问题
I have a very huge file in which I need to obtain every nth line and print it into a row.
My data:
1 937 4.320194
2 667 4.913314
3 934 1.783326
4 940 -0.299312
5 939 2.309559
6 936 3.229496
7 611 -1.41808
8 608 -1.154019
9 606 2.159683
10 549 0.767828
I want my data to look like this:
1 937 4.320194
3 934 1.783326
5 939 2.309559
7 611 -1.41808
9 606 2.159683
This is of course an example, I want every 10th line for my huge data file. I tried this so far:
NF == 6 {
if(NR%10) {print;}
}
回答1:
To print every second line, starting with the first:
awk 'NR%2==1' file.txt
To print every tenth line, starting with the tenth line:
awk 'NR%10==0' file.txt
To use this in a script, add the following to a file called script.awk:
BEGIN {
print "Processing file"
}
NR%10==0
END {
print "Finished processing"
}
Then execute:
awk -f script.awk file.txt
回答2:
With sed, you can do a lot of variations on this quite easily with the first~step command. For instance:
# Odd lines
sed -n 1~2p file
# Every tenth line (10, 20, 30, ...)
sed -n 10~10p file
# Every tenth line (1, 11, 21, ...)
sed -n 1~10p file
# First plus every tenth (1, 10, 20, 30, ...)
sed -n -e 1p -e 10~10p file
回答3:
Piece of cake: cat test.txt | awk 'NR % 10 == 1'
回答4:
It's not (g)awk, but it'll work:
cat myfile | grep ^[[:digit:]]*0[[:blank:]] should do the trick.
回答5:
Doing it directly in command Prompt (Windows).
Put the gawk.exe file in the folder where the file is and start a command Prompt in the folder, and write
gawk "NR%n==x" oldfile.txt>newfile.txt
n is every n'th line you want to print and x is the starting line.
E.g n=10 and x=1, printing line 1,11,21,31,41......end line from the original file into the new file.
E.g n=20 and x=5, printing line 5,25,45,65......end line from the original file into the new file.
来源：https://stackoverflow.com/questions/9968916/print-every-nth-line-into-a-row-using-gawk