I could manage following code to replaces items in a list using 2 other lists. Orilist and newlist have original and new terms in order. The replacement is done using orilist and newlist- if orilist items are present in slist, slist is changed to have corresponding new items from newlist:
(define (list-replace-from-lists slist orilist newlist)
(define replaced #f)
(define outl '())
(for ((item slist))
(set! replaced #f)
(for ((ori_ orilist) (i (in-naturals)) #:when (equal? item ori_))
(set! outl (cons (list-ref newlist i) outl))
(set! replaced #t))
(when (not replaced)
(set! outl (cons item outl))))
(reverse outl))
To replace 2 and 5 to 12 and 15, respectively, in (list 1 2 3 4 5 6) :
(list-replace-from-lists (list 1 2 3 4 5 6) (list 2 5) (list 12 15))
Output is:
'(1 12 3 4 15 6)
However, above code does not look functional and has many set! statements. How can this be converted to functional code? Should I use structures or some other data-types for above purpose?
Edit: items may recur in original list, e.g. (list 1 2 3 4 5 2 6)
You can still use lists and keep everything functional. :-) Here's my solution:
(define (replace-all haystack needles new-needles)
(define replace-alist (map cons needles new-needles))
(define (replace-one item)
(cond ((assoc item replace-alist) => cdr)
(else item)))
(map replace-one haystack))
Explanation of the code:
First, we build a replacement association list (alist). This is a list of pairs, of which the keys correspond to the needles and the values correspond to new-needles.
Then we define a replace-one function that takes an item, and sees if it matches any of the keys in the alist. If so, we return the corresponding value; otherwise, we return the original item.
Finally, we map the haystack through replace-one. Yay higher-order functions!
Note that this code is O(m*n) where m is the size of haystack and n is the size of needles, which is the same runtime as your version. If needles is large, you will want to use a hashtable instead of an alist, which will amortise the runtime of the function to O(m).
This is a functional solution that uses hash to keep the associations. That makes this solution O(haystack-length log needle-length) since immutable hashes are implemented with trees.
(define (list-replace-all haystack needles new-values)
;; make a dictionary of elements to be replaced
(define hash
(foldl (λ (needle new-value hash)
(hash-set hash needle new-value))
#hash()
needles
new-values))
;; do the replace. If not in hash the actual key is default
(map (λ (e) (hash-ref hash e e)) haystack))
(list-replace-all '(1 2 3 4 5 6) '(2 5) '(12 15))
; ==> (1 12 3 4 15 6)
Related
According to the book, this is what the function definition is,
The function scramble takes a non-empty tuple in which no argument is greater than its own index and returns a tuple of same length. Each number in the argument is treated as a backward index from its own position to a point earlier in tuple. The result at each position is obtained by counting backward from the current position according to this index.
And these are some examples,
; Examples of scramble
(scramble '(1 1 1 3 4 2 1 1 9 2)) ; '(1 1 1 1 1 4 1 1 1 9)
(scramble '(1 2 3 4 5 6 7 8 9)) ; '(1 1 1 1 1 1 1 1 1)
(scramble '(1 2 3 1 2 3 4 1 8 2 10)) ; '(1 1 1 1 1 1 1 1 2 8 2)
Here is the implementation,
(define pick
(λ (i lat)
(cond
((eq? i 1) (car lat))
(else (pick (sub1 i)
(cdr lat))))))
(define scramble-b
(lambda (tup rev-pre)
(cond
((null? tup) '())
(else
(cons (pick (car tup) (cons (car tup) rev-pre))
(scramble-b (cdr tup)
(cons (car tup) rev-pre)))))))
(define scramble
(lambda (tup)
(scramble-b tup '())))
This is a case where using a very minimal version of the language means that the code is verbose enough that understanding the algorithm is not perhaps easy.
One way of dealing with this problem is to write the program in a much richer language, and then work out how the algorithm, which is now obvious, is implemented in the minimal version. Let's pick Racket as the rich language.
Racket has a function (as does Scheme) called list-ref: (list-ref l i) returns the ith element of l, zero-based.
It also has a nice notion of 'sequences' which are pretty much 'things you can iterate over' and a bunch of constructs whose names begin with for for iterating over sequences. There are two functions which make sequences we care about:
in-naturals makes an infinite sequence of the natural numbers, which by default starts from 0, but (in-naturals n) starts from n.
in-list makes a sequence from a list (a list is already a sequence in fact, but in-list makes things clearer and there are rumours also faster).
