I have following File (wishlist.txt):
Alligatoah Musik_ist_keine_lösung;https:///uhfhf
Alligatoah STRW;https:///uhfhf?i
Amewu Entwicklungshilfe;https:///uhfhf?i
and want to have the first word of line n.
so for n = 1:
Alligatoah
What i have so far is:
sed -e 's/\s.*//g' wishlist.txt
is there a elegant way to get rid of all lines except n?
Edit:
How to pass a bash variable "$i" to sed since
sed -n '$is/ .*//p' $wishlist
and
sed -n "\`${i}\`s/ .*//p" $wishlist
doesn't work
A couple of other techniques to get the first word of the 3rd line:
awk -v line=3 'NR == line {print $1; exit}' file
or
head -n 3 file | tail -n 1 | cut -d ' ' -f 1
Something like this. For the 1st word of the 3rd line.
sed -n '3s/\s.*//p' wishlist.txt
To use a variable: Note: Double quotes.
line=3; sed -n "${line}s/\s.*//p" wishlist.txt
sed supports "addresses", so you can tell it what lines to operate on. To print only the first line, you can use
sed -e '1!d; s/\s.*//'
where 1!d means: on lines other then 1, delete the line.
Related
I am having trouble saving variables from file using grep/sed/awk.
The text in file.txt is on the form:
NUM_ITER = 1000 # Number of iterations
NUM_STEP = 1000
And I would like to save these to bash variables without the comments.
So far, I have attempted this:
grep -oP "^NUM_ITER[ ]*=\K.*#" file.txt
which yields
1000 #
Any suggestions?
I would use awk, like this:
awk -F'[=[:blank:]#]+' '$1 == "NUM_ITER" {print $2}' file
To store it in a variable:
NUM_ITER=$(awk -F'[=[:blank:]#]+' '$1 == "NUM_ITER" {print $2}' file)
As long as a line can only contain a single match, this is easy with sed.
sed -n '# Remove comments
s/[ ]*#.*//
# If keyword found, remove keyword and print value
s/^NUM_ITER[ ]*=[ ]*//p' file.txt
This can be trimmed down to a one-liner if you remove the comments.
sed -n 's/[ ]*#.*//;s/^NUM_ITER[ ]*=[ ]*//p' file.txt
The -n option turns off printing, and the /p flag after the final substitution says to print that line after all only if the substitution was successful.
I have several fasta files with the following headers:
M01498:408:000000000-BLBYD:1:1101:11790:1823 1:N:0:1
I want to remove all symbols (colon, dash, and space), and add "barcodelabel=FILENAME;"
I can do it for one file using:
cat A1.fasta |sed s/-//g | sed s/://g| sed s/\ //g|sed 's/^>/>barcodelabel=A1;/g' >A1.renamed.fasta
How can I do this but for all of my files at once? I tried the code below but it didn't work:
for i in {A..H}{1..6}; do cat ${i}.fasta |sed s/-//g | sed s/://g| sed s/\ //g | sed 's/^>/>barcodelabel=${i};/g' >${i}.named.fasta; done
any help would be appreciated !
Considering that you want to substitute -,: or space with null and want to add string at last of the first line then following may help you on same:
awk 'FNR==1{gsub(/:|-| +/,"");print $0,"barcodelabel=FILENAME";next} 1' Input_file
In case you want to save output in to same Input_file then add following in above code too > temp_file && mv temp_file Input_file
I figured it out. First, I reduced the number of sed to simplify the code. The mistake was in the final sed I had simple quotation marks and it should have been double so it can read the ${i}. final code is:
for i in {A..H}{1..6}; do cat ${i}.fasta |
sed 's/[-: ]//g' |
sed "s/^>/>barcodelabel=${i};/g" > ${i}.final4.fasta; done
I have this text file:
some text A=10 some text
some more text A more text
some other text A=30 other text
I'm trying to use sed to capture only the numeric value of A. Using this
cat textfile | sed -r 's/.*A=(\S+).*/\1/'
I get:
10
some more text A more text
30
But what i really need is:
10
0
30
If the string A= does not exist output a 0. How can I accomplish this?
