Getting the first n elements from a list

[Suresh Jeevanandam]

Hi all,
Is there a skill function to get the first n elements of a list.

Thanks in advance,

--
Suresh

 

[Suresh Jeevanandam]

Suresh Jeevanandam wrote:

Hi all,
Is there a skill function to get the first n elements of a list.

Thanks in advance,

--
Suresh

I could come up with this. Your suggestion to make it more efficient is
most welcome.

procedure( nElements(theList, n)
let( (x)
x = nil
for(i 1 n
x = tconc(x car(theList))
theList = cdr(theList)
)
car(x)
);let
);procedure

regards,
Suresh

 

[Andrew Beckett]

Suresh,

If you don't mind it being destructive, you can use:

procedure(nElements(theList n)
rplacd(nthcdr(sub1 theList) nil)
theList
)

You could use a little trick to destructively modify it, copy, and then
modify back again:

procedure(nElements(theList n)
let((tail tailcdr)
tail=nthcdr(sub1 theList)
tailcdr=cdr(tail)
rplacd(tail nil)
prog1(
copy(theList)
rplacd(tail tailcdr)
)
)
)

However, I don't think this is any more efficient than what you're doing. It
may even be a little less efficient because it traverses the list once in the
nthcdr, and again in the copy (just the first n elements).

The function below may be a little more efficient, because you're
using a built-in loop (forall) to do the loop traversal:

procedure(nElements(theList n)
let((new)
forall(elem theList n-->0 && (new=tconc(new elem)))
car(new)
)
)

Note - in your code, there's no need to initialise x to nil, since that is the
default value for anything in a let().

Probably I'd do some profiling to be certain... just wanted to give a few
options for the benefit of you and the audience ;-)

Finally, why are you getting the first n elements in a list? Are you approaching
the bigger problem in the right way?

Andrew.

On Thu, 19 Aug 2004 12:15:50 +0530, Suresh Jeevanandam
<sureshj@DELETETHISti.com> wrote:

Suresh Jeevanandam wrote:
Hi all,
Is there a skill function to get the first n elements of a list.

Thanks in advance,

--
Suresh

I could come up with this. Your suggestion to make it more efficient is
most welcome.

procedure( nElements(theList, n)
let( (x)
x = nil
for(i 1 n
x = tconc(x car(theList))
theList = cdr(theList)
)
car(x)
);let
);procedure

regards,
Suresh

--
Andrew Beckett
Senior Technical Leader
Custom IC Solutions
Cadence Design Systems Ltd

 

[Suresh Jeevanandam]

Andrew Beckett wrote:
[....]

procedure(nElements(theList n)
let((tail tailcdr)
tail=nthcdr(sub1 theList)
tailcdr=cdr(tail)
rplacd(tail nil)
prog1(
copy(theList)
rplacd(tail tailcdr)
)
)
)

However, I don't think this is any more efficient than what you're doing. It
may even be a little less efficient because it traverses the list once in the
nthcdr, and again in the copy (just the first n elements).
Andrew,

I think the code snippet above would be the most efficient:
Reasons:
1)Locating the tail is done using 'nthcdr'. Mine uses for loop which
would be less efficient.

2) Copy is done by the standard copy function. Mine does copying by calling

x = tconc(x car(theList))
theList = cdr(theList)
n times, which would be less efficient.

thanks,
Suresh

----
A little experiment on this,

procedure(lrange(num)
;return a list of numbers from 1 to num
let((rlist)
rlist = nil
while(num > 0
rlist = cons(num rlist)
num = num-1
)
rlist
))

a = lrange(100000)

procedure( nElements(theList, n)
let( (x)
x = nil
for(i 1 n
x = tconc(x car(theList))
theList = cdr(theList)
)
car(x)
);let
);procedure

procedure(nElements2(theList n)
let((tail tailcdr)
tail=nthcdr(sub1 theList)
tailcdr=cdr(tail)
rplacd(tail nil)
prog1(
copy(theList)
rplacd(tail tailcdr)
)
)
)



x = measureTime(nElements(a 50000))
y = measureTime(nElements2(a 50000))