Remove list from list of lists Clojure - clojure

I have a list such as:
(def lst '((a b c) (d e) (f g h)))
I need to create a new list having removed one of the inner lists,
I've tried
(remove '(d e) lst)
which returns
java.lang.ClassCastException: clojure.lang.PersistentList cannot be cast to clojure.lang.IFn
I've also tried
(filter (not= '(d e)) lst)
which returns
java.lang.ClassCastException: null

If you want to use filter, note that the first argument has to be a function, so you could use
(filter #(not= '(d e) %) lst)
That's why you get the ClassCastException.
The same is true for remove. You could also use partial instead of an anonymous function:
(remove (partial = '(d e)) lst)

Both filter and remove functions expect first argument to be a predicate function. You could use partial function to transform operator = into a predicate:
(remove (partial = '(d e)) lst)

Related

How to use swap to get result in clojure

I want to write the function collect which can combine the sublists into a list, like:
user=> (collect '(a b c d e) 1)
((a)(b)(c)(d)(e))
user=> (collect '(a b c d e) 2)
((a b)(c d)(e))
user=> (collect '(a b c d e) 5)
(a b c d e))
this is my code:
(defn collect [lst num]
(loop [l lst res (atom ())]
(if (<= (count l) num) #res
(recur (drop num l) (swap! res conj (take num (drop num l)))))))
But when I run
user=> (collect '(a b c d e) 1)
I got the error:
ClassCastException clojure.lang.PersistentList cannot be cast to clojure.lang.IAtom clojure.core/swap!
why I cannot get the res when I use "swap!" ? Thank you.
It's blowing up in the second pass through the loop.
swap returns the value that was put into the atom, not the atom it's self. So the first pass is updating the atom, and then passing the value it just put into the atom to the second pass through the loop. in the second pass it's trying to use the value as the atom, and getting the exception above.
To "fix" this use a do to update the atom, then pass the atom to the next pass through the loop once it contains the correct value.
user> (defn collect [lst num]
(loop [l lst res (atom ())]
(if (<= (count l) num) #res
(recur (drop num l)
(do (swap! res conj (take num (drop num l)))
res)))))
#'user/collect
user> (collect '(a b c d e) 2)
((e) (c d))
You can also in this case, just remove the atom completely and get exactly the same result (I fixed on ordering problem from your example by using a [] instead of () in the initial value of res):
user> (defn collect [lst num]
(loop [l lst res []]
(if (<= (count l) num) res
(recur (drop num l)
(conj res (take num (drop num l)))))))
#'user/collect
user> (collect '(a b c d e) 2)
[(c d) (e)]
and of course you can also use partition-all as glts mentions above.
;; this would be a correct way to do it
(defn collect [coll n]
(partition-all n coll))
;; this would be a clumsy way to do it
(defn collect
"using a loop (there is not point to do that but at least you can see the logic working as in your example)"
[coll n]
(lazy-seq
(loop [res []
coll coll]
(if (empty? coll)
res
(recur (conj res (take n coll)) (drop n coll))))))
Regarding your error, on the second loop, res is a list-like value, not an atom anymore. That would lead us to :
(defn collect [coll n]
(lazy-seq (loop [res (atom [])
coll coll]
(if (empty? coll)
#res
(recur (do (swap! res conj (take n coll))
;; return the atom instead of the value'
res)
(drop n coll))))))
Note that in order to preserve the order in the solution, I use a vector (litteral []) instead of a list (litteral '()). This is because of the behaviour of conj described here.

How to transform a list into a set in LISP?

