Deleting an element

[nour]

Hi everybody!
Please how can I delete an elelment of a skill list?

 

[Riad KACED]

Nour,

The following document will bring you from darkness to ... Nour
The Cadence SKILL Language User Guide which you can check out from
your UNIX terminal at the following location :
$CDSHOME/doc/sklanguser/sklanguser.pdf
-> Chapter 8 : Advanced List Operations
-> Section 9 : Removing Elements from a List
-> Line ... I'm too lazy to count the lines ;-)

Enjoy yourself !

PS : For those who are still wondering what : "from darkness to ...
Nour" means.
Nour in Arabic means light ;-)

 

[danniel]

On May 5, 6:37 pm, nour <nour.laou...@gmail.com> wrote:

Hi everybody!
Please how can I delete an elelment of a skill list?

Hi,
you can use the "remove" option:
for example -
lst=list(1 2 3 4)
lst2=remove(3 lst)

now

lst=(1 2 3 4) ; and
lst2=(1 2 4)

or you can use "remd":
remd(3 lst)

now

lst=(1 2 4)

 

[S. Badel]

Just a few more comments

you can use the "remove" option:
for example -
lst=list(1 2 3 4)
lst2=remove(3 lst)

note that this'll remove all elements equal to 3, so that
lst=list(1 2 3 2 3)
lst2=remove(3 lst)

will result in lst2='(1 2 2)

or you can use "remd":
remd(3 lst)

now

lst=(1 2 4)

remd is destructive, i.e. it modifies the list itself (while remove works on a copy of the list).
But there's a catch :

lst=list(1 2 3 4)
remd(1 lst)
=> (2 3 4)
lst
=> (1 2 3 4)

remd can not modify where lst is "pointing to", so it can't remove the car of a list. Therefore it's
safer to use

lst=remd(1 lst)


also, be careful with destructive operations ; consider that symbols are "pointing" to lists, and
therefore when two symbols are pointing to the same list, modifying one also affects the other :

lst=list(1 2 3 4)
lst2=lst ;; <= both are pointing to the same list here
remd(2 lst)
=> (1 3 4)
lst2
=> (1 3 4)


one last word : to destructively remove the element at index i in the list lst,

rplacd(rplacd(nthcdr(i-1 lst) nthcdr(i+1 lst))



Cheers,
Stéphane

 

[S. Badel]

rplacd(rplacd(nthcdr(i-1 lst) nthcdr(i+1 lst))

woops, typo

rplacd(nthcdr(i-1 lst) nthcdr(i+1 lst))

 

[Andrew Beckett]

S. Badel wrote, on 05/06/08 13:38:

rplacd(rplacd(nthcdr(i-1 lst) nthcdr(i+1 lst))

woops, typo

rplacd(nthcdr(i-1 lst) nthcdr(i+1 lst))

Also, you should consider whether a list is the right data structure if you're
deleting stuff from it - especially if removing elements based on a position in
the list as in Stéphane's example above.

If it's unordered, perhaps a hash (an "association table" in SKILL terminology)
makes more sense - as created by makeTable. You can also remove() elements from
tables.

Andrew.