strange mapping

[eddier]

Suppose

Code Select Expand


(define data '((1 2) (2 3) (3 4)))
(println (map first data))

=>

Code Select Expand

(1,2,3)

Okydoky! That's what I expect. But,

Code Select Expand

(println (map 0 data))
=>

Code Select Expand

((1 2) (2 3) (3 4))

and

Code Select Expand

(println (map 1 data))
=>

Code Select Expand

((2) (3) (4))

Looks like (map n list) means appling (n sublist) instead of (sublist n).

Shouldn't this be the other way around? That

Code Select Expand

(map 0 data) => (1 2 3) and

Code Select Expand

(map 1 data) => (2 3 4)?



Eddie[/i]

[Lutz]

The implicit indexing form (idx lst) doesn't mean 'nth' but is taking the 'slice' starting at index:

Code Select Expand


(0 '(1 2 3 4)) => (1 2 3 4)
(1 '(1 2 3 4)) => (2 3 4)

;; or

(1 2 '(1 2 3 4)) => (2 3)


and (map idx lst) behaves accordingly (the first part of the example)



what you mean is the implicit indexing form of 'nth' with the index behind the list:

Code Select Expand


(set 'data '(1 2 3 4 5))
(data 0) => 1
(data 1) = 2


Lutz



ps: see http://newlisp.org/downloads/newlisp_manual.html#indexing">http://newlisp.org/downloads/newlisp_ma ... l#indexing">http://newlisp.org/downloads/newlisp_manual.html#indexing

[eddier]

Yes,



I was wondering why doesn't map take implicit indexing instead of implicit slicing?



Would it not be more appropriate for map to take indexing?  Why would we ever want map to be slicing?



Thanks!



Eddie

[Lutz]

'map' always applies the first parameter to each element of the followoing list(s) parameter(s). You cannot suddenly change that sequence for the case of implicit indexing.



But you still could do:

Code Select Expand


(set 'data '(a b c d))

(map 'data '(3 2 1 0)) => (d c b a)


so everything works consistent. Note that data is quoted because 'map' evaluates the 1st parameter before applying it and so does implicit indexing.



mapping slice is useful for stripping all lists in a list of leading elements.



Lutz