;;; <- can one use cons to do ((a . b) . (c . d))?
(define x (cons a b)); nil -- should it be error
(define x (cons 'a 'b)); (a . b)
(define y (cons 'c 'd)); (c . d)
(define z00 (cons x y)) ; (((a . b) c . d) <- cannot use cons to do ((a . b) . (c . d))?
(define z01 (cons x 'y)) ; ((a . b) . y)
(define z10 (cons 'x y)) ; (x c . d)
(define z11 (cons 'x 'y)); (x . y))
(define z (list x y z00 z01 z10 z11))
; ((a . b) (c . d) ((a . b) c . d) ((a . b) . y) (x c . d) (x . y))
;;; and if not any other means or dot pair cannot have 2nd element like this?
Yes, you can. And the language has a wonderful predicate called equal? which will allow you to test this:
> (equal? (cons (cons 'a 'b) (cons 'c 'd))
'((a . b) . (c . d)))
#t
> (equal? '((a . b) . (c . d))
'((a . b) c . d))
#t
And you can even write a little display function which will confirm this:
(define (display-thing thing)
(if (cons? thing)
(begin
(display "(")
(display-thing (car thing))
(display " . ")
(display-thing (cdr thing))
(display ")"))
(display thing)))
And now
> (display-thing (cons (cons 'a 'b) (cons 'c 'd)))
((a . b) . (c . d))
> (display-thing '((a . b) . (c . d)))
((a . b) . (c . d))
> (display-thing '((a . b) c . d))
((a . b) . (c . d))
What this should all be telling you is that ((a . b) . (c . d)) and ((a . b) c . d) are merely different ways of writing a structurally identical object.
Pairs visualize differently based on their content. If the cdr of a pair contains the empty list it is a proper list an dthe dot and extra empty list is not shown:
(cons 'a '())
'(a . ())
; ==> (a)
A pair that has pair as it's cdr can be visualized as a list element without the . and extra parenthesis:
(cons 'b '(a))
'(b . (a))
; ==> (b a)
(cons 'b '(a . c))
'(b . (a . c))
; ==> (b a . c)
These are just made so that we can have (1 2 3) displayed instead of (1 . (2 . (3 . ()))) which is how it really is made.
If you were to not have a pair or a empty list in the cdr then it falls back to showing the dotted pair:
(cons 'a 'b)
'(a . b)
; ==> (a . b)
In your example '((a . b) . (c . d)) because there is a pair after a dot (eg. the cdr if the pair the visualization will remove the dot and one pair of parentheses and show it like ((a . b) c . d). This is the only acceptable correct way for a REPL to display this even though both your example and the display will be read in as the same structure.
There is a similar issue with numbers. In code you can use 10, #xa and #o12 to get the number 10 and the value will have no idea what format is was read in as and only show the base 10 in the REPL.
;;; ```
;;; <- can one use cons to do ((a . b) . (c . d))?
(define x (cons a b)); nil -- should it be error
(define x (cons 'a 'b)); (a . b)
(define y (cons 'c 'd)); (c . d)
(define z00 (cons x y)) ; (((a . b) c . d) <- cannot use cons to do ((a . b) . (c . d))?
(define z01 (cons x 'y)) ; ((a . b) . y)
(define z10 (cons 'x y)) ; (x c . d)
(define z11 (cons 'x 'y)); (x . y))
(define z22 (cons '(f g) '(h i)))
(define z2c (cons (cons 'f 'g) (cons 'h 'i)))
(define fgc (cons ('f 'g))); should it be error (nil)
; actually not as 'f is an exoression (quote f) and f for some reason is nil it becomes nil from quote of nil abd so is the second one. now (nil . nil) is (nil)
(define z (list x y z00 z01 z10 z11 z22 z2c fgc))
; ((a . b) (c . d) ((a . b) c . d) ((a . b) . y) (x c . d) (x . y))
;;; ```
;;; and if not any other means or dot pair cannot have 2nd element like this?
;;; possibly not
;;; as the cons join 2 pairs of dotted pairs and can generate one dotted pair
;;; ((a . b) . (c . d)) but the printing rule is reflected the list bias
;;; this new dotted pair will have the first element as (( a . b) ... print as
;;; ((a . b) ...
;;; the 2nd element it will consider whether it is an atom or another dotted pair
;;; (other possibilities like loop back or something else ... not sure)
;;; as (c . d) is a dotted pair the "printing" continues as a list would
;;;
;;; ((a . b) c ...
;;; however the second element of (c . d) is not a dotted pair but an atom and print as
;;; . d) will it becomes
;;; hence even though you form the binary tree head the dot pair would display as a partial list
;;; you can have a list of dotted pairs like z
;;; but not dotted pair of dotted pairs
Related
Is there any built-in function in LISP languages (or Racket in particular) that would work like map, but pass index of the element as one of the arguments to the mapping function?
