!d
more commands follow...
' "$1"
or run it not for first line:
1!{
/regex_pattern/!d
more commands follow...
}
This might work for you (GNU sed):
sed -E '1!{/^[0-9]+,[0-9]+$/!d}' file
If it is not the first line, delete any line that does not match one set of comma separated natural numbers.
Alternative:
sed -E '1b;/^[0-9]+,[0-9]+$/!d' file
Or:
sed -nE '1p;1b;/^[0-9]+,[0-9]+$/p' file
27211;18:05:03479;20161025;0;0;0;0;10991;0;10991;000;0;0;000;1000000;0;0;000;0;0;0;82
Second string after ; is time. gg:mm:sssss:. I just want to be gg:mm:ss:
Like so:
27211;18:05:03;20161025;0;0;0;0;10991;0;10991;000;0;0;000;1000000;0;0;000;0;0;0;82
I tried with cut but it deletes everything after n'th occurance of character, and for now I am stuck, please help.
give this one liner a try:
awk -F';' -v OFS=";" 'sub(/...$/,"",$2)+1' file
It removes the last 3 chars from column 2.
update with sed one liner
If you are a fan of sed:
sed -r 's/(;[^;]*)...;/\1;/' file
With sed:
sed -r 's/^([^;]+;[^;]+)...;/\1;/' file
(Or)
sed -r 's/^([^;]+;[0-9]{2}:[0-9]{2}:[0-9]{2})...;/\1;/' file
It also can be something like sed 's/(.*)([0-9]{2}\:){2}([0-9]{3})[0-9]*\;(.*)/\1\2\3\4/g'
It is not very clean, but at least is more clear for me.
Regards
I'd use perl for this:
perl -pe 's/(?<=:\d\d)\d+(?=;)//' file
That removes any digits between "colon-digit-digit" and the semicolon (first match only, not globally in the line).
If you want to edit the file in-place: perl -i -pe ...
With sed:
sed -E 's/(:[0-9]{2})[0-9]{3}/\1/' file
or perl:
perl -pe's/:\d\d\B\K...//' file
I have a file in which some lines start by a >
For these lines, and only these ones, I want to keep the first eleven characters.
How can I do that using sed ?
Or maybe something else is better ?
Thanks !
Muriel
Let's start with this test file:
$ cat file
line one with something or other
>1234567890abc
other line in file
To keep only the first 11 characters of lines starting with > while keeping all other lines:
$ sed -r '/^>/ s/(.{11}).*/\1/' file
line one with something or other
>1234567890
other line in file
To keep only the first eleven characters of lines starting with > and deleting all other lines:
$ sed -rn '/^>/ s/(.{11}).*/\1/p' file
>1234567890
The above was tested with GNU sed. For BSD sed, replace the -r option with -E.
Explanation:
/^>/ is a condition. It means that the command which follows only applies to lines that start with >
s/(.{11}).*/\1/ is a substitution command. It replaces the whole line with just the first eleven characters.
-r turns on extended regular expression format, eliminating the need for some escape characters.
-n turns off automatic printing. With -n in effect, lines are only printed if we explicitly ask them to be printed. In the second case above, that is done by adding a p after the substitute command.
Other forms:
$ sed -r 's/(>.{10}).*/\1/' file
line one with something or other
>1234567890
other line in file
And:
$ sed -rn 's/(>.{10}).*/\1/p' file
>1234567890
I'm on Linux command line and I have file with
127.0.0.1
128.0.0.0
121.121.33.111
I want
127.0.0.1:80
128.0.0.0:80
121.121.33.111:80
I remember my colleagues were using sed for that, but after reading sed manual still not clear how to do it on command line?
You could try using something like:
sed -n 's/$/:80/' ips.txt > new-ips.txt
Provided that your file format is just as you have described in your question.
The s/// substitution command matches (finds) the end of each line in your file (using the $ character) and then appends (replaces) the :80 to the end of each line. The ips.txt file is your input file... and new-ips.txt is your newly-created file (the final result of your changes.)
Also, if you have a list of IP numbers that happen to have port numbers attached already, (as noted by Vlad and as given by aragaer,) you could try using something like:
sed '/:[0-9]*$/ ! s/$/:80/' ips.txt > new-ips.txt
So, for example, if your input file looked something like this (note the :80):
127.0.0.1
128.0.0.0:80
121.121.33.111
The final result would look something like this:
127.0.0.1:80
128.0.0.0:80
121.121.33.111:80
Concise version of the sed command:
sed -i s/$/:80/ file.txt
Explanation:
sed stream editor
-i in-place (edit file in place)
s substitution command
/replacement_from_reg_exp/replacement_to_text/ statement
$ matches the end of line (replacement_from_reg_exp)
:80 text you want to add at the end of every line (replacement_to_text)
file.txt the file name
How can this be achieved without modifying the original file?
