I know how to take e.g the first 2 lines from a .txt data and appending it to the end of a .txt data. But how should I add the last 2 lines of a .txt data to the 1st line of a .txt data
I've tried :
tail -n 2 test1.txt >> head test1.txt # takes last 2 lines of text and adding
it to the head
Looks awfully wrong but I can't find the answer anywhere, doing it with tail and head.
tail n 2 test1.txt >> head test1.txt
cat test1.txt
Someone please correct my code so I get my expected result.
Just run the two commands one after the other -- the stdout resulting from doing so will be exactly the same as what you'd get by concatenating their output together, without needing to do an explicit/extra concatenation step:
tail -n 2 test1.txt
head -n 1 test1.txt
If you want to redirect their output together, put them in a brace group:
{
tail -n 2 test1.txt
head -n 1 test1.txt
} >out.txt
What about:
$ cat file1.txt
file 1 line 1
file 1 line 2
file 1 line 3
file 1 line 4
$ cat file2.txt
file 2 line 1
file 2 line 2
file 2 line 3
file 2 line 4
$ tail -n 2 file1.txt > output.txt
$ head -n 1 file2.txt >> output.txt
$ cat output.txt
file 1 line 3
file 1 line 4
file 2 line 1
Related
I am concatenating a large number of files into a single one with the following command:
$ cat num_*.dat > dataset.dat
However, due to the structure of the files, I'd like to omit concatenating the first two and last two lines of each file. Those lines contain file information which is not important for my necesities.
I know the existence of head and tail, but I don't now how to combine them in a UNIX instruction to solve my issue.
The head command has some odd parameter usage.
You can use the following to list all of the lines except the last two.
$ cat num_*.dat | head -n-2 > dataset.dat
Next, take that and run the following tail command on it
$ tail dataset.dat -n+3 >> dataset.dat
I believe the following will work as one command.
$ cat num_*.dat | head -n-2 | tail -n+3 > dataset.dat
I tested on a file that had lines like the following:
Line 1 Line 2 Line 3 Line 4 Line 5 Line 6 Line
7
This one will get you started:
cat test.txt | head -n-2 | tail -n+3
From the file above it prints :
Line 3 Line 4 Line 5
The challenge is that when you use cat filename*.dat or whatever is that the command cats all of the files then runs the command one time so it becomes one large file with only removing the first two lines of the first catted file and the two lines of that last catted file.
Final Answer - Need to Write a Bash Script
I wrote a bash script that will do this for you.
This one will iterate through each file in your directory and run the command.
Notice that it appends (>>) to the dataset.dat file.
for file in num_*.dat; do
if [ -f "$file" ]; then
cat $file | head -n-2 | tail -n+3 >> dataset.dat
echo "$file"
fi
done
I had two files that looked like the following:
line 1 line 2 line 3 line 4 line 5 line 6 line
7 2 line 1 2 line 2 2 line 3 2 line 4 2 line 5
2 line 6 2 line 7
The final output was:
line 3 line 4 line 5 2 line 3 2 line 4 2 line
5
for i in num_*.dat; do # loop through all files concerned
cat $i | tail -n +3 | head -n -2 >> dataset.dat
done
For example:how can I print specific lines of a .txt file between line 5 and line 8 using only tail and head
Copied from here
infile.txt contains a numerical value on each line.
➜ X=3
➜ Y=10
➜ < infile.txt tail -n +"$X" | head -n "$((Y - X))"
3
4
5
6
7
8
9
➜
I have one zip folder with 5 text files inside. I have to check the row count of each file without unzipping the zip folder.
I tried zcat file.zip | wc -l but it gives the count of the first file only.
Can you guys help me to get the result as mentioned below:
File_Name Rowcount
file1 100
file2 100
file3 100
file4 100
file5 100
If your file is a gzipped tar archive, then you can simply loop over each filename in the archive to get the number of lines in each. For example if you archive contains:
$ tar -tzf /tmp/tmp-david/zipfile.tar.gz
yon.c
yourmachinecode.s
zeroonect.c
zeros
zz
You can loop over the filenames with:
$ for i in $(tar -tzf /tmp/tmp-david/zipfile.tar.gz); do
printf "%8d lines - %s\n" $(wc -l <"$i") "$i"
done
61 lines - yon.c
5 lines - yourmachinecode.s
63 lines - zeroonect.c
0 lines - zeros
1 lines - zz
You can keep a sum and increment with each count as required to get the total.
If your file is a .zip (MS-DOS) archive, then you can do the same thing, but the parsing of the individual filenames from the output of unzip -l takes a bit more work, e.g.
$ unzip -l /tmp/tmp-david/zipfile.zip | grep -v '^-' | \
tail -n+3 | head -n-1 | awk '{print $4}'
(you would use the above as in your command substitution to drive the for loop)
I need to count all lines of an unix file. The file has 3 lines but wc -l gives only 2 count.
