How do you get a random element from a list in racket, the equivalent of Clojure
rand-nth?
How do you get the key from a hash-map in Racket, the equivalent of Clojure keys?
How do you get the value corresponding to a key in Racket, equivalent of get from Clojure?
This function will pick the random element:
(define (rand-nth list)
(list-ref list (random (length list))))
This function will pick all the keys from a hashmap:
(define (keys array)
(map car array))
This function will get the value if you give as arguments the hash-map and the key:
(define (get array k)
(if (eq? (caar array) k)
(car (cdar array))
(get (cdr array) k)))
Related
I want to write a recursive function that checks the list and either returns true if the list is in ascending order or NIL otherwise. If the list is empty it is still true. I am completely new to Lisp, so its still very confusing.
(defun sorted (x)
(if (null x)
T
(if (<= car x (car (cdr x)))
(sorted (cdr x))
nil)))
The recursive version:
(defun sorted (list)
(or (endp list)
(endp (cdr list))
(and (<= (first list) (second list))
(sorted (cdr list)))))
The more idiomatic loop-based predicate accepting a :test argument:
(defun sortedp (list &key (test #'<=))
(loop for (a b) on list
while b
always (funcall test a b)))
The version accepting a :key; we only call the key function once per visited element:
(defun sortedp (list &key (test #'<=) (key #'identity))
(loop for x in list
for old = nil then new
for new = (funcall key x)
for holdp = T then (funcall test old new)
always holdp))
Some tests:
(loop for k in '(()
((1))
((1) (2))
((2) (1))
((1) (2) (3))
((3) (2) (1)))
collect (sortedp k :test #'> :key #'car))
=> (T T NIL T NIL T)
This one also works with other kinds of sequences:
(defun sortedp (sequence &key (test #'<=) (key #'identity))
(reduce (lambda (old x &aux (new (funcall key x)))
(if (or (eq old t)
(funcall test old new))
new
(return-from sortedp nil)))
sequence
:initial-value t))
The above test gives:
(T 1 NIL 1 NIL 1)
... which is a correct result thanks to generalized booleans.
If you are doing your homework (seems so), then the above answers are fine. If you are just learning Lisp, and don't have constraints about recursivity, then the following might give you a glimpse about the power of Lisp:
(defun sorted (l)
(or (null l) (apply #'< l)))
The first problem with your solution is the base case You need to stop not at the end of the list, but when looking at the last to elements, as you need to elements to do the comparison. Also the parens are missing in the call to (car x)
(defun sorted (list)
(if (endp (cddr list))
(<= (car list) (cadr list))
(and (<= (car list) (cadr list))
(sorted (cdr list)))))
Bare in mind that recursive solutions are discouraged in CL
I'm trying to implement deep-reverse in clojure. If lst is (1 (2 (3 4 5)) (2 3)), it should return ((3 2) ((5 4 3) 2) 1). This is what I have so far:
defn dRev [lst]
( if (= lst ())
nil
( if (list? (first lst))
( dRev (first lst) )
( concat
( dRev (rest lst)) (list (first lst))
)
)
)
)
However, my implementation only works if the nested list is the last element, but the resulted list is also flattened.
For eg: (dRev '(1 2 (3 4)) will return (4 3 2 1).
Otherwise, for eg: (dRev '(1 (2 3) 4)) will return (3 2 1) only.
I hit this brick wall for a while now, and I can't find out the problem with my code. Can anyone please help me out?
The other answer gave you the best possible implementation of a deep-reverse in Clojure, because it uses the clojure.walk/postwalk function which generalizes the problem of deep-applying a function to every element of a collection. Here I will instead walk you through the problems of the implementation you posted.
First, the unusual formatting makes it hard to spot what's going on. Here's the same just with fixed formatting:
(defn dRev [lst]
(if (= lst ())
nil
(if (list? (first lst))
(dRev (first lst))
(concat (dRev (rest lst))
(list (first lst))))))
Next, some other small fixes that don't yet fix the behaviour:
change the function name to conform to Clojure conventions (hyphenation instead of camel-case),
use the usual Clojure default name for collection parameters coll instead of lst,
use empty? to check for an empty collection,
return () in the default case because we know we want to return a list instead of some other kind of seq,
and use coll? instead list? because we can just as well reverse any collection instead of just lists:
(If you really want to reverse only lists and leave all other collections as is, reverse the last change.)