And the iteration construct we care about is for/list which iterates over some sequences and collects the result from its body into a list.
Given these, then the algorithm is almost trivial: we want to iterate along the list, keeping track of the current index and then do the appropriate subtraction to pick a value further back along the list. The only non-trivial bit is dealing with zero- vs one-based indexing.
(define (scramble l)
(for/list ([index (in-naturals)]
[element (in-list l)])
(list-ref l (+ (- index element) 1))))
And in fact if we cause in-naturals to count from 1 we can avoid the awkward adding-1:
(define (scramble l)
(for/list ([index (in-naturals 1)]
(element (in-list l)))
(list-ref l (- index element))))
Now looking at this code, even if you don't know Racket, the algorithm is very clear, and you can check it gives the answers in the book:
> (scramble '(1 1 1 3 4 2 1 1 9 2))
'(1 1 1 1 1 4 1 1 1 9)
Now it remains to work out how the code in the book implements the same algorithm. That's fiddly, but once you know what the algorithm is it should be straightforward.
If the verbal description looks vague and hard to follow, we can try following the code itself, turning it into a more visual pseudocode as we go:
pick i [x, ...ys] =
case i {
1 --> x ;
pick (i-1) ys }
==>
pick i xs = nth1 i xs
(* 1 <= i <= |xs| *)
scramble xs =
scramble2 xs []
scramble2 xs revPre =
case xs {
[] --> [] ;
[x, ...ys] -->
[ pick x [x, ...revPre],
...scramble2 ys
[x, ...revPre]] }
Thus,
scramble [x,y,z,w, ...]
=
[ nth1 x [x] (*x=1..1*)
, nth1 y [y,x] (*y=1..2*)
, nth1 z [z,y,x] (*z=1..3*)
, nth1 w [w,z,y,x] (*w=1..4*)
, ... ]
Thus each element in the input list is used as an index into the reversed prefix of that list, up to and including that element. In other words, an index into the prefix while counting backwards, i.e. from the element to the left, i.e. towards the list's start.
So we have now visualized what the code is doing, and have also discovered requirements for its input list's elements.
It's very simple program which just return the input as a list shuffled. I wrote this program in python. Now I want to convert this program to lisp code. but I couldn't. How do I write down this program in lisp?
def my_shuffle(a, b, c, d):
return [b, c, d, a]
I tried the following code but an error occur.
(defun my_shuffle (a b c d) (list b c d a))
Thee are several things here that I think that need to be pointing out. First the code that you presented is correct but do shuffle a list, present a new list of four algorithms that you pass, allways with the same order. First of all shuffle a sequence is:
generating a random permutation of a finite sequence
From wikipedia you can find several algorithms for that:
https://en.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle
Also in the rosseta code there is an implementation of the knuth shuffle:
(defun nshuffle (sequence)
(loop for i from (length sequence) downto 2
do (rotatef (elt sequence (random i))
(elt sequence (1- i))))
sequence)
Then if you apply this in the repl:
CL-USER> (nshuffle (list 1 2 3 4))
(3 1 4 2)
CL-USER> (nshuffle (list 1 2 3 4))
(3 1 2 4)
Note Two different results on the same list!!! (also the same can happen, because is a random order)
In python there are build algorithms for that:
https://docs.python.org/3/library/random.html#random.shuffle
also in the Common lisp library Alexandria:
CL-USER> (ql:quickload :alexandria)
To load "alexandria":
Load 1 ASDF system:
alexandria
; Loading "alexandria"
(:ALEXANDRIA)
CL-USER> (alexandria:shuffle (list 1 2 3 4))
(3 2 4 1)
(defun my_shuffle (a b c d) (list b c d a))
The above code defines a function which will take 4 items and return a rearranged list of those 4 items. It can take input of 4 lists, 4 atoms, 4 numbers, 4 anything, but it cannot separate sublists present inside a single list.
What you can do is:
(defun my_shuffle (myList)
(list (second myList) (third myList) (fourth myList) (first myList)))
or
(defun my_shuffle (myList)
(list (cadr myList) (caddr myList) (cadddr myList) (car myList)))
or
(defun my_shuffle (myList)
(list (nth 1 myList) (nth 2 myList) (nth 3 myList) (nth 1 myList)))
car returns the first element of a list
cdr returns the tail of a list (part of the list following car of the list)
I have used combinations of car and cdr to extract the different elements of the list. Find that in your textbook.
first, second, third, fourth are relatively easy to use and do the same thing as car, cadr, caddr and cadddr
(nth x list) returns the (x+1)th item of the list, counting from zero.