I understand that it is not counting last line because it does not have end of line character
Could any one please tell me how to count that line as well ?
grep -c returns the number of matching lines. Just use an empty string "" as your matching expression:
$ echo -n $'a\nb\nc' > 2or3.txt
$ cat 2or3.txt | wc -l
2
$ grep -c "" 2or3.txt
3
It is better to have all lines ending with EOL \n in Unix files. You can do:
{ cat file; echo ''; } | wc -l
Or this awk:
awk 'END{print NR}' file
This approach will give the correct line count regardless of whether the last line in the file ends with a newline or not.
awk will make sure that, in its output, each line it prints ends with a new line character. Thus, to be sure each line ends in a newline before sending the line to wc, use:
awk '1' file | wc -l
Here, we use the trivial awk program that consists solely of the number 1. awk interprets this cryptic statement to mean "print the line" which it does, being assured that a trailing newline is present.
Examples
Let us create a file with three lines, each ending with a newline, and count the lines:
$ echo -n $'a\nb\nc\n' >file
$ awk '1' f | wc -l
3
The correct number is found.
Now, let's try again with the last new line missing:
$ echo -n $'a\nb\nc' >file
$ awk '1' f | wc -l
3
This still provides the right number. awk automatically corrects for a missing newline but leaves the file alone if the last newline is present.
Respect
I respect the answer from John1024 and would like to expand upon it.
Line Count function
I find myself comparing line counts A LOT especially from the clipboard, so I have defined a bash function. I'd like to modify it to show the filenames and when passed more than 1 file a total. However, it hasn't been important enough for me to do so far.
# semicolons used because this is a condensed to 1 line in my ~/.bash_profile
function wcl(){
if [[ -z "${1:-}" ]]; then
set -- /dev/stdin "$#";
fi;
for f in "$#"; do
awk 1 "$f" | wc -l;
done;
}
Counting lines without the function
# Line count of the file
$ cat file_with_newline | wc -l
3
# Line count of the file
$ cat file_without_newline | wc -l
2
# Line count of the file unchanged by cat
$ cat file_without_newline | cat | wc -l
2
# Line count of the file changed by awk
$ cat file_without_newline | awk 1 | wc -l
3
# Line count of the file changed by only the first call to awk
$ cat file_without_newline | awk 1 | awk 1 | awk 1 | wc -l
3
# Line count of the file unchanged by awk because it ends with a newline character
$ cat file_with_newline | awk 1 | awk 1 | awk 1 | wc -l
3
Counting characters (why you don't want to put a wrapper around wc)
# Character count of the file
$ cat file_with_newline | wc -c
6
# Character count of the file unchanged by awk because it ends with a newline character
$ cat file_with_newline | awk 1 | awk 1 | awk 1 | wc -c
6
# Character count of the file
$ cat file_without_newline | wc -c
5
# Character count of the file changed by awk
$ cat file_without_newline | awk 1 | wc -c
6
Counting lines with the function
# Line count function used on stdin
$ cat file_with_newline | wcl
3
# Line count function used on stdin
$ cat file_without_newline | wcl
3
# Line count function used on filenames passed as arguments
$ wcl file_without_newline file_with_newline
3
3
I need to move the last 4 lines of a text file and move them to the second row in the text file.
I'm assuming that tail and sed are used but, I haven't much luck so far.
Here is a head and tail solution. Let us start with the same sample file as Glenn Jackman:
$ seq 10 >file
Apply these commands:
$ head -n1 file ; tail -n4 file; tail -n+2 file | head -n-4
1
7
8
9
10
2
3
4
5
6
Explanation:
head -n1 file
Print first line
tail -n4 file
Print last four lines
tail -n+2 file | head -n-4
Print the lines starting with line 2 and ending before the fourth-to-last line.
If I'm assuming correctly, ed can handle your task:
seq 10 > file
ed file <<'COMMANDS'
$-3,$m1
w
q
COMMANDS
cat file
1
7
8
9
10
2
3
4
5
6
lines 7,8,9,10 have been moved to the 2nd line
$-3,$m1 means, for the range of lines from "$-3" (3 lines before the last line) to "$" (the last line, move them ("m") below the first line ("1")
Note that the heredoc has been quoted so the shell does not try to interpret the strings $- and $m1 as variables
If you don't want to actually modify the file, but instead print to stdout:
ed -s file <<'COMMANDS'
$-3,$m1
%p
Q
COMMANDS
Here is an awk solution:
seq 10 > file
awk '{a[NR]=$0} END {for (i=1;i<=NR-4;i++) if (i==2) {for (j=NR-3;j<=NR;j++) print a[j];print a[i]} else print a[i]}' file
1
7
8
9
10
2
3
4
5
6