What would be a more correct functional way to write the following code which checks if a number is prime or not:
(defn prime? [n]
(loop [k 2]
(cond
(< k n) (if (not= 0 (mod n k))
(recur (inc k))
(println n "is not prim"))
:else (println n "is prim"))))
Regardless of which algorithm you use to test primality, the "correct functional way" would be for your prime? function to return true or false. As it stands, your function returns nil and has side effects (prints something out).
You could then do (println (prime? x)) to check a particular number, and have the side effects constrained to that single statement.
A simpler way using standard library functions such as every? and range:
(defn divisible? [a b]
(zero? (mod a b)))
(defn prime? [n]
(and (> n 1) (not-any? (partial divisible? n) (range 2 n))))
and refactoring I/O into a separate function for greater reuse:
(defn format-primality [n]
(str n " " (if (prime? n) "is prim" "is not prim")))
(def print-primality
(comp println format-primality))
Example:
user=> (map (fn [n] [n (prime? n)]) (range 1 15))
([1 false] [2 true] [3 true] [4 false] [5 true] [6 false] [7 true]
[8 false] [9 false] [10 false] [11 true] [12 false] [13 true] [14 false])
Related
I am trying to get into Lisps and FP by trying out the 99 problems.
Here is the problem statement (Problem 15)
Replicate the elements of a list a given number of times.
I have come up with the following code which simply returns an empty list []
I am unable to figure out why my code doesn't work and would really appreciate some help.
(defn replicateList "Replicates each element of the list n times" [l n]
(loop [initList l returnList []]
(if (empty? initList)
returnList
(let [[head & rest] initList]
(loop [x 0]
(when (< x n)
(conj returnList head)
(recur (inc x))))
(recur rest returnList)))))
(defn -main
"Main" []
(test/is (=
(replicateList [1 2] 2)
[1 1 2 2])
"Failed basic test")
)
copying my comment to answer:
this line: (conj returnList head) doesn't modify returnlist, rather it just drops the result in your case. You should restructure your program to pass the accumulated list further to the next iteration. But there are better ways to do this in clojure. Like (defn replicate-list [data times] (apply concat (repeat times data)))
If you still need the loop/recur version for educational reasons, i would go with this:
(defn replicate-list [data times]
(loop [[h & t :as input] data times times result []]
(if-not (pos? times)
result
(if (empty? input)
(recur data (dec times) result)
(recur t times (conj result h))))))
user> (replicate-list [1 2 3] 3)
;;=> [1 2 3 1 2 3 1 2 3]
user> (replicate-list [ ] 2)
;;=> []
user> (replicate-list [1 2 3] -1)
;;=> []
update
based on the clarified question, the simplest way to do this is
(defn replicate-list [data times]
(mapcat (partial repeat times) data))
user> (replicate-list [1 2 3] 3)
;;=> (1 1 1 2 2 2 3 3 3)
and the loop/recur variant:
(defn replicate-list [data times]
(loop [[h & t :as data] data n 0 res []]
(cond (empty? data) res
(>= n times) (recur t 0 res)
:else (recur data (inc n) (conj res h)))))
user> (replicate-list [1 2 3] 3)
;;=> [1 1 1 2 2 2 3 3 3]
user> (replicate-list [1 2 3] 0)
;;=> []
user> (replicate-list [] 10)
;;=> []
Here is a version based on the original post, with minimal modifications:
;; Based on the original version posted
(defn replicateList "Replicates each element of the list n times" [l n]
(loop [initList l returnList []]
(if (empty? initList)
returnList
(let [[head & rest] initList]
(recur
rest
(loop [inner-returnList returnList
x 0]
(if (< x n)
(recur (conj inner-returnList head) (inc x))
inner-returnList)))))))
Please keep in mind that Clojure is mainly a functional language, meaning that most functions produce their results as a new return value instead of updating in place. So, as pointed out in the comment, the line (conj returnList head) will not have an effect, because it's return value is ignored.
The above version works, but does not really take advantage of Clojure's sequence processing facilities. So here are two other suggestions for solving your problem:
;; Using lazy seqs and reduce
(defn replicateList2 [l n]
(reduce into [] (map #(take n (repeat %)) l)))
;; Yet another way using transducers
(defn replicateList3 [l n]
(transduce
(comp (map #(take n (repeat %)))
cat
)
conj
[]
l))
One thing is not clear about your question though: From your implementation, it looks like you want to create a new list where each element is repeated n times, e.g.
playground.replicate> (replicateList [1 2 3] 4)
[1 1 1 1 2 2 2 2 3 3 3 3]
But if you would instead like this result
playground.replicate> (replicateList [1 2 3] 4)
[1 2 3 1 2 3 1 2 3 1 2 3]
the answer to your question will be different.
If you want to learn idiomatic Clojure you should try to find a solution without such low level facilities as loop. Rather try to combine higher level functions like take, repeat, repeatedly. If you're feeling adventurous you might throw in laziness as well. Clojure's sequences are lazy, that is they get evaluated only when needed.
