I have more than 50000 files in a directory such as file1.txt, file2.txt, ....., file50000.txt. I would like to concatenate of some files whose file numbers are listed in the following text file (need.txt).
need.txt
1
4
35
45
71
.
.
.
I tried with the following. Though it is working, but I look for more simpler and short way.
n1=1
n2=$(wc -l < need.txt)
while [ $n1 -le $n2 ]
do
f1=$(awk 'NR=="$n1" {print $1}' need.txt)
cat file$f1.txt >> out.txt
(( n1++ ))
done
This might also work for you:
sed 's/.*/file&.txt/' < need.txt | xargs cat > out.txt
Something like this should work for you:
sed -e 's/.*/file&.txt/' need.txt | xargs cat > out.txt
It uses sed to translate each line into the appropriate file name and then hands the filenames to xargs to hand them to cat.
Using awk it could be done this way:
awk 'NR==FNR{ARGV[ARGC]="file"$1".txt"; ARGC++; next} {print}' need.txt > out.txt
Which adds each file to the ARGV array of files to process and then prints every line it sees.
It is possible do it without any sed or awk command. Directly using bash built-in functions and cat (of course).
for i in $(cat need.txt); do cat file${i}.txt >> out.txt; done
And as you want, it is quite simple.
Related
I have a simple question. I am trying to check the 3rd line of multiple files in a folder, so I used this:
head -n 3 MiseqData/result2012/12* | tail -n 1
but this doesn't work obviously, because it only shows the third line of the last file. But I actually want to have last line of every file in the result2012 folder.
Does anyone know how to do that?
Also sorry just another questions, is it also possible to show which file the particular third line belongs to?
like before the third line is shown, is it also possible to show the filename of each of the third line extracted from?
because if I used head or tail command, the filename is also shown.
thank you
With Awk, the variable FNR is the number of the "record" (line, by default) in the current file, so you can simply compare it to 3 to print the third line of each input file:
awk 'FNR == 3' MiseqData/result2012/12*
A more optimized version for long files would skip to the next file on match, since you know there's only that one line where the condition is true:
awk 'FNR == 3 { print; nextfile }' MiseqData/result2012/12*
However, not all Awks support nextfile (but it is also not exclusive to GNU Awk).
A more portable variant using your head and tail solution would be a loop in the shell:
for f in MiseqData/result2012/12*; do head -n 3 "$f" | tail -n 1; done
Or with sed (without GNU extensions, i.e., the -s argument):
for f in MiseqData/result2012/12*; do sed '3q;d' "$f"; done
edit: As for the additional question of how to print the name of each file, you need to explicitly print it for each file yourself, e.g.,
awk 'FNR == 3 { print FILENAME ": " $0; nextfile }' MiseqData/result2012/12*
for f in MiseqData/result2012/12*; do
echo -n `basename "$f"`': '
head -n 3 "$f" | tail -n 1
done
for f in MiseqData/result2012/12*; do
echo -n "$f: "
sed '3q;d' "$f"
done
With GNU sed:
sed -s -n '3p' MiseqData/result2012/12*
or shorter
sed -s '3!d' MiseqData/result2012/12*
From man sed:
-s: consider files as separate rather than as a single continuous long stream.
You can do this:
awk 'FNR==3' MiseqData/result2012/12*
If you like the file name as well:
awk 'FNR==3 {print FILENAME,$0}' MiseqData/result2012/12*
This might work for you (GNU sed & parallel):
parallel -k sed -n '3p\;3q' {} ::: file1 file2 file3
Parallel applies the sed command to each file and returns the results in order.
N.B. All files will only be read upto the 3rd line.
Also,you may be tempted (as I was) to use:
sed -ns '3p;3q' file1 file2 file3
but this will only return the first file.
Hi bro I am answering this question as we know FNR is used to check no of lines so we can run this command to get 3rd line of every file.
awk 'FNR==3' MiseqData/result2012/12*
I have a file containing list of 4000 words (A.txt). Now I want to grep lines from another file (sentence_per_line.txt) containing those 4000 words mentioned in the file A.txt.
The shell script I wrote for the above problem is
#!/bin/bash
file="A.txt"
while IFS= read -r line
do
# display $line or do somthing with $line
printf '%s\n' "$line"
grep $line sentence_per_line.txt >> output.txt
# tried printing the grep command to check its working or not
result=$(grep "$line" sentence_per_line.txt >> output.txt)
echo "$result"
done <"$file"
And A.txt looks like this
applicable
available
White
Black
..
The code is neither working nor does it shows any error.
Grep has this built in:
grep -f A.txt sentence_per_line.txt > output.txt
Remarks to your code:
Looping over a file to execute grep/sed/awk on each line is typically an antipattern, see this Q&A.
If your $line parameter contains more than one word, you have to quote it (doesn't hurt anyway), or grep tries to look for the first word in a file named after the second word:
grep "$line" sentence_per_line.txt >> output.txt
If you write output in a loop, don't redirect within the loop, do it outside:
while read -r line; do
grep "$line" sentence_per_line.txt
done < "$file" > output.txt
but remember, it's usually not a good idea in the first place.
If you'd like to write to a file and at the same time see what you're writing, you can use tee:
grep "$line" sentence_per_line.txt | tee output.txt
writes to output.txt and stdout.
If A.txt contains words which you want to match only if the complete word matches, i.e., pattern should not match longerpattern, you can use grep -wf – the -w matches only complete words.
If the words in A.txt aren't regular expressions, but fixed strings, you can use grep -fF – the -F option looks for fixed strings and is faster. These two can be combined: grep -WfF
I have a lot of text files and I want to make a bash script in linux to print the name of file in each lines of file. For example I have file lenovo.txt and I want that every line in the file to start with lenovo.txt.
