I want to write a function in lisp that reverses all elements from the list using map functions but I don't have any idea how to start this.. I think I have to use the built in reverse function somehow..
For example if I have the list (1 2 3 (4 5 6 (7 8 9))) I would get (((9 8 7) 6 5 4) 3 2 1)
or if I had the list(1 2 3 (4 5) (6 7)) I would get ((7 6) (5 4) 3 2 1) ..
Any help is appreciated!
Just a quick answer, not sure about efficiency/elegancy:
(defun reverse-deeply (list)
(mapcar #'(lambda (li)
(cond
((consp li) (reverse-deeply li))
(t li)))
(reverse list)))
(defun reverse-list (list)
(let ((result nil))
(dolist (e list result)
(push e result))))
Here is a version that works for me in Common-Lisp.
(defun reverse-list (list)
(if (atom list)
list ;; Not actually a list, return the atom
(reverse (mapcar #'reverse-list list)))
;; Testing it out
(reverse-list '((1 2 3) (4 5 (3 6))))
Output:
(((6 3) 5 4) (3 2 1))
Mapcar is a function that takes another function as its first parameter and a list as its second parameter. It then calls that function on each element of the list. It returns a list of all of the answers. So after I use 'mapcar' to reverse all of the sublists, I call 'reverse' again to reverse the bigger list.
The function that it calls on each sublist is 'reverse-list'. This checks if the list is an atom. If it is, then it returns itself. If it's a list, then it calls mapcar again on each element in the list, then reverses the result.
Related
This question already has answers here:
Delete element from List in Scheme
(3 answers)
Closed 5 years ago.
In Racket using plai-typed, I am trying to find an element in a list and remove the first one it finds. It will then return the list with the first element it sees removed.
Here is an example:
'(1 2 3 2 5)
Filter out the first 2 and you should get:
'(1 3 2 5)
This is what I am currently doing, but it returns: '(1 3 5)
(filter (lambda (x) (not (equal? x 2))) '(1 2 3 2 5))
So what I am currently doing removes all the occurrences when I just want to remove the first one.
[edit]If you did not want to use that built in function you can do
(define (remove-1st-helper l n)
(cond [(empty? l) empty]
[(= 1 n) l]
[(= 2 (first l)) (remove-1st-helper (rest l)(add1 n))]
[else (cons (first l) (remove-1st-helper (rest l) n))]))
(define (remove-1st-occurence l)
(if (empty? l) empty (remove-1st-helper l 0)))
Don't know Racket, but maybe you can define it as a piece-wise function like this:
-- Define myFilter of an empty list to be empty
myFilter [] = []
-- Decomposes list into first element and rest of list
myFilter (x:xs)
-- Ignore `x` but include the rest of the list
| x == 2 = xs
-- Include x and apply myFilter on xs
| otherwise = x:(myFilter xs)
(Haskell code).
So i was asked to do a function i LISP that calculates the average of any given numbers. The way i was asked to do this was by using the &rest parameter. so i came up with this :
(defun average (a &rest b)
(cond ((null a) nil)
((null b) a)
(t (+ (car b) (average a (cdr b))))))
Now i know this is incorrect because the (cdr b) returns a list with a list inside so when i do (car b) it never returns an atom and so it never adds (+)
And that is my first question:
How can i call the CDR of a &rest parameter and get only one list instead of a list inside a list ?
Now there is other thing :
When i run this function and give values to the &rest, say (average 1 2 3 4 5) it gives me stackoverflow error. I traced the funcion and i saw that it was stuck in a loop, always calling the function with the (cdr b) witch is null and so it loops there.
My question is:
If i have a stopping condition: ( (null b) a) , shouldnt the program stop when b is null and add "a" to the + operation ? why does it start an infinite loop ?
EDIT: I know the function only does the + operation, i know i have to divide by the length of the b list + 1, but since i got this error i'd like to solve it first.
(defun average (a &rest b)
; ...
)
When you call this with (average 1 2 3 4) then inside the function the symbol a will be bound to 1 and the symbol b to the proper list (2 3 4).
So, inside average, (car b) will give you the first of the rest parameters, and (cdr b) will give you the rest of the rest parameters.
But when you then recursively call (average a (cdr b)), then you call it with only two arguments, no matter how many parameters where given to the function in the first place. In our example, it's the same as (average 1 '(3 4)).