I cannot think on a one-liner, so this is my approach:
while read line
do
grep -Po '(?<=A=)\d+' <<< "$line" || echo "0"
done < file
I am using the look-behind grep to get any number after A=. In case there is none, the || (else) will print a 0.
I love code-golf!
sed -e 's/^/A=0 /; s/.*\<A=\(\d\+\).*/\1/'
This prepends A=0 to the line before substituting.
try this one-liner:
awk -F'A=' 'NF==1{print "0";next}{sub(/ .*/,"",$2);print $2}' file
with your data:
kent$ echo "some text A=10 some text
some more text A more text
some other text A=30 other text"|awk -F'A=' 'NF==1{print "0";next}{sub(/.*/,"",$2);print $2}'
10
0
30
gawk
awk '{$0=gensub(/^.*A=?([[:digit:]]+).*$/, "\\1", "g"); print($0+0)}' file.txt
This might work for you (GNU sed):
sed '/.*A=\([0-9][0-9]*\).*/s//\1/;t;s/.*/0/' file
Look for the string A= followed by one or more numbers and if it occurs replace the whole line by the back reference. Otherwise replace the whole of the line by 0.
I think the best way is to do two different commands - the first replaces lines without 'A=' with the line 'A=0', the second does what you did.
So
cat textfile | sed -r 's/^([^A]|A[^=)*$/A=0/' | sed -r 's/.*A=(\S+).*/\1/'
How about:
sed -r -e 's/.*A=(\S+).*/\1/' -e 's/.*A.*/0/'
Some grep-sed-cut combination:
grep -o 'A=\?[0-9]*' input | sed 's/A$/A=0/' | cut -d= -f2
Produces:
10
0
30
Is there a unix one liner to do this?
head -n 3 test.txt > out_dir/test.head.txt
grep hello test.txt > out_dir/test.tmp.txt
cat out_dir/test.head.txt out_dir/test.tmp.txt > out_dir/test.hello.txt
rm out_dir/test.head.txt out_dir/test.tmp.txt
I.e., I want to get the header and some grep lines from a given file, simultaneously.
Use awk:
awk 'NR<=3 || /hello/' test.txt > out_dir/test.hello.txt
You can say:
{ head -n 3 test.txt ; grep hello test.txt ; } > out_dir/test.hello.txt
Try using sed
sed -n '1,3p; /hello/p' test.txt > out_dir/test.hello.txt
The awk solution is the best, but I'll add a sed solution for completeness:
$ sed -n test.txt -e '1,3p' -e '4,$s/hello/hello/p' test.txt > $output_file
The -n says not to print out a line unless specified. The -e are the commands '1,3p prints ou the first three lines 4,$s/hello/hello/p looks for all lines that contain the word hello, and substitutes hello back in. The p on the end prints out all lines the substitution operated upon.
There should be a way of using 4,$g/HELLO/p, but I couldn't get it to work. It's been a long time since I really messed with sed.
Of course, I would go awk but here is an ed solution for the pre-vi nostalgics:
ed test.txt <<%
4,$ v/hello/d
w test.hello.txt
%
I'm want to read a string from file
this string is for example
&0001 = 1234 5678 9abc
now I want to take this string and build another string from it which is
123456789abc
I succeeded to read the the string from the end of the file by
read_addr="`awk "END {print}" file.txt`"
echo ${read_addr}
how should I continue to create the string 123456789abc out of the above?
How about this instead:
tail -n 1 file.txt | sed 's/ //g' | sed 's/.*=//'
The tail -n 1 gives you the last line of the file and the sed 's/ //g' removes the spaces.
you can just change your awk line a little bit:
awk -F= 'END{gsub(/ /,"",$2);print $2}' file.txt
this awk line will do the simple task with single process.