I have been trying to transform a linear list into a set but with no avail. Everytime I run this, I get some weird compilation errors like "badly formed lambda" which points to the way I use append. Here is my code:
(defun mem(e l)
(cond
((null l) nil)
((equal e (car l)) t)
((listp (car l)) (mem e (car l)))
(t(mem e (cdr l)))
)
)
(defun st(l k)
(cond
((null l) nil)
(( mem '(car l) 'k) (st (cdr l) k))
((listp (car l)) (st (car l) k))
( t (st (cdr l) (append((car l) k)) ))
(t(mem e (cdr l)))
)
)
EDIT: frankly I just want to remove the duplicates from list l
Prefer Standard Library Functions
EDIT: frankly I just want to remove the duplicates from list l
Common Lisp has a remove-duplicates function. The documentation inclues examples:
Examples:
(remove-duplicates "aBcDAbCd" :test #'char-equal :from-end t) => "aBcD"
(remove-duplicates '(a b c b d d e)) => (A C B D E)
(remove-duplicates '(a b c b d d e) :from-end t) => (A B C D E)
(remove-duplicates '((foo #\a) (bar #\%) (baz #\A))
:test #'char-equal :key #'cadr) => ((BAR #\%) (BAZ #\A))
(remove-duplicates '((foo #\a) (bar #\%) (baz #\A))
:test #'char-equal :key #'cadr :from-end t) => ((FOO #\a) (BAR #\%))
Are you trying to flatten the list too?
From your code for mem, where you do:
((listp (car l)) (mem e (car l)))
it looks like you want your member function to also recurse into sublists. That's a bit questionable, even when working with sets, since sets can traditionally include other sets. E.g., {{3},{4},5} is a set containing 5, the set {3}, and the set {4}. It's not the same as the set {3,4,5}. Your st function also looks like it's trying to recurse into lists, which makes it seem like you want to flatten you lists, too. Again, that's a bit questionable, but if you want to do that, then your conversion to a set would be easier as a "flatten, then remove duplicates" process:
(defun flatten (list)
"Returns a fresh list containing the leaf elements of LIST."
(if (listp list)
(mapcan 'flatten list)
(list list)))
;; CL-USER> (flatten '(1 2 (3 4) 5 ((6))))
;; (1 2 3 4 5 6)
(defun to-set (list)
"Returns a set based on the elements of LIST. The result
is a flat list containing the leaf elements of LIST, but
with any duplicate elements removed."
(delete-duplicates (flatten list)))
;; CL-USER> (to-set '(1 3 (3 4) ((4) 5)))
;; (1 3 4 5)
Notes
I get some weird compilation errors like "badly formed lambda" which points to the way I use append.
Yes, you're trying to call append like: (append((car l) k)). That's actually not a problem for append. Remember, the syntax for a function call in Lisp is (function argument…). That means that you've got:
(append ((car l) k))
<function> <argument1>
But your argument1 is also a function call:
((car l) k )
<function> <argument1>
In Common Lisp, you can't use (car l) as a function. The only thing that can appear for a function is a symbol (e.g., car, append) or a lambda expression (e.g., (lambda (x) (+ x 1)).
You want to call (append (car l) k) instead.
First, CL does not have a set data type.
Lists, however, can be used as sets, you do not need to write any special code for that.
Second, I don't understand what your st function is supposed to do, but I bet that in the second cond clause you should not quote (car l) and k. You should use meaningful names for your functions and avoid abbreviations. As per your explanation in the comment, you should use pushnew instead.
Third, your mem function is quite weird, I am pretty sure you do not mean what you wrote: e is searched along a path in the tree l, not in the list l. As per your explanation in the comment, you should check both car and cdr:
(defun tree-member (tree element &key (test #'eql))
(if (consp tree)
(or (tree-member (car tree) element :test test)
(tree-member (cdr tree) element :test test))
(funcall test element tree)))

Abstract List Functions in Racket/Scheme - Num of element occurrences in list

So I'm currently stuck on a "simple?" function in Racket. It's using the Intermediate Student with lambda language.
Some restrictions on this are that NO recursion is allowed, neither are local functions. It's plain and simple abstract list functions.
What this function is supposed to do is to take in a list of numbers, and output a list of pairs in which each pair has the first element as the number with the second element being the number it has occurred in the list.
Examples:
(1 1 2 3) => ((1 2) (2 1) (3 1))
(2 3 4 3) => ((2 1) (3 2) (4 1))
I have a function that produces the number of occurrences by inputting a list of numbers and a number which is:
(define (occurrences lon n)
(length (filter (lambda (x) (= x n)) lon)))
My approach, which was clearly wrong was:
(define (num-pairs-occurrences lon)
(list (lambda (x) (map (occurrences lon x) (remove x lon)) x))
I thought the above would work, but apparently my lambda isn't placed properly. Any ideas?
It's a bit trickier than you imagine. As you've probably noticed, we must remove duplicate elements in the output list. For this, is better that we define a remove-duplicates helper function (also using abstract list functions) - in fact, this is so common that is a built-in function in Racket, but not available in your current language settings:
(define (remove-duplicates lst)
(foldr (lambda (e acc)
(if (member e acc)
acc
(cons e acc)))
'()
lst))
Now it's easy to compose the solution using abstract list functions:
(define (num-pairs-occurrences lon)
(map (lambda (e) (list e (occurrences lon e)))
(remove-duplicates lon)))
The above might return and output list in a different order, but that's all right. And before you ask: yes, we do need that helper function. Please don't ask for a solution without it...
An easy, self-contained solution would be:
(define (num-pairs-occurences lst)
(foldl (lambda (e r)
(if (or (null? r) (not (= (caar r) e)))
(cons (list e 1) r)
(cons (list e (add1 (cadar r))) (cdr r))))
null
(sort lst >)))
Basically, you sort the list first, and then you fold over it. If the element (e) you get is the same as the first element of the result list (r), you increment the count, otherwise you add a new sublist to r.
If you sort by > (descending), you can actually use foldl which is more memory-efficient. If you sort by < (ascending), you need to use foldr which is less efficient.