More importantly, the second argument is now a list. Thus, in the second call to average, the symbols will be bound as follows:
a = 1
b = ((3 4))
b is a list with only a single element: Another list. This is why you'll get an error when passing (car b) as argument to +.
Now there is other thing : When i run this function and give values to the &rest, say (average 1 2 3 4 5) it gives me stackoverflow error. I traced the funcion and i saw that it was stuck in a loop, always calling the function with the (cdr b) witch is null and so it loops there. My question is:
If i have a stopping condition: ( (null b) a) , shouldnt the program stop when b is null and add "a" to the + operation ? why does it start an infinite loop ?
(null b) will only be truthy when b is the empty list. But when you call (average a '()), then b will be bound to (()), that is a list containing the empty list.
Solving the issue that you only pass exactly two arguments on the following calls can be done with apply: It takes the function as well as a list of parameters to call it with: (appply #'average (cons a (cdr b)))
Now tackling your original goal of writing an average function: Computing the average consists of two tasks:
Compute the sum of all elements.
Divide that with the number of all elements.
You could write your own function to recursively add all elements to solve the first part (do it!), but there's already such a function:
(+ 1 2) ; Sum of two elements
(+ 1 2 3) ; Sum of three elements
(apply #'+ '(1 2 3)) ; same as above
(apply #'+ some-list) ; Summing up all elements from some-list
Thus your average is simply
(defun average (&rest parameters)
(if parameters ; don't divide by 0 on empty list
(/ (apply #'+ parameters) (length parameters))
0))
As a final note: You shouldn't use car and cdr when working with lists. Better use the more descriptive names first and rest.
If performance is critical to you, it's probably best to fold the parameters (using reduce which might be optimized):
(defun average (&rest parameters)
(if parameters
(let ((accum
(reduce #'(lambda (state value)
(list (+ (first state) value) ;; using setf is probably even better, performance wise.
(1+ (second state))))
parameters
:initial-value (list 0 0))))
(/ (first accum) (second accum)))
0))
(Live demo)
#' is a reader macro, specifically one of the standard dispatching macro characters, and as such an abbreviation for (function ...)
Just define average*, which calls the usual average function.
(defun average* (&rest numbers)
(average numbers))
I think that Rainer Joswig's answer is pretty good advice: it's easier to first define a version that takes a simple list argument, and then define the &rest version in terms of it. This is a nice opportunity to mention spreadable arglists, though. They're a nice technique that can make your library code more convenient to use.
In most common form, the Common Lisp function apply takes a function designator and a list of arguments. You can do, for instance,
(apply 'cons '(1 2))
;;=> (1 . 2)
If you check the docs, though, apply actually accepts a spreadable arglist designator as an &rest argument. That's a list whose last element must be a list, and that represents a list of all the elements of the list except the last followed by all the elements in that final list. E.g.,
(apply 'cons 1 '(2))
;;=> (1 . 2)
because the spreadable arglist is (1 (2)), so the actual arguments (1 2). It's easy to write a utility to unspread a spreadable arglist designator:
(defun unspread-arglist (spread-arglist)
(reduce 'cons spread-arglist :from-end t))
(unspread-arglist '(1 2 3 (4 5 6)))
;;=> (1 2 3 4 5 6)
(unspread-arglist '((1 2 3)))
;;=> (1 2 3)
Now you can write an average* function that takes one of those (which, among other things, gets you the behavior, just like with apply, that you can pass a plain list):
(defun %average (args)
"Returns the average of a list of numbers."
(do ((sum 0 (+ sum (pop args)))
(length 0 (1+ length)))
((endp args) (/ sum length))))
(defun average* (&rest spreadable-arglist)
(%average (unspread-arglist spreadable-arglist)))
(float (average* 1 2 '(5 5)))
;;=> 3.25
(float (average* '(1 2 5)))
;;=> 2.66..
Now you can write average as a function that takes a &rest argument and just passes it to average*:
(defun average (&rest args)
(average* args))
(float (average 1 2 5 5))
;;=> 3.5
(float (average 1 2 5))
;;=> 2.66..
I am practicing for my programming paradigms exam and working through problem sets I come to this problem. This is the first problem after reversing and joining lists recursively, so I suppose there is an elegant recursive solution.
I am given a list of lists and a permutation. I should permute every list including a list of lists with that specified permutation.