So,
(nth 3 (list a b c d)) => d
(nth 0 (list a b c d)) => a
and so on.
I am trying to find the length of a list using Map/Foldl/Foldr
(define (suml lst)
(length lst))
Input : (suml '(1 2 3))
Output : 3
Input : (suml '(((((2)))) (1)))
Output: 2
How can I modify it work with foldl/map/foldr?
As has been mentioned in the comments, map takes a function and applies it elementwise. A function that uses map will create a list of the same length. To create a length function, we are condensing a list to a single value. This is the purpose of fold.
(define (length l)
(foldr (lambda (_ cur-length) (+ 1 cur-length)) 0 l))
When you think about foldr, you should think about it just replacing cons in a list with the function and the empty list with the base case argument. Take the following example:
'(1 2 3 4)
= (cons 1 (cons 2 (cons 3 (cons 4 '()))))
(foldr f base '(1 2 3 4))
= (f 1 (f 2 (f 3 (f 4 base))))
It turns out, foldl also works in this case because we're just adding one for every element, it doesn't matter if we go left to right or right to left.
(define (length l)
(foldl (lambda (_ cur-length) (+ 1 cur-length)) 0 l))
I am working on program related to the different of dealing with even numbers in C and lisp , finished my c program but still having troubles with lisp
isprime function is defined and I need help in:
define function primesinlist that returns unique prime numbers in a lis
here what i got so far ,
any help with that please?
(defun comprimento (lista)
(if (null lista)
0
(1+ (comprimento (rest lista)))))
(defun primesinlist (number-list)
(let ((result ()))
(dolist (number number-list)
(when (isprime number)
( number result)))
(nreverse result)))
You need to either flatten the argument before processing:
(defun primesinlist (number-list)
(let ((result ()))
(dolist (number (flatten number-list))
(when (isprime number)
(push number result)))
(delete-duplicates (nreverse result))))
or, if you want to avoid consing up a fresh list, flatten it as you go:
(defun primesinlist (number-list)
(let ((result ()))
(labels ((f (l)
(dolist (x l)
(etypecase x
(integer (when (isprime x)
(push x result)))
(list (f x))))))
(f number-list))
(delete-duplicates (nreverse result))))
To count distinct primes, take the length of the list returned by primesinlist.
Alternatively, you can use count-if:
(count-if #'isprime (delete-duplicates (flatten number-list)))
It sounds like you've already got a primality test implemented, but for sake of completeness, lets add a very simple one that just tries to divide a number by the numbers less than it up to its square root:
(defun primep (x)
"Very simple implementation of a primality test. Checks
for each n above 1 and below (sqrt x) whether n divides x.
Example:
(mapcar 'primep '(2 3 4 5 6 7 8 9 10 11 12 13))
;=> (T T NIL T NIL T NIL NIL NIL T NIL T)
"
(do ((sqrt-x (sqrt x))
(i 2 (1+ i)))
((> i sqrt-x) t)
(when (zerop (mod x i))
(return nil))))
Now, you need a way to flatten a potentially nested list of lists into a single list. When approaching this problem, I usually find it a bit easier to think in terms of trees built of cons-cells. Here's an efficient flattening function that returns a completely new list. That is, it doesn't share any structure with the original tree. That can be useful, especially if we want to modify the resulting structure later, without modifying the original input.
(defun flatten-tree (x &optional (tail '()))
"Efficiently flatten a tree of cons cells into
a list of all the non-NIL leafs of the tree. A completely
fresh list is returned.
Examples:
(flatten-tree nil) ;=> ()
(flatten-tree 1) ;=> (1)
(flatten-tree '(1 (2 (3)) (4) 5)) ;=> (1 2 3 4 5)
(flatten-tree '(1 () () 5)) ;=> (1 5)
"
(cond
((null x) tail)
((atom x) (list* x tail))
((consp x) (flatten-tree (car x)
(flatten-tree (cdr x) tail)))))
Now it's just a matter of flatting a list, removing the number that are not prime, and removing duplicates from that list. Common Lisp includes functions for doing these things, namely remove-if-not and remove-duplicates. Those are the "safe" versions that don't modify their input arguments. Since we know that the flattened list is freshly generated, we can use their (potentially) destructive counterparts, delete-if-not and delete-duplicates.
