NetNews Usenet Archive 1992 #27/NN_1992_27.iso/spool/comp/unix/question/13816

Text File | 1992-11-22 | 2.8 KB | 77 lines

Newsgroups: comp.unix.questions
Path: sparky!uunet!spool.mu.edu!uwm.edu!linac!att!princeton!fish.Princeton.EDU!lhjensen
From: [e-mail] (Leif Jensen)
Subject: Re: How do I get a particualr line of a file?
Message-ID: <[e-mail]>
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Organization: Princeton University
References: <[e-mail]> <[e-mail]>
Date: Mon, 23 Nov 1992 01:09:21 GMT
Lines: 63

In article <[e-mail]> [e-mail] (Paul E. Carroll) writes:
>This seems extremely simple, but I've run into the problem several times
>and gotten stumped each time.  So how do I view a particualr line of a
>file?  Head looks at the top, tail the end, cat - echo - more - or less-
>look at the whole thing... but what if I want to see just line 30?
>or JUST line 210?  I'm stumped.

There are lots of ways to do this.  Some are better than others.  Since
there are so many correct answers, I think that for efficiency's sake,
it is important to use a method that does not read more of the file
than it has to. Here's a short list.

sed -n '200{;p;q;}' /etc/termcap
awk 'NR==200 {print;exit}' /etc/termcap
sed 200q /etc/termcap | tail -1
head -200 /etc/termcap | tail -1
perl -ne 'if ($. == 200) {print; exit}' /etc/termcap

In article <[e-mail]> [e-mail] (Mr J L Saunders) writes:
>Have a look at the manual for awk...
>
>awk '{if (NR == line) print}' line=linenumber filename
>
>Where linenumber is the line you want, and filename is the file you're looking
>at. eg.
>
>>awk '{if (NR == line) print}' line=5000 /usr/dict/words
>concrete

I don't know why you would use the if(...) method to choose the line.
The awk command I list above seems much more intuitive.  Your solution
also suffers from reading much more of the file than it has to.  Even
on a fast sun workstation with no other users, the effect is dramatic:

% ls -l /tmp/foo
-rw-------  1 lhjensen  3138380 Nov 22 19:46 /tmp/foo
% time awk '{if (NR == line) print}' line=200 /tmp/foo
hn|2621-nl|hp2621nl|2621nl|hp2621-nl|hp 2621 with no labels:\
22.810u 0.780s 0:23.72 99.4% 0+162k 0+0io 0pf+0w
%

In article <[e-mail]> [e-mail] (Rouben Rostamian) writes:
>To print line 30:
>sed -n 30p filename
>
>To print lines 30 to 35:
>sed -n 30,35p filename

Ditto for this command.  The relevant times are

% time sed -n 200p /tmp/foo
hn|2621-nl|hp2621nl|2621nl|hp2621-nl|hp 2621 with no labels:\
2.810u 1.070s 0:03.96 97.9% 0+114k 0+0io 0pf+0w
%

which isn't bad, but is much worse than it has to be.

In summary, there's much more to things than just finding a solution
that works.  Try to find one that works well before you post it.

--
Leif Jensen
[e-mail]