I am given an example:
->(permute '((1 2 3) (a b c) (5 6 7)) '(1 3 2))
->((1 3 2) (5 7 6) (a c b))
I have no idea even how to start. I need to formulate the problem in recursive interpretation to be able to solve it, but I can not figure out how.
Well, let's see how we can break this problem down. We are given a list of lists, and a list of numbers, and we want to order each list according to the order specified by the list of numbers:
=>(permute '((1 2 3) (4 5 6)) '(3 2 1))
'((3 2 1) (6 5 4))
We can see that each list in the list of lists can be handled separately, their solutions are unrelated to each other. So we can have a helper permute1 that handles the case of one list, then use map to apply this function to each of the lists (with the same ordering each time):
(define (permute lists ordering)
(map (lambda (xs) (permute1 xs ordering))
lists))
(define (permute1 items ordering)
...)
Now, to calculate (permute1 '(4 5 6) '(3 2 1)), what we mean is:
The first item of the new list will be the 3rd item of items, because the first number in ordering is 3.
The rest of the items of the new list will be determined by using the rest of the numbers in the ordering.
If the ordering is the empty list, return the empty list.
This forms the base case (3), the recursive case (1), and the steps to recur deeper (2). So a sketch of our solution would look like:
(define (permute1 items ordering)
(if (empty? ordering)
'()
(let ([next-item ???])
(??? next-item
(permute1 items (rest ordering))))))
Where the ???s represent getting the item based on the first number in ordering and combining this item with the remainder of the calculation, respectively.
Here's another option, using higher-order functions. This is the idiomatic way to think about a solution in a functional language - we split the problem in sub-problems, solve each one using existing procedures and finally we compose the answer:
(define (atom? x)
(and (not (null? x))
(not (pair? x))))
(define (perm lst order)
(foldr (lambda (idx acc)
(cons (list-ref lst (sub1 idx)) acc))
'()
order))
(define (permute lst order)
(if (atom? lst)
lst
(perm (map (lambda (x) (permute x order)) lst)
order)))
We start by defining atom?, a generic predicate and perm, a helper that will reorder any given list according to the ordering specified in one of its parameters. It uses foldr to build the output list and list-ref to access elements in a list, given its 0-based indexes (that's why we subtract one from each index).
The main permute function takes care of (recursively) mapping perm on each element of an arbitrarily nested input list, so we can obtain the desired result:
(permute '((1 2 3) (a b c) (5 6 7)) '(1 3 2))
=> '((1 3 2) (5 7 6) (a c b))
I am given an example:
(permute ('(1 2 3) '(a b c) '(5 6 7)) '(1 3 2))
((1 3 2) (5 7 6) (a c b))
The syntax you've given isn't correct, and will cause an error, but it's fairly clear what you mean. You want that
(permute '((1 2 3) (a b c) (5 6 7)) '(1 3 2))
;=> ((1 3 2) (5 7 6) (a c b))
Now, it's not clear how you're indicating the permutation. Is '(1 3 2) a permutation because it has some (1-based) indices, and indicates the way to rearrange elements, or is it because it is actually a permutation of the elements of the first list of the first list? E.g., would
(permute '((x y z) (a b c) (5 6 7)) '(1 3 2))
;=> ((x z y) (5 7 6) (a c b))
work too? I'm going to assume that it would, because it will make the problem much easier.
I have no idea even how to start. I need to formulate the problem in
recursive interpretation to be able to solve it, but I can not figure
out how.
You need to write a function that can take a list of indices, and that returns a function that will perform the permutation. E.g,.
(define (make-permutation indices)
…)
such that
((make-permutation '(3 1 2)) '(a b c))
;=> (c a b)
One you have that, it sounds like your permute function is pretty simple:
(define (permute lists indices)
(let ((p (make-permutation indices)))
(p (map p lists))))
That would handle the case you've given in your example, since (map p lists) will return ((1 3 2) (a b c) (5 7 6)), and then calling p with that will return ((1 3 2) (5 7 6) (a c b)). If you need to be able to handle more deeply nested lists, you'll need to implement a recursive mapping function.