(defn d-rev [coll]
(if (empty? coll)
()
(if (coll? (first coll))
(d-rev (first coll))
(concat (d-rev (rest coll))
(list (first coll))))))
Now, the formatting fix makes it obvious what's the main problem with your implementation: in your recursive call ((d-rev (first coll)) resp. (dRev (first lst))), you return only the result of that recursion, but you forget to handle the rest of the list. Basically, what you need to do is handle the rest of the collection always the same and only change how you handle the first element based on whether that first element is a list resp. collection or not:
(defn d-rev [coll]
(if (empty? coll)
()
(concat (d-rev (rest coll))
(list (if (coll? (first coll))
(d-rev (first coll))
(first coll))))))
This is a working solution.
It is terribly inefficient though, because the concat completely rebuilds the list for every element. You can get a much better result by using a tail-recursive algorithm which is quite trivial to do (because it's natural for tail-recursion over a sequence to reverse the order of elements):
(defn d-rev [coll]
(loop [coll coll, acc ()]
(if (empty? coll)
acc
(recur (rest coll)
(cons (if (coll? (first coll))
(d-rev (first coll))
(first coll))
acc)))))
As one final suggestion, here's a solution that's halfways towards the one from the other answer by also solving the problem on a higher level, but it uses only the core functions reverse and map that applies a function to every element of sequence but doesn't deep-recurse by itself:
(defn deep-reverse [coll]
(reverse (map #(if (coll? %) (deep-reverse %) %) coll)))
You can build what you are writing with clojure.walk/postwalk and clojure.core/reverse. This does a depth-first traversal of your tree input and reverses any seq that it finds.
(defn dRev [lst]
(clojure.walk/postwalk #(if (seq? %) (reverse %) %) lst))
Here is my version of the problem, if you enter something like this:
(deep-reverse '(a (b c d) 3))
It returns
=> '(3 (d c b) a)
The problem is taken from Ninety-Nine Lisp Problems
My code ended up like this, though, they might be better implementations, this one works fine.
(defn deep-reverse
"Returns the given list in reverse order. Works with nested lists."
[lst]
(cond
(empty? (rest lst)) lst
(list? (first lst)) (deep-reverse(cons (deep-reverse (first lst)) (deep-reverse (rest lst))))
:else
(concat (deep-reverse (rest lst)) (list (first lst)))))
Hope this is what you were looking for!
I have written this function to 'check' if the list contains only 2-elements pairs like these: ((1 2)(3 1)(6 2)) --- (sorted based on first elements & no repetition of 1st elements). But I am getting errors, can anyone please give an idea:
(define Bag?
(lambda setlist
(cond ((null? setlist) '())
((and (pair? (caar setlist)) (= (length (caar setlist)) 2)))
((> (caar setlist) 0) (< (caar setlist) (car (cdr (car setlist) ))))
(else(Bag? (cdr setlist))))
))
The list traversal doesn't look right. The lambda is incorrectly declaring its parameter, and the second condition is wrong, you're not advancing the recursion there.
It'd better to start from scratch. I'll give you a high-level solution in Racket stating what needs to be checked, it's up to you to rewrite it in terms of simpler procedures:
(define (bag? setlist)
(and (apply < (map car setlist))
(andmap (lambda (e) (and (pair? e) (= 2 (length e))))
setlist)))
The above verifies that the first-elements in each pair appear sorted in ascending order, and that each pair in the list contains exactly two elements.
I have been working on a call to accumulate which goes as follows:
(define (accumulate op initial sequence)
(if (null? sequence)
initial
(op (car sequence)
(accumulate op initial (cdr sequence)))))
However when I try to square something by slecting it through filter the answer doesn't work. What I have so far is this:
(define (f2b items)
(accumulate (lambda (x y)
(cons (append
(map square (filter negative? (filter number? x))) x) y)) () items)
)
The Input I give is:
(f2a '(("sdas" 89) (-53 "sad")))
The output I get is:
((sdas 89) (2809 -53 sad))
I can't seem to get the negative number to go away.
It would be much easier to use filter and map. Filter is predefined but it looks like this.
(define (filter1 predicate sequence)
(cond
((null? sequence) null)
((predicate (car sequence))
(cons (car sequence)
(filter predicate (cdr sequence))))
(else (filter predicate (cdr sequence)))))
map is also predefined, it just runs a function over a list.
This should be pretty simple to write, but incase you need help you should just write a lamdba for the predicate in filter.
Actually, the functionality you describe is not usually the job of an accumulator. Instead, squaring negative numbers in a list seems like the perfect job for something like a map.
