How do I get the line count of a file from the 2nd line of the file, as the first line is header?
wc -l filename
Is there a way to set some condition into it?
Use the tail command:
tail -n +2 file | wc -l
-n +2 would print the file starting from line 2
You can use awk to count from 2nd line onwards:
awk 'NR>1{c++} END {print c}' file
Or simply use NR variable in the END block:
awk 'END {print NR-1}' file
Alternatively using BASH arithmetic subtract 1 from wc output:
echo $(( $(wc -l < file) -1 ))
Delete first line with GNU sed:
sed '1d' file | wc -l
There is no way to tweak the wc command itself. You should whether process the result of the command, or use another tool.
As suggested in other answers, if you are running Bash, a good way is to put the result of the command into an arithmetic expression like $(( $(command) - 1 )).
In case if you are searching for a portable solution, here is a Perl version:
perl -e '1 while <>; print $. - 1' < file
The variable $. holds the number of lines read since a file handle was last closed. The while loop reads all the lines from the file.
Alternately, you could just subtract 2.
echo $((`cat FILE | wc -l`-2))
Please try this one. It will be solved your problem
$ tail -n +2 filename | wc -l
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,$0}' MiseqData/result2012/12*
This might work for you (GNU sed & parallel):
parallel -k sed -n '3p\;3q' {} ::: file1 file2 file3
Parallel applies the sed command to each file and returns the results in order.
N.B. All files will only be read upto the 3rd line.
Also,you may be tempted (as I was) to use:
sed -ns '3p;3q' file1 file2 file3
but this will only return the first file.
Hi bro I am answering this question as we know FNR is used to check no of lines so we can run this command to get 3rd line of every file.
awk 'FNR==3' MiseqData/result2012/12*
I can't tail the last n lines of a file on linux, likewise i can grep
e.g grep "2015-09-29 04:" filename.ext
But how can i combine both such that i display from a certain grep to the end of the file.
You don't use grep or tail any more. You use sed:
sed -n '/^2015-09-29 04:/,$p'
Don't print by default (-n). From the first line starting 2015-09-29 04: to the end of file ($), print the lines.
If you absolutely must use grep and you have GNU grep, then you could consider:
grep -A 999999999 -e '^2015-09-29 04:'
That prints the first billion or so lines after the first line that matches the pattern (and the counter resets if the pattern appears during that trailing material). Of course, if your file is 2 billion lines long and the pattern occurs after a million lines (and never again), then you'll be missing a lot of data.
#!/bin/bash
line_no=`grep -n -m1 "$1" $2 | cut -d: -f1`
echo "Starting at line: " $line_no
tail -n +$line_no $2
Usage: script 'text to hunt for' filename
e.g.
./grep.sh 'Sep 29 13:14' /var/log/syslog
-n issue line numbers
-m1 stop reading after 1 hit
cut out the line number
tail -n +K output lines starting with the Kth
In a very large file I need to find the position (line number) of a string, then extract the 2 lines above and below that string.
To do this right now - I launch vi, find the string, note it's line number, exit vi, then use sed to extract the lines surrounding that string.
Is there a way to streamline this process... ideally without having to run vi at all.
Maybe using grep like this:
grep -n -2 your_searched_for_string your_large_text_file
Will give you almost what you expect
-n : tells grep to print the line number
-2 : print 2 additional lines (and the wanted string, of course)
You can do
grep -C 2 yourSearch yourFile
To send it in a file, do
grep -C 2 yourSearch yourFile > result.txt
Use grep -n string file to find the line number without opening the file.
you can use cat -n to display the line numbers and then use awk to get the line number after a grep in order to extract line number:
cat -n FILE | grep WORD | awk '{print $1;}'
although grep already does what you mention if you give -C 2 (above/below 2 lines):
grep -C 2 WORD FILE
You can do it with grep -A and -B options, like this:
grep -B 2 -A 2 "searchstring" | sed 3d
grep will find the line and show two lines of context before and after, later remove the third one with sed.
If you want to automate this, simple you can do a Shell Script. You may try the following:
#!/bin/bash
VAL="your_search_keyword"
NUM1=`grep -n "$VAL" file.txt | cut -f1 -d ':'`
echo $NUM1 #show the line number of the matched keyword
MYNUMUP=$["NUM1"-1] #get above keyword
MYNUMDOWN=$["NUM1"+1] #get below keyword
sed -n "$MYNUMUP"p file.txt #display above keyword
sed -n "$MYNUMDOWN"p file.txt #display below keyword
The plus point of the script is you can change the keyword in VAL variable as you like and execute to get the needed output.