One example I came up with would be
(defn repeat-list-items [l n]
(lazy-seq
(when-let [s (seq l)]
(concat (repeat n (first l))
(repeat-list-items (next l) n)))))
Please also note the common naming with kebab-case
This seems to do what you want pretty well and works for an unlimited input (see the call (range) below), too:
experi.core> (def l [:a :b :c])
#'experi.core/
experi.core> (repeat-list-items l 2)
(:a :a :b :b :c :c)
experi.core> (repeat-list-items l 0)
()
experi.core> (repeat-list-items l 1)
(:a :b :c)
experi.core> (take 10 (drop 10000 (repeat-list-items (range) 4)))
(2500 2500 2500 2500 2501 2501 2501 2501 2502 2502)
If I have a vector [[[1 2 3] [4 5 6] [7 8 9]] [[10 11] [12 13]] [[14] [15]]]
How can I return the positions of each element in the vector?
For example 1 has index [0 0 0], 2 has index [0 0 1], etc
I want something like
(some-fn [[[1 2 3] [4 5 6] [7 8 9]] [[10 11] [12 13]] [[14] [15]]] 1)
=> [0 0 0]
I know that if I have a vector [1 2 3 4], I can do (.indexOf [1 2 3 4] 1) => 0 but how can I extend this to vectors within vectors.
Thanks
and one more solution with zippers:
(require '[clojure.zip :as z])
(defn find-in-vec [x data]
(loop [curr (z/vector-zip data)]
(cond (z/end? curr) nil
(= x (z/node curr)) (let [path (rseq (conj (z/path curr) x))]
(reverse (map #(.indexOf %2 %1) path (rest path))))
:else (recur (z/next curr)))))
user> (find-in-vec 11 data)
(1 0 1)
user> (find-in-vec 12 data)
(1 1 0)
user> (find-in-vec 18 data)
nil
user> (find-in-vec 8 data)
(0 2 1)
the idea is to make a depth-first search for an item, and then reconstruct a path to it, indexing it.
Maybe something like this.
Unlike Asthor's answer it works for any nesting depth (until it runs out of stack). Their answer will give the indices of all items that match, while mine will return the first one. Which one you want depends on the specific use-case.
(defn indexed [coll]
(map-indexed vector coll))
(defn nested-index-of [coll target]
(letfn [(step [indices coll]
(reduce (fn [_ [i x]]
(if (sequential? x)
(when-let [result (step (conj indices i) x)]
(reduced result))
(when (= x target)
(reduced (conj indices i)))))
nil, (indexed coll)))]
(step [] coll)))
(def x [[[1 2 3] [4 5 6] [7 8 9]] [[10 11] [12 13]] [[14] [15]]])
(nested-index-of x 2) ;=> [0 0 1]
(nested-index-of x 15) ;=> [2 1 0]
Edit: Target never changes, so the inner step fn doesn't need it as an argument.
Edit 2: Cause I'm procrastinating here, and recursion is a nice puzzle, maybe you wanted the indices of all matches.
You can tweak my first function slightly to carry around an accumulator.
(defn nested-indices-of [coll target]
(letfn [(step [indices acc coll]
(reduce (fn [acc [i x]]
(if (sequential? x)
(step (conj indices i) acc x)
(if (= x target)
(conj acc (conj indices i))
acc)))
acc, (indexed coll)))]
(step [] [] coll)))
(def y [[[1 2 3] [4 5 6] [7 8 9]] [[10 11] [12 13]] [[14] [15 [16 17 4]]]])
(nested-indices-of y 4) ;=> [[0 1 0] [2 1 1 2]]
Vectors within vectors are no different to ints within vectors:
(.indexOf [[[1 2 3] [4 5 6] [7 8 9]] [[10 11] [12 13]] [[14] [15]]] [[14] [15]])
;;=> 2
The above might be a bit difficult to read, but [[14] [15]] is the third element.
Something like
(defn indexer [vec number]
(for [[x set1] (map-indexed vector vec)
[y set2] (map-indexed vector set1)
[z val] (map-indexed vector set2)
:when (= number val)]
[x y z]))
Written directly into here so not tested. Giving more context on what this would be used for might make it easier to give a good answer as this feels like something you shouldn't end up doing in Clojure.
You can also try and flatten the vectors in some way
An other solution to find the path of every occurrences of a given number.