Example of such function would be:
(define map-index (lambda (func list)
(map func list (build-list (length list) (lambda (i) i)))))
;usage:
> (map-index cons '(a b c d))
;output:
'((a . 0) (b . 1) (c . 2) (d . 3))
Obviously this is not a very efficient implementation and doesn't support multiple lists as arguments, like regular map does.
Racket
You can write a dead-simple version of map-index using the Racket range procedure and mapping over the result:
(define (map-index-1 f xs)
(map f xs (range (length xs))))
In some situations you might want the indices first:
(define (map-index-2 f xs)
(map f (range (length xs)) xs))
If you want to be able to use map-index over multiple lists, you can pass the list arguments to an optional parameter. Here, apply applies the map procedure to a list constructed from the function f, the input lists, and a range list:
(define (map-index-3 f . xs)
(apply map (cons f
(append xs
(list (range (length (car xs))))))))
But it might make more sense to place the indices first when mapping over multiple lists:
(define (map-index-4 f . xs)
(apply map (cons f
(cons (range (length (car xs)))
xs))))
scratch.rkt> (map-index-1 cons '(a b c d))
'((a . 0) (b . 1) (c . 2) (d . 3))
scratch.rkt> (map-index-2 cons '(a b c d))
'((0 . a) (1 . b) (2 . c) (3 . d))
scratch.rkt> (map-index-3 list '(a b c d) '(one two three four) '(w x y z))
'((a one w 0) (b two x 1) (c three y 2) (d four z 3))
scratch.rkt> (map-index-4 list '(a b c d) '(one two three four) '(w x y z))
'((0 a one w) (1 b two x) (2 c three y) (3 d four z))
Scheme
Standard Scheme does not have a built-in range procedure, but it is easy enough to write a simple version. These solutions will work on any R4RS, R5RS, R6RS, or R7RS Scheme implementation. This version of range does more than is required for the current application, taking a step argument which can be positive or negative:
(define (range start stop step)
(if (or (and (> step 0)
(>= start stop))
(and (<= step 0)
(<= start stop)))
'()
(cons start (range (+ start step) stop step))))
Having defined a range procedure, the same approach used for the Racket solutions above can be used in Scheme:
(define (map-index-5 f xs)
(map f xs (range 0 (length xs) 1)))
(define (map-index-6 f xs)
(map f (range 0 (length xs) 1) xs))
(define (map-index-7 f . xs)
(apply map (cons f
(append xs
(list (range 0 (length (car xs)) 1))))))
(define (map-index-8 f . xs)
(apply map (cons f
(cons (range 0 (length (car xs)) 1)
xs))))
> (map-index-5 cons '(a b c d))
((a . 0) (b . 1) (c . 2) (d . 3))
> (map-index-6 cons '(a b c d))
((0 . a) (1 . b) (2 . c) (3 . d))
> (map-index-7 list '(a b c d) '(one two three four) '(w x y z))
((a one w 0) (b two x 1) (c three y 2) (d four z 3))
> (map-index-8 list '(a b c d) '(one two three four) '(w x y z))
((0 a one w) (1 b two x) (2 c three y) (3 d four z))
A map-range Procedure for Standard Scheme
This method can be extended to use numbers in a more complex range that may not represent indices by taking advantage of the capabilities of a range function. Using the range definition from above, this map-range procedure still works on R4RS to R7RS Scheme implementations:
(define (map-range f start step . xs)
(let ((stop (+ start (* step (length (car xs))))))
(apply map (cons f
(cons (range start stop step)
xs)))))
> (map-range cons 2 2 '(a b c d))
((2 . a) (4 . b) (6 . c) (8 . d))
> (map-range list 5 5 '(a b c d) '(one two three four) '(w x y z))
((5 a one w) (10 b two x) (15 c three y) (20 d four z))
> (map-range cons 2 -2 '(a b c d))
((2 . a) (0 . b) (-2 . c) (-4 . d))
> (map-range list 5 -5 '(a b c d) '(one two three four) '(w x y z))
((5 a one w) (0 b two x) (-5 c three y) (-10 d four z))
Not exactly, but there are other things for similar goals, like for/list with in-naturals or in-indexed.
For example instead of (map-index f lst), the pattern would be
(for/list ([x lst] [i (in-naturals)]) (f x i))
or
(for/list ([(x i) (in-indexed lst)]) (f x i))
And either of those patterns could be used to implement map-index as well as your combination of map and build-list.
Racket's iteration forms like for/list are more flexible than a fixed set of map-like functions.
Concrete Examples:
> (for/list ([x '(a b c d)] [i (in-naturals)]) (cons x i))
'((a . 0) (b . 1) (c . 2) (d . 3))
> (for/list ([(x i) (in-indexed '(a b c d))]) (cons x i))
'((a . 0) (b . 1) (c . 2) (d . 3))
Or if you still want a map-index function you could define it more succinctly using this.
> (define (map-index f lst)
(for/list ([(x i) (in-indexed lst)]) (f x i)))
> (map-index cons '(a b c d))
'((a . 0) (b . 1) (c . 2) (d . 3))
Given a list of lists as an input, I want to execute a procedure such that the final result would be:
(define (thing . lists) ; list of lists (l1 l2 ... lN)
;returns ...f(f(f(l1 l2) l3) lN)...
