Re: how to drop the nth element of a list

Re: how to drop the nth element of a list

[Rich Salz]

I don't have seq.  This seems to work okay:
	x=(a b c d e f)
	n=3
	{ for (i in $x ) echo $i } | sed -e $n^d
	a
	b
	d
	e

Re: how to drop the nth element of a list

[DaviD W. Sanderson]

> Packaging this into an rc function is left as an excercise for the
> reader...

One (simple) solution is the 'chop' function I contributed to the
rc 1.2 EXAMPLES file.

DaviD Sanderson (dws@cs.wisc.edu)

Re: how to drop the nth element of a list

[Byron Rakitzis]

You can do this in at most the cost of 3 forks and 2 execs; suppose that
$foo contains the list you want to operate on, and $n contains the number
of the element you wish to remove:

	foo=$foo(`{seq 1 $#foo | grep -v '^'$n'$'})

You could also probably do the stuff in backquotes in a single awk job,
but it involves more messing around with quotes, and takes longer to
type. Still, it will run faster:

	awk 'BEGIN{
		for (i = 1; i < '$#foo'; i++)
			if (i != '$n')
				print i
		exit
	}'

Packaging this into an rc function is left as an excercise for the
reader...

how to drop the nth element of a list

[malte]

Does anyone have a good idea to drop the nth element from a list?
I can only think of something using head(1) and tail(1), which is
not too efficient.


_______________________________________________________________________________
malte@techfak.uni-bielefeld.de

send mail reply to:	Universitaet Bielefeld, Technische Fakultaet
		z. Hd. Malte Uhl
		Postfach 8640
		4800 Bielefeld 1