How can I get second line in a file using SED
#SRR005108.1 :3:1:643:216
GATTTCTGGCCCGCCGCTCGATAATACAGTAATTCC
+
IIIIII/III*IIIIIIIIII+IIIII;IIAIII%>
With the data that looks like above I want only to get
GATTTCTGGCCCGCCGCTCGATAATACAGTAATTCC
You don't really need Sed, but if the pourpose is to learn... you can use -n
n read the next input line and starts processing the newline with the command rather than the first command
sed -n 2p somefile.txt
Edit: You can also improve the performance using the tip that manatwork mentions in his comment:
sed -n '2{p;q}' somefile.txt
You always want the second line of a file? No need for SED:
head -2 file | tail -1
This will print the second line of every file:
awk 'FNR==2'
and this one only the second line of the first file:
awk 'NR==2'
This might work for you:
sed '2q;d' file
cat your_file | head -2 | tail -1
This question already has answers here:
How can I remove the first line of a text file using bash/sed script?
(19 answers)
Closed 3 years ago.
I have a very long file which I want to print, skipping the first 1,000,000 lines, for example.
I looked into the cat man page, but I did not see any option to do this. I am looking for a command to do this or a simple Bash program.
You'll need tail. Some examples:
$ tail great-big-file.log
< Last 10 lines of great-big-file.log >
If you really need to SKIP a particular number of "first" lines, use
$ tail -n +<N+1> <filename>
< filename, excluding first N lines. >
That is, if you want to skip N lines, you start printing line N+1. Example:
$ tail -n +11 /tmp/myfile
< /tmp/myfile, starting at line 11, or skipping the first 10 lines. >
If you want to just see the last so many lines, omit the "+":
$ tail -n <N> <filename>
< last N lines of file. >
Easiest way I found to remove the first ten lines of a file:
$ sed 1,10d file.txt
In the general case where X is the number of initial lines to delete, credit to commenters and editors for this:
$ sed 1,Xd file.txt
If you have GNU tail available on your system, you can do the following:
tail -n +1000001 huge-file.log
It's the + character that does what you want. To quote from the man page:
If the first character of K (the number of bytes or lines) is a
`+', print beginning with the Kth item from the start of each file.
Thus, as noted in the comment, putting +1000001 starts printing with the first item after the first 1,000,000 lines.
If you want to skip first two line:
tail -n +3 <filename>
If you want to skip first x line:
tail -n +$((x+1)) <filename>
A less verbose version with AWK:
awk 'NR > 1e6' myfile.txt
But I would recommend using integer numbers.
Use the sed delete command with a range address. For example:
sed 1,100d file.txt # Print file.txt omitting lines 1-100.
Alternatively, if you want to only print a known range, use the print command with the -n flag:
sed -n 201,300p file.txt # Print lines 201-300 from file.txt
This solution should work reliably on all Unix systems, regardless of the presence of GNU utilities.
Use:
sed -n '1d;p'
This command will delete the first line and print the rest.
If you want to see the first 10 lines you can use sed as below:
sed -n '1,10 p' myFile.txt
Or if you want to see lines from 20 to 30 you can use:
sed -n '20,30 p' myFile.txt
Just to propose a sed alternative. :) To skip first one million lines, try |sed '1,1000000d'.
Example:
$ perl -wle 'print for (1..1_000_005)'|sed '1,1000000d'
1000001
1000002
1000003
1000004
1000005
You can do this using the head and tail commands:
head -n <num> | tail -n <lines to print>
where num is 1e6 + the number of lines you want to print.
This shell script works fine for me:
#!/bin/bash
awk -v initial_line=$1 -v end_line=$2 '{
if (NR >= initial_line && NR <= end_line)
print $0
}' $3
Used with this sample file (file.txt):
one
two
three
four
five
six
The command (it will extract from second to fourth line in the file):
edu#debian5:~$./script.sh 2 4 file.txt
Output of this command:
two
three
four
Of course, you can improve it, for example by testing that all argument values are the expected :-)
cat < File > | awk '{if(NR > 6) print $0}'
I needed to do the same and found this thread.
I tried "tail -n +, but it just printed everything.
The more +lines worked nicely on the prompt, but it turned out it behaved totally different when run in headless mode (cronjob).
I finally wrote this myself:
skip=5
FILE="/tmp/filetoprint"
tail -n$((`cat "${FILE}" | wc -l` - skip)) "${FILE}"