)
So for example:
(thing '(a b) '(c d) '(e f))
...would result in f(f((a b) (c d)) (e f))
I am fighting with folding, lambda, apply and map, but I can't figure out right way.
Assuming that the input has at least two lists and that f was previously defined:
(define (thing . lists)
(foldr (lambda (lst acc)
(f acc lst))
(f (car lists) (cadr lists))
(cddr lists)))
For example:
(define f append)
(thing '(a b) '(c d) '(e f))
=> '(a b c d e f)
I'm new in scheme programming and I just want to extract the Items of the pair List. Example my variable
val==> ((a . 1) (b . 0.026279533) (c . 0.026616231) (d . 0.0060348272) (e . 0.00070986058) (f . 0.050281039) (g . 0.12546714) (h . 0.00014385414))
And I want to print into two list type:
a b c d e f g h
1 0.026279533 0.026616231 0.0060348272 0.00070986058 0.050281039 0.12546714 0.00014385414
(display (map car val))
(newline)
(display (map cdr val))
(newline)
Hi guys i'm wanted to know if i have the correct expression for this picture, if not why please
(a((f(b c))(g h))e)
You're close, but not quite right. It'll be more clear if we build the list structure explicitly using cons; this is more like it:
(cons 'a
(cons (cons (cons 'f
(cons 'b 'c))
(cons 'g
(cons 'h '())))
(cons 'e '())))
=> '(a ((f b . c) g h) e)
Notice that in this part: (f b . c) we have an improper list, because the sublist doesn't end in null.
You answer is incorrect as it doesn't properly express the improper list (f b . c). Also the parentheses around g h are an error.
With dotted pairs the full expression would be:
'(a ((f b . c) g h) e)
Note that '(f b . c) is not the same as '(f (b c)).
See that '(f (b c)) is:
(cons 'f (cons (cons 'b (cons 'c '())) '()))
Rather than what '(f b . c) is:
(cons 'f (cons (cons 'b 'c) '()))
Note the improper list.
Given an s-expression '((a . b) . (c . d)) and a list '(e f g h), how can I traverse the s-expression create an s-expression with the same shape, but with elements taken from the list? E.g., for the s-expression and list above, the result would be '((e . f) g . h)?
Traversing a tree of pairs in left to right order isn't particularly difficult, as car and cdr let you get to both sides, and cons can put things back together. The tricky part in a problem like this is that to "replace" elements in the right hand side of a tree, you need to know how many of the available inputs you used when processing the left hand side of the tree. So, here's a procedure reshape that takes a template (a tree with the shape that you want) and a list of elements to use in the new tree. It returns as multiple values the new tree and any remaining elements from the list. This means that in the recursive calls for a pair, you can easily obtain both the new left and right subtrees, along with the remaining elements.
(define (reshape template list)
;; Creates a tree shaped like TEMPLATE, but with
;; elements taken from LIST. Returns two values:
;; the new tree, and a list of any remaining
;; elements from LIST.
(if (not (pair? template))
(values (first list) (rest list))
(let-values (((left list) (reshape (car template) list)))
(let-values (((right list) (reshape (cdr template) list)))
(values (cons left right) list)))))
(reshape '((a . b) . (c . d)) '(e f g h))
;=> ((e . f) g . h)
;=> ()
(reshape '((a . b) . (c . d)) '(e f g h i j k))
;=> ((e . f) g . h)
;=> (i j k) ; leftovers
I'll assume that you want to create a new s-expression with the same shape of the s-expression given as the first parameter, but with the elements of the list from the second parameter.
If that's right, here's one possible solution using a list to save the point where we are in the replacement list and Racket's begin0 to keep the list updated (if that's not available in you interpreter use a let, as suggested by Chris and Joshua in the comments):
(define (transform sexp lst)
(let loop ((sexp sexp)) ; the s-expression list to be traversed
(cond ((null? sexp) '()) ; if it's empty, we're finished
((not (pair? sexp)) ; if it's an atom
(begin0 ; then (alternatively: use a `let`)
(car lst) ; return first element in replacements list
(set! lst (cdr lst)))) ; and update replacements to next element
(else ; otherwise advance recursion
(cons (loop (car sexp)) ; over both the `car` part of input
(loop (cdr sexp))))))) ; and the `cdr` part
For example:
(transform '((a . b) . (c . d)) '(e f g h))
=> '((e . f) g . h)
(transform '((a . b) (c d (x y) . z) . t) '(e f g h i j k m))
=> '((e . f) (g h (i j) . k) . m)
The solution is similar to my previous answer:
(define (transform sxp lst)
(let loop ((sxp sxp))
(cond ((null? sxp) sxp)
((pair? sxp) (cons (loop (car sxp)) (loop (cdr sxp))))
(else (begin0 (car lst) (set! lst (cdr lst)))))))
then
> (transform '((a . b) . (c . d)) '(e f g h))
'((e . f) g . h)