Usually with functional programming you can go for broader, general, elegant, bite size solution. You will always be able to optimize using language constructs or techniques as you need (tail recursion, use of accumulator, use of lazy-seq, etc)
(defn indexes-of-value [v coll]
(into []
(comp (map-indexed #(if (== v %2) %1))
(remove nil?))
coll))
(defn coord' [v path coll]
(cond
;; node is a leaf: empty or coll of numbers
(or (empty? coll)
(number? (first coll)))
(when-let [indexes (seq (indexes-of-value v coll))]
(map #(conj path %) indexes))
;; node is branch: a coll of colls
(coll? (first coll))
(seq (sequence (comp (map-indexed vector)
(mapcat #(coord' v (conj path (first %)) (second %))))
coll))))
(defn coords [v coll] (coord' v [] coll))
Execution examples:
(def coll [[2 1] [] [7 8 9] [[] [1 2 2 3 2]]])
(coords 2 coll)
=> ([0 0] [3 1 1] [3 1 2] [3 1 4])
As a bonus you can write a function to test if paths are all valid:
(defn valid-coords? [v coll coords]
(->> coords
(map #(get-in coll %))
(remove #(== v %))
empty?))
and try the solution with input generated with clojure.spec:
(s/def ::leaf-vec (s/coll-of nat-int? :kind vector?))
(s/def ::branch-vec (s/or :branch (s/coll-of ::branch-vec :kind vector?
:min-count 1)
:leaf ::leaf-vec))
(let [v 1
coll (first (gen/sample (s/gen ::branch-vec) 1))
res (coords v coll)]
(println "generated coll: " coll)
(if-not (valid-coords? v coll res)
(println "Error:" res)
:ok))
Here is a function that can recursively search for a target value, keeping track of the indexes as it goes:
(ns tst.clj.core
(:use clj.core tupelo.test)
(:require [tupelo.core :as t] ))
(t/refer-tupelo)
(defn index-impl
[idxs data tgt]
(apply glue
(for [[idx val] (zip (range (count data)) data)]
(let [idxs-curr (append idxs idx)]
(if (sequential? val)
(index-impl idxs-curr val tgt)
(if (= val tgt)
[{:idxs idxs-curr :val val}]
[nil]))))))
(defn index [data tgt]
(keep-if not-nil? (index-impl [] data tgt)))
(dotest
(let [data-1 [1 2 3]
data-2 [[1 2 3]
[10 11]
[]]
data-3 [[[1 2 3]
[4 5 6]
[7 8 9]]
[[10 11]
[12 13]]
[[20]
[21]]
[[30]]
[[]]]
]
(spyx (index data-1 2))
(spyx (index data-2 10))
(spyx (index data-3 13))
(spyx (index data-3 21))
(spyx (index data-3 99))
))
with results:
(index data-1 2) => [{:idxs [1], :val 2}]
(index data-2 10) => [{:idxs [1 0], :val 10}]
(index data-3 13) => [{:idxs [1 1 1], :val 13}]
(index data-3 21) => [{:idxs [2 1 0], :val 21}]
(index data-3 99) => []
If we add repeated values we get the following:
data-4 [[[1 2 3]
[4 5 6]
[7 8 9]]
[[10 11]
[12 2]]
[[20]
[21]]
[[30]]
[[2]]]
(index data-4 2) => [{:idxs [0 0 1], :val 2}
{:idxs [1 1 1], :val 2}
{:idxs [4 0 0], :val 2}]
I’m having trouble writing an elegant drop-last-by or butlast-by function.
(drop-last-by odd? [2 1 9 4 7 7 3]) ; => (2 1 9 4)
(drop-last-by odd? [2 4]) ; => (2 4)
(drop-last-by odd? [9]) ; => ()
What I have so far works but seems a little clumsy and I wonder if it can be done in just two or three lines.
(defn drop-last-by [pred coll]
(let [p (partition-by pred coll)]
(apply concat (if (and (seq p) (pred (first (last p))))
(butlast p)
p))))
Since drop-while already does basically what you need, and since your current solution is already not lazy, I'd write drop-last-by like this:
(defn drop-last-by [pred coll]
(reverse (drop-while pred (reverse coll))))
The version below is lazy to the degree permitted by the problem specification:
any elements that do not satisfy the predicate are immediately passed through without reading any additional elements from the source;
any elements that do satisfy the predicate are passed through as soon as an element that does not satisfy the predicate is read in from the source;
any elements that satisfy the predicate and are not followed by further elements that do not satisfy the predicate are dropped.
Additionally, it can be used as a (stateful) transducer; indeed the lazy seq version is implemented in terms of the transducer and clojure.core/sequence.
(defn drop-last-by
([pred]
(fn [rf]
(let [xs (volatile! [])]
(fn
([] (rf))
([result] (rf result))
([result input]
(if-not (pred input)
(do
(reduce rf result #xs)
(vreset! xs [])
(rf result input))
(do
(vswap! xs conj input)
result)))))))
([pred coll]
(sequence (drop-last-by pred) coll)))
At the REPL:
(drop-last-by odd? [2 1 9 4 7 7 3])
;= (2 1 9 4)
(drop-last-by odd? [2 4])
;= (2 4)
(drop-last-by odd? [9])
;= ()
Composed with other transducers:
(into []
(comp (drop-while even?)
(drop-last-by odd?)
(map #(str "foo " %)))
[0 1 2 3 4 5])
;= ["foo 1" "foo 2