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Is there a function to get first several elements of a list?

richardyuan

I need to get part of a list. Is there a function in skill? It's better return a list.

I tried some coding, but failed:

a = list((0:0) (1:1) (2:2) (3:3) (4:2) (5:0))

b = nthcdr(3 a) 

c = remove(b a)

As I suppose, c should be a list of ((0:0) (1:1) (2:2)). But actually c is equal to a. 

What's wrong?

 

Parents
  • Andrew Beckett

    You can do this destructively (which means it is quite efficient) using the following:

    rplacd(nthcdr(2 a) nil)

    a =>  ((0 0) (1 1) (2 2))

    If you wanted a non-destructive list range function, you could implement it like this:

    procedure(CCFlistRange(lst start end)
      let((tail)
        tail=copy(nthcdr(start lst))
        rplacd(nthcdr(end-start tail) nil)
        tail
      )
    )

    Whether this would be more efficient than doing something like:

    procedure(CCFlistRange(lst start end)
      let(((count 0))
        setof(elem lst prog1(count>=start && count<=end count++))
      )
    )

    Would need some profiling and testing with various list sizes.

    Regards,

    Andrew.

     

Reply
  • Andrew Beckett

    You can do this destructively (which means it is quite efficient) using the following:

    rplacd(nthcdr(2 a) nil)

    a =>  ((0 0) (1 1) (2 2))

    If you wanted a non-destructive list range function, you could implement it like this:

    procedure(CCFlistRange(lst start end)
      let((tail)
        tail=copy(nthcdr(start lst))
        rplacd(nthcdr(end-start tail) nil)
        tail
      )
    )

    Whether this would be more efficient than doing something like:

    procedure(CCFlistRange(lst start end)
      let(((count 0))
        setof(elem lst prog1(count>=start && count<=end count++))
      )
    )

    Would need some profiling and testing with various list sizes.

    Regards,

    Andrew.

     

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