Topic: 2D Point car/cdr

[David Bethel]

Can anyone think of a combination of car/cdr that could extract the X & Y values of a point group

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(setq p '(10 2 4 6))

-> (2 4)



Instead of the kludge :

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(setq p2 (list (nth 1 p) (nth 2 p))

Thanks  -David

[Lee Mac]

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(reverse (cdr (reverse (cdr p))))
Not sure its any less of a kludge.

Others:

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(list (cadr p) (caddr p))

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(mapcar 'nth '(1 2) (list p p))

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(mapcar 'progn '(0 0) (cdr p))

[David Bethel]

LOL

I looked at the double reverse - way ugly 

[ribarm]

I don't understand :

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(mapcar 'progn '(0 0) (cdr p))

I thought it should be :

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(mapcar '+ '(0 0) (cdr p))


[Lee Mac]

LOL

I looked at the double reverse - way ugly 

 

[Lee Mac]

I don't understand :

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(mapcar 'progn '(0 0) (cdr p))

I thought it should be :

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(mapcar '+ '(0 0) (cdr p))


Why?

I think it should be:

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(mapcar 'progn '(Lee Mac) (cdr p))

 

[ymg]

David,

FWIW,

(benchmark '((butlast (cdr p)) (list (nth 1 p) (nth 2 p))  (list (cadr p) (caddr p))))
Benchmarking ...................Elapsed milliseconds / relative speed for 65536 iteration(s):

    (BUTLAST (CDR P))..............1591 / 1.19 <fastest>
    (LIST (CADR P) (CADDR P))......1794 / 1.05
    (LIST (NTH 1 P) (NTH 2 P)).....1888 / 1.00 <slowest>

ymg

[MickD]

I'm a lisp neophyte but I prefer clarity over conciseness/speed for maintenance and comprehension reasons and I would do it something like this if I knew the point group list was always the same.

(note: this is in scheme but you get the picture)

(define (x-point pgroup)
  (cadr p))

(define (y-point pgroup)
  (caddr p))

output:
> (x-coord '(10 2 4 6))
2
> (y-coord '(10 2 4 6))
4

If more sample code was written like this (i.e. with simple helper functions) I think the logic behind the function would be easier to understand.
Don't mind me though, just having a play and learning from you guys