Every other letter in a list? LISP

I am relatively new to LISP and am trying some new things for a Lisp program I am trying to create for a presentation.
I need to be able to print every other character in a list, for example, (A B C D E F) would return (A C E) .. but I am getting easily confused...
I am normally program Java, so this comes kind of different to me.
I am trying to program this using purely recursion.. So something along the lines of....
(defun every-other (lst)
(cond ((null lst) 0)
(( **** now this is where I get confused as to what I should do..
I've tried adding a counter to only remove even numbered elements, but I think I implemented the counter wrong, I also tried remove(cadr lst) lst, but that would only return zeros...
any help would be greatly appreciated..
Thanks!
Since you say you want it to be done recursively, just think it through case by case.
The list is null -> return the empty list [the empty list is '()].
Otherwise the list is not null -> In this case you want to build a new list containing
the first element, skip the second element, then grab
every-other element of the remaining list.
Turning this case analysis into code looks something like this:
(defun every-other (lst)
(cond
;; If the list is null return the empty list.
((null lst) '())
;; If the list is not null, construct [cons] a new list with the first element of lst
;; and every-other element of the list after the first two elements [rest returns the
;; list without the first element, so we can just use it twice].
(t (cons (first lst) (every-other (rest (rest lst)))))))
Now going through the evaluation of this code should look something like this:
(every-other '(a b c d e f))
=> (cons 'a (every-other '(c d e f)))
=> (cons 'a (cons 'c (every-other '(e f))))
=> (cons 'a (cons 'c (cons 'e (every-other '())))
=> (cons 'a (cons 'c (cons 'e '())))
=> (cons 'a (cons 'c '(e)))
=> (cons 'a '(c e))
=> '(a c e)
For fun, a loop-based solution:
(defun every-other (lst)
(loop
for i in lst
for keep = t then (not keep)
if keep collect i))
Just use a loop.
(loop :for c :in '(a b c d e f) :by #'cddr
:collect c)
:By in a for-in clause sets the stepping function (default is #'cdr). In order to get every other element, step two steps each time. Cddr is a shortcut for applying cdr two times.
(defun aaa (x)
(aa (length x) x))
(defun aa (n x)
(cond ((null x) nil)
((evenp (- n (length x))) (cons (car x) (aa n (cdr x))))
(t (aa n (cdr x)))))
This is a stupid case lol~
shorter recursive solution:
(defun every-other (l)
(unless (null l)
(cons (first l) (every-other (cddr l)))))

Finding the position of an object in a sequence in Clojure

Essentially, I want a function that works like this:
user=> (pos 'c '(a b c d e f g) =)
2
user=> (pos 'z '(a b c d e f g) =)
nil
And I came up with this:
(defn pos
"Gets position of first object in a sequence that satisfies match"
[object sequence match]
(loop [aseq sequence position 0]
(cond (match object (first aseq)) position
(empty? aseq) nil
:else (recur (rest aseq) (inc position)))))
So my question is, is there some built-in function that would allow us to do this, or would there be a better, more functional/Clojure-ish way to write the pos function?
Well, if you really want to look for a particular item you can use .indexOf on the collection; if you're looking to do something more general with predicates you don't need a function and an item, just a function is plenty.
(defn pos [pred coll]
(->> coll
(map-indexed #(when (pred %2) %1))
(remove nil?)
(first)))
user> (pos #{'c} '(a b c d e f g))
2
On the other hand, there's a reason this isn't included in clojure.core: it's not very efficient, and you very rarely care about indices in a collection - if you do, you should usually rethink your algorithm.