There's a caveat when you're removing duplicate elements, though. If you have a list like (1 3 5 3), there are two possible results that could be returned (assuming you keep all the other elements in order): (1 3 5) and (1 5 3). That is, you can either remove the the later duplicate or the earlier duplicate. In general, you have the question of "which one should be left behind?" Common Lisp, by default, removes the earlier duplicate and leaves the last occurrence. That behavior can be customized by the :from-end keyword argument. It can be nice to duplicate that behavior in your own API.
So, here's a function that puts all those considerations together.
(defun primes-in-tree (tree &key from-end)
"Flatten the tree, remove elements which are not prime numbers,
using FROM-END to determine whether earlier or later occurrences
are kept in the list.
Examples:
(primes-in-list '(2 (7 4) ((3 3) 5) 6 7))
;;=> (2 3 5 7)
(primes-in-list '(2 (7 4) ((3 3) 5) 6 7) :from-end t)
;;=> (2 7 3 5)"
;; Because FLATTEN-TREE returns a fresh list, it's OK
;; to use the destructive functions DELETE-IF-NOT and
;; DELETE-DUPLICATES.
(delete-duplicates
(delete-if-not 'primep (flatten-tree list))
:from-end from-end))
If I have the following list:
(define thelist '(0 1 0 0 7 7 7))
How can I write a function that returns a new list in which the value in the requested cell is replaced.
Example:
(set-cell thelist 2 4)
This would return a new list with the same values, but in cell (2) there will be the value 4 instead of 1:
(0 4 0 0 7 7 7)
HtDP provides a really concrete methodology for solving this kind of problem. For this problem, your job is going to be to write down the template for lists, then to stare at it until you can see what the arguments to the recursive call should be, and what the results will be. I'm hoping that you've solved a bunch of warm-up problems on lists--compute the length of a list, count the number of 6's in the list, etc.
Although you can implement the requested functionality with lists, the natural way to solve that problem is to use a vector, and remember that in Scheme indexes start in 0 (that's why the second argument for vector-set! is a 1 and not a 2):
(define thevector (vector 0 1 0 0 7 7 7))
; thevector is #(0 1 0 0 7 7 7)
(vector-set! thevector 1 4)
; thevector is #(0 4 0 0 7 7 7)
Now, if you definitely need to use a list, something like this would work:
(define (set-cell lst idx val)
(cond ((null? lst) '())
((zero? idx) (cons val (cdr lst)))
(else (cons (car lst)
(set-cell (cdr lst) (sub1 idx) val)))))
And you'd call it like this:
(define thelist '(0 1 0 0 7 7 7))
(set-cell thelist 1 4)
> (0 4 0 0 7 7 7)
Again, I'm using 0-based indexing, as is the convention. Also notice that thelist was not modified, instead, set-cell returns a new list with the modification.
Other people have mentioned the 0-based indexes convention. I'll assume 0-based indexes, then.
The other answers you've gotten are wrong, given how you stated your problem. I'll cite the key point in your question (my emphasis):
This would return a new list with the same values, but in cell (2) there will be the value 4 instead of 1.
The answers you've been given are modifying a list in place, not returning a newly constructed list while leaving the original intact!
Here's a correct (though suboptimal) solution:
(define (set-list index new-value list)
(if (= index 0)
(cons new-value (cdr list))
(cons (car list)
(set-list (- index 1) new-value (cdr list))))))
I didn't bother to put in error checking here (e.g., if somebody passes in the empty list, or a list with too few elements relative to index, or a negative index).
Here is an approach using an auxiliary function that starts the count from 0. On each recursion, the count is incremented, and the list is shortened.
If the desired index is hit, the recursion can stop. If it gets to the end of the list without hitting the index, the original list is returned.
(define (set-cell ls ind val)
(define (aux c ls)
(cond ((null? ls) ls)
((= c ind) (cons val (cdr ls)))
(else (cons (car ls) (aux (+ c 1) (cdr ls))))))
(aux 0 ls))
What a mess above!! Keep it simple: Scheme has list-set! for this.
Example:
(define ls (list 1 2 3 4 5))
(list-set! ls 0 "new")
(list-set! ls 3 "changed")
(list-ref ls 0)
=> "new"
ls
=> '("new" 2 3 "changed" 5)
However, you might want to limit assignments on lists. Vectors are faster for this since you don't need to traverse the previous elements. Use lists whenever you can't specify the length before construction or when the length is continously changed.