First, let's do:
(define (make-positive x)
(if (and (number? x) (negative? x))
(square x)
x))
Now suppose we want to operate on a list called lst. If it was just a flat list, like '(1 "2" -5 -4 6), then we could just
(map make-positive lst)
Since we need to operate on lists which are nested two levels deep, we could do:
(map (lambda (x)
(map make-positive x))
lst)
If we wanted to operate on lists which are nested arbitrarily deep, we could do:
(define (nested-map fn elm)
(if (list? elm)
(map (lambda (x) (nested-map fn x)) elm)
(fn elm)))
(nested-map make-positive lst)
PS - we can define map like this:
(define (map fn lst)
(if (empty? lst)
'()
(cons (fn (car lst))
(map fn (cdr lst)))))
How do I get the index of any of the elements on a list of strings as so:
(list "a" "b" "c")
For example, (function "a") would have to return 0, (function "b") 1, (function "c") 2 and so on.
and... will it be better to use any other type of collection if dealing with a very long list of data?
Christian Berg's answer is fine. Also it is possible to just fall back on Java's indexOf method of class String:
(.indexOf (apply str (list "a" "b" "c")) "c")
; => 2
Of course, this will only work with lists (or more general, seqs) of strings (of length 1) or characters.
A more general approach would be:
(defn index-of [e coll] (first (keep-indexed #(if (= e %2) %1) coll)))
More idiomatic would be to lazily return all indexes and only ask for the ones you need:
(defn indexes-of [e coll] (keep-indexed #(if (= e %2) %1) coll))
(first (indexes-of "a" (list "a" "a" "b"))) ;; => 0
I'm not sure I understand your question. Do you want the nth letter of each of the strings in a list? That could be accomplished like this:
(map #(nth % 1) (list "abc" "def" "ghi"))
The result is:
(\b \e \h)
Update
After reading your comment on my initial answer, I assume your question is "How do I find the index (position) of a search string in a list?"
One possibility is to search for the string from the beginning of the list and count all the entries you have to skip:
(defn index-of [item coll]
(count (take-while (partial not= item) coll)))
Example: (index-of "b" (list "a" "b" "c")) returns 1.
If you have to do a lot of look-ups, it might be more efficient to construct a hash-map of all strings and their indices:
(def my-list (list "a" "b" "c"))
(def index-map (zipmap my-list (range)))
(index-map "b") ;; returns 1
Note that with the above definitions, when there are duplicate entries in the list index-of will return the first index, while index-map will return the last.
You can use the Java .indexOf method reliably for strings and vectors, but not for lists. This solution should work for all collections, I think:
(defn index-of
"Clojure doesn't have an index-of function. The Java .indexOf method
works reliably for vectors and strings, but not for lists. This solution
works for all three."
[item coll]
(let [v (if
(or (vector? coll) (string? coll))
coll
(apply vector coll))]
(.indexOf coll item)))
Do you mean, how do you get the nth element of a list?
For example, if you want to get the 2nd element on the list (with zero-based index):
(nth (list "a" "b" "c") 2)
yields
"c"
Cat-skinning is fun. Here's a low-level approach.
(defn index-of
([item coll]
(index-of item coll 0))
([item coll from-idx]
(loop [idx from-idx coll (seq (drop from-idx coll))]
(if coll
(if (= item (first coll))
idx
(recur (inc idx) (next coll)))
-1))))
This is a Lispy answer, I suspect those expert in Clojure could do it better:
(defn position
"Returns the position of elt in this list, or nil if not present"
([list elt n]
(cond
(empty? list) nil
(= (first list) elt) n
true (position (rest list) elt (inc n))))
([list elt]
(position list elt 0)))
You seem to want to use the nth function.
From the docs for that function:
clojure.core/nth
([coll index] [coll index not-found])
Returns the value at the index. get returns nil if index out of
bounds, nth throws an exception unless not-found is supplied. nth
also works for strings, Java arrays, regex Matchers and Lists, and,
in O(n) time, for sequences.
That last clause means that in practice, nth is slower for elements "farther off" in sequences, with no guarantee to work quicker for collections that in principle support faster access (~ O(n)) to indexed elements. For (clojure) sequences, this makes sense; the clojure seq API is based on the linked-list API and in a linked list, you can only access the nth item by traversing every item before it. Keeping that restriction is what makes concrete list implementations interchangeable with lazy sequences.
Clojure collection access functions are generally designed this way; functions that do have significantly better access times on specific collections have separate names and cannot be used "by accident" on slower collections.
As an example of a collection type that supports fast "random" access to items, clojure vectors are callable; (vector-collection index-number) yields the item at index index-number - and note that clojure seqs are not callable.
I know this question has been answered a million time but here is a recursive solution that leverages deconstructing.
(defn index-of-coll
([coll elm]
(index-of-coll coll elm 0))
([[first & rest :as coll] elm idx]
(cond (empty? coll) -1
(= first elm) idx
:else (recur rest elm (inc idx)))))
(defn index-of [item items]
(or (last (first (filter (fn [x] (= (first x) item))
(map list items (range (count items))))))
-1))
seems to work - but I only have like three items in my list