Here's my take, which seems to be shorter than the previous examples:
(define (permute lst ord)
(define ord-1 (map sub1 ord)) ; change from 1-based to 0-based indexes
(define (perm elts) ; recursive sub-procedure
(if (list? elts)
(map perm (map (curry list-ref elts) ord-1)) ; list -> recurse
elts)) ; else return unchanged
(perm lst)) ; initial call
testing
> (permute '((1 2 3) (a b c) (5 6 7)) '(1 3 2))
'((1 3 2) (5 7 6) (a c b))
> (permute '((1 (i permute did) 3) (a b (scheme cool is)) (5 6 7)) '(1 3 2))
'((1 3 (i did permute)) (5 7 6) (a (scheme is cool) b))
I have a small exercise in Lisp:
Write a function test-delta with parameters delta and lst, which will
check if the difference between successive elements in lst is smaller than
delta. Write the function in two ways:
recursively
using a mapping function
I have no problem writing that function recursively, but I don't know which mapping function I should use. All the standard mapping functions work with only one element of the list at a time. reduce cannot be used either, because I do not have some operation to use between successive elements. What function could I use here?
All standard functions are working only with one element at time.
Reduce function cannot be use either
because i do not have some operation to use between to elements.
There's already an answer by uselpa showing that you can do this with reduce, but it feels a bit awkward to me to bend reduce to this case.
It's much more natural, in my opinion, to recognize that the standard mapping functions actually let you work with multiple lists. I'll show mapcar and loop first, and then every, which I think is the real winner here. Finally, just for completeness, I've also included maplist.
mapcar
The standard mapcar can take more than one list, which means that you can take elements from two different lists at once. Of particular note, it could take a list and (rest list). E.g.,
(let ((list '(1 2 3 4 5 6)))
(mapcar 'cons
list
(rest list)))
;=> ((1 . 2) (2 . 3) (3 . 4) (4 . 5) (5 . 6))
loop
You can use loop to do the same sort of thing:
(loop
with l = '(1 2 3 4 5 6)
for a in l
for b in (rest l)
collect (cons a b))
;=> ((1 . 2) (2 . 3) (3 . 4) (4 . 5) (5 . 6))
There are some other variations on loop that you can use, but some of them have less conventient results. E.g., you could loop for (a b) on list, but then you get a (perhaps) unexpected final binding of your variables:
(loop for (a b) on '(1 2 3 4 5 6)
collect (list a b))
;=> ((1 2) (2 3) (3 4) (4 5) (5 6) (6 NIL))
This is similar to what maplist will give you.
every
I think the real winners here, though, are going to the be every, some, notevery, and notany functions. These, like mapcar can take more than one list as an argument. This means that your problem can simply be:
(let ((delta 4)
(lst '(1 2 4 7 9)))
(every (lambda (x y)
(< (abs (- x y)) delta))
lst
(rest lst)))
;=> T
(let ((delta 2)
(lst '(1 2 4 7 9)))
(every (lambda (x y)
(< (abs (- x y)) delta))
lst
(rest lst)))
;=> NIL
maplist
You could also do this with maplist, which works on successive tails of the list, which means you'd have access to each element and the one following. This has the same 6 NIL at the end that the second loop solution did, though. E.g.:
(maplist (lambda (tail)
(list (first tail)
(second tail)))
'(1 2 3 4 5 6))
;=> ((1 2) (2 3) (3 4) (4 5) (5 6) (6 NIL))
reduce can be used:
(defun testdelta (delta lst)
(reduce
(lambda (r e)
(if (< (abs (- r e)) delta)
e
(return-from testdelta nil)))
lst)
t)
or, without return-from (but possibly slower):
(defun testdelta (delta lst)
(and
(reduce
(lambda (r e)
(and r (if (< (abs (- r e)) delta) e nil)))
lst)
t))
I am having a problem using list-ref.
Is there a way to use list-ref to obtain size of an unknown list?
(... (if (number? (deref(+ array-ref index)))
(array-len array-ref (+ index 1))
0) )
#|(define (deref ref)
(list-ref the-store ref))
|#
If you want the size of a list use length, not list-ref. For example, in a list such as this:
(define lst '(1 2 3 4 5))
… The valid indexes will be between 0 and list's length minus one:
(list-ref lst 0)
=> 1
(list-ref lst (- (length lst) 1))
=> 5
However: in Scheme it's unusual to write code that depends on the index of an element in a list, that's how you'd think about a solution in a C-like language using an array, but Scheme lists are different, and normally you traverse a list using recursion - forget about indexes!
If I understand correctly, the list is a list of lists such as:
(define lst '((a b c)
(1 "hi")
((lambda(x)(+ 5 x)) 42 'a (14 7 12))
("hello" " world")))
A quick and dirty way to get the size of an element would be
(define third-of-a-size (third (map length lst)))