If you want to leave the original file unchanged and have the results in another file, then give up -i option and add the redirection (>) to another file:
sed s/$/:80/ file.txt > another_file.txt
sed 's/.*/&:80/' abcd.txt >abcde.txt
If you'd like to add text at the end of each line in-place (in the same file), you can use -i parameter, for example:
sed -i'.bak' 's/$/:80/' foo.txt
However -i option is non-standard Unix extension and may not be available on all operating systems.
So you can consider using ex (which is equivalent to vi -e/vim -e):
ex +"%s/$/:80/g" -cwq foo.txt
which will add :80 to each line, but sometimes it can append it to blank lines.
So better method is to check if the line actually contain any number, and then append it, for example:
ex +"g/[0-9]/s/$/:80/g" -cwq foo.txt
If the file has more complex format, consider using proper regex, instead of [0-9].
You can also achieve this using the backreference technique
sed -i.bak 's/\(.*\)/\1:80/' foo.txt
You can also use with awk like this
awk '{print $0":80"}' foo.txt > tmp && mv tmp foo.txt
Using a text editor, check for ^M (control-M, or carriage return) at the end of each line. You will need to remove them first, then append the additional text at the end of the line.
sed -i 's|^M||g' ips.txt
sed -i 's|$|:80|g' ips.txt
sed -i 's/$/,/g' foo.txt
I do this quite often to add a comma to the end of an output so I can just easily copy and paste it into a Python(or your fav lang) array
how to remove comment lines (as # bal bla ) and empty lines (lines without charecters) from file with one sed command?
THX
lidia
If you're worried about starting two sed processes in a pipeline for performance reasons, you probably shouldn't be, it's still very efficient. But based on your comment that you want to do in-place editing, you can still do that with distinct commands (sed commands rather than invocations of sed itself).
You can either use multiple -e arguments or separate commands with a semicolon, something like (just one of these, not both):
sed -i 's/#.*$//' -e '/^$/d' fileName
sed -i 's/#.*$//;/^$/d' fileName
The following transcript shows this in action:
pax> printf 'Line # with a comment\n\n# Line with only a comment\n' >file
pax> cat file
Line # with a comment
# Line with only a comment
pax> cp file filex ; sed -i 's/#.*$//;/^$/d' filex ; cat filex
Line
pax> cp file filex ; sed -i -e 's/#.*$//' -e '/^$/d' filex ; cat filex
Line
Note how the file is modified in-place even with two -e options. You can see that both commands are executed on each line. The line with a comment first has the comment removed then all is removed because it's empty.
In addition, the original empty line is also removed.
#paxdiablo has a good answer but it can be improved.
(1) The '/^$/d' clause only matches 100% blank lines.
If you want to also match lines that are entirely whitespace (spaces, tabs etc.) use this instead:
'/^\s*$/d'
(2) The 's/#.*$//' clause only matches lines that start with the # character in column 0.
If you want to also match lines that have only whitespace before the first # use this instead:
'/^\s*#.*$/d'
The above criteria may not be universal (e.g. within a HEREDOC block, or in a Python multi-line string the different approaches could be significant), but in many cases the conventional definition of "blank" lines include whitespace-only, and "comment" lines include whitespace-then-#.
(3) Lastly, on OSX at least, the #paxdiablo solution in which the first clause turns comment lines into blank lines, and the second clause strips blank lines (including what were originally comments) doesn't work. It seems to be more portable to make both clauses /d delete actions as I've done.
The revised command incorporating the above is:
sed -e '/^\s*#.*$/d' -e '/^\s*$/d' inputFile
This tiny jewel removes all # comments, no matter where they begin in a line (see caution below):
sed -e 's/\s*#.*$//'
Example:
text="
this is a # test
#this is a test
#this is a #test
this is # another #test
"
$echo "$text" | sed -e 's/\s*#.*$//'
this is a
this is
Next this removes any resulting blank lines:
$echo "$text" | sed -e 's/\s*#.*$//' | sed -e '/^\s*$/d'
Caution: Depending on the syntax and/or interpretation of the lines your processing, this might not be an appropriate solution, as it just stupidly removes end of lines, even if the '#' is part of your data or code. However, for use cases where you'll never use a hash except for as an end of line comment then it works fine. So just as with all coding, context must be taken into consideration.
Alternative variant, using grep:
cat file.txt | grep -Ev '(#.*$)|(^$)'
you can use awk
awk 'NF{gsub(/^[ \t]*#/,"");print}' file
First example(paxdiablo) is very good except its not change file, just output result. If you want to change it inline:
sudo sed -i 's/#.*$//;/^$/d' inputFile
On (one of) my linux boxes, sed understands extended regular expressions with the -r option, so:
sed -r '/(^\s*#)|(^\s*$)/d' squid.conf.installed
is very useful for showing all non-blank, non comment lines.
The regex matches either start of line followed by zero or more spaces or tabs followed by either a hash or end of line, and deletes those matching lines from the input.