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
Related
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.
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 want to get only one word from this txt file: http://pastebin.com/jFDu0Le5 . The word is from last row: WER: 45.67% Correct: 65.87% Acc: 54.33%
I want to get only the value: 45.67 to save it to the file value.txt..I want to create BASH script to get this value. Can you give me an example how to do it??? I am new in Bash and I need it for school. The whole .txt file is saved on my server as text file file.txt.
Try this:
grep WER file.txt | awk '{print $2}' | uniq | sed -e 's/%//' > value.txt
Note that this will overwrite value.txt each time you run the command.
You want grep "WER:" value.txt | cut -???
I have ??? because I do not know the structure of the file. Tab delimited? Fixed Width?
Do man cut an you can get the arguments you need.
There a many ways and instruments to do the task:
sed
tac file.txt | sed -n '/^WER: /{s///;s/%.*//;p;q}' > value.txt
awk
tac file.txt | awk -F'[ %]' '/^WER:/{print $2;exit}' > value.txt
bash
while read a b c
do
if [ $a = "WER:" ]
then
b=${b%\%*}
echo ${b#* }
break
fi
done < <(tac file.txt) > value.txt
If the format is as you said, then this also works
awk -F'[: %]' '/^WER/{print $3}' file.txt > value.txt
Explanation
-F specifies the field separator as one of [: %]
/<PATTERN>/ {<ACTION>} refers to: if a line matches some PATTERN, then do some ACTION
in my case,
the PATTERN is: starts with ^ the string WER
the ACTION is: print field $3 (as split by the -F field separators)
> sends the output to value.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.
How would I remove the first word from each line of text in a stream?
For example,
$ cat myfile
some text 1
some text 2
some text 3
I want:
$ cat myfile | magiccommand
text 1
text 2
text 3
How would I go about this using Bash? I could use awk '{print $2 $3 $4 $5 ....}', but that's messy and would result in extra spaces for all null arguments. I was thinking that sed might be able to do this, but I could not find any examples of this.
Based on your example text,
cut -d' ' -f2- yourFile
should do the job.
That should work:
$ cat test.txt
some text 1
some text 2
some text 3
$ sed -e 's/^\w*\ *//' test.txt
text 1
text 2
text 3
Here is a solution using awk
awk '{$1= ""; print $0}' yourfile
Run this:
sed "s/^some\s//g" myfile
You even don't need to use a pipe.
To remove the first word, until space no matter how many spaces exist, use: sed 's/[^ ]* *//'
Example:
$ cat myfile
some text 1
some text 2
some text 3
$ cat myfile | sed 's/[^ ]* *//'
text 1
text 2
text 3