What is the best way (better performance) to read a specific line of a file? Currently, I'm using the following command line:
head -line_number file_name | tail -1
ps.: preferentially, using shell tools.
+1
A:
You could use sed.
# print line number 10
$ sed -n '10p' file_name
$ sed '10!d' file_name
$ sed '10q;d' file_name
The MYYN
2010-09-22 01:04:08
On my system, your last example is generally faster than the awk, head/tail or ruby versions unless the line is near the end of the file. Only the tail/head version begins to get a little faster as the line approaches the end of the file.
Dennis Williamson
2010-09-22 05:43:41
If the file is huge, you don't want to read the rest of it needlessly, so exit.
glenn jackman
2010-09-23 13:01:30
+1
A:
If you know the lines are the same length, then a program could directly index in to that line without reading all the preceeding ones: something like od might be able to do that, or you could code it up in half a dozen lines in most-any language. Look for a function called seek() or fseek().
Otherwise, perhaps...
tail +N | head -1
...as this asks tail to skip to the Nth line, and there are less lines put needlessly through the pipe than with your head to tail solution.
Tony
2010-09-22 04:16:53
@Dennis: are you sure? With all the head implementations I've ever seen `head 1` would try to find a file called "1". I've double-checked on GNU/Linux and it's definitely `head -1`. Which version of head do you use?
Tony
2010-09-22 06:56:13
Oh, sorry, GNU `head` insists on `-n` so `head -n 1`. I typoed my comment. With a `-n -1` GNU `head` outputs all but the last line rather than just the first. Version: head (GNU coreutils) 7.4
Dennis Williamson
2010-09-22 10:41:52