I try to make a "for" for this but didn't work.
for i in *.txt
do
awk '{print '$i' $0}' /var/SambaShare/$i > /var/SambaShare/new_$i
done
Thanks!
It doesn't work because you need to pass $i to awk with the -v option. But you can also use the FILENAME built-in variable in awk :
ls *txt
file.txt file2.txt
cat *txt
A
B
C
A2
B2
C2
for i in *txt; do
awk '{print FILENAME,$0}' $i;
done
file.txt A
file.txt B
file.txt C
file2.txt A2
file2.txt B2
file2.txt C2
An to redirect into a new file :
for i in *txt; do
awk '{print FILENAME,$0}' $i > ${i%.txt}_new.txt;
done
As for your corrected version :
for i in *.txt
do
awk -v i=$i '{print i,$0}' $i > new_$i
done
Hope this helps.
Using grep you can make use of the --with-filename (alias -H) option and use an empty pattern that always matches:
for i in *.txt
do
grep -H "" $i > new_$i
done
Awk and Bash don't share the same variables as they are different languages with separate interpreters. You should pass Bash variables to Awk with the -v option.
You should also quote your file name variables to ensure they don't get expanded as separate arguments if they contain whitespace.
for i in *.txt
do
awk -v i="$i" '{print i,$0}' "$i" > "$i"
done
I have file.txt with names one per line as shown below:
ABCB8
ABCC12
ABCC3
ABCC4
AHR
ALDH4A1
ALDH5A1
....
I want to grep each of these from an input.txt file.
Manually i do this one at a time as
grep "ABCB8" input.txt > output.txt
Could someone help to automatically grep all the strings in file.txt from input.txt and write it to output.txt.
You can use the -f flag as described in Bash, Linux, Need to remove lines from one file based on matching content from another file
grep -o -f file.txt input.txt > output.txt
Flag
-f FILE, --file=FILE:
Obtain patterns from FILE, one per line. The empty file
contains zero patterns, and therefore matches nothing. (-f is
specified by POSIX.)
-o, --only-matching:
Print only the matched (non-empty) parts of a matching line, with
each such part on a separate output line.
for line in `cat text.txt`; do grep $line input.txt >> output.txt; done
Contents of text.txt:
ABCB8
ABCC12
ABCC3
ABCC4
AHR
ALDH4A1
ALDH5A1
Edit:
A safer solution with while read:
cat text.txt | while read line; do grep "$line" input.txt >> output.txt; done
Edit 2:
Sample text.txt:
ABCB8
ABCB8XY
ABCC12
Sample input.txt:
You were hired to do a job; we expect you to do it.
You were hired because ABCB8 you kick ass;
we expect you to kick ass.
ABCB8XY You were hired because you can commit to a rational deadline and meet it;
ABCC12 we'll expect you to do that too.
You're not someone who needs a middle manager tracking your mouse clicks
If You don't care about the order of lines, the quick workaround would be to pipe the solution through a sort | uniq:
cat text.txt | while read line; do grep "$line" input.txt >> output.txt; done; cat output.txt | sort | uniq > output2.txt
The result is then in output.txt.
Edit 3:
cat text.txt | while read line; do grep "\<${line}\>" input.txt >> output.txt; done
Is that fine?
So here is the task which I can't solve. I have a directory with .h files and a directory with .i files, which have the same names as the .h files. I want just by typing a command to have all .h files which are not found as .i files. It's not a hard problem, I can do it in some programming language, but I'm just curious how it will look like in cmd :). To be more specific here is the algo:
get file names without extensions from ls *.h
get file names without extensions from ls *.i
compare them
print all names from 1 that are not met in 2
Good luck!
diff \
<(ls dir.with.h | sed 's/\.h$//') \
<(ls dir.with.i | sed 's/\.i$//') \
| grep '$<' \
| cut -c3-
diff <(ls dir.with.h | sed 's/\.h$//') <(ls dir.with.i | sed 's/\.i$//') executes ls on the two directories, cuts off the extensions, and compares the two lists. Then grep '$<' finds the files that are only in the first listing, and cut -c3- cuts off the "< " characters that diff inserted.
ls ./dir_h/*.h | sed -r -n 's:.*dir_h/([^.]*).h$:dir_i/\1.i:p' | xargs ls 2>&1 | \
grep "No such file or directory" | awk '{print $4}' | sed -n -r 's:dir_i/([^:]*).*:dir_h/\1:p'
ls -1 dir1/*.hh dir2/*.ii | awk -F"/" '{print $NF}' |awk -F"." '{a[$1]++;b[$0]}END{for(i in a)if(a[i]==1 && b[i".hh"]) print i}'
explanation:
ls -1 dir1/*.hh dir2/*.ii
above will list all the files *.hh and *.ii files in both the directories.
awk -F"/" '{print $NF}'
above will just print the file name excluding the complete path of the file.
awk -F"." '{a[$1]++;b[$0]}END{for(i in a)if(a[i]==1 && b[i".hh"]) print i}'
above will create two associative arrays one with file name and one with excluding the extension.
if both hh and ii files exist the value in the assosciative array will 2 if there is only one file then the value will be 1.so we need array item whose value is 1 and it should be a header file (.hh).
this can be checked using the asso..array b which is done in the END block.
Assuming bash is your shell:
for file in $( ls dir_with_h/*.h ); do
name=${file%\.h}; # trim trailing ".h" file extension
name=${name#dir_with_h/}; # trim leading folder name
if [ ! -e dir_with_i/${name}.i ]; then
echo ${name};
fi
done
Undoubtedly this can be ported to virtually all other shells. I find this less cryptic than some other approaches (although this is surely my problem) but it is a little wordy. As such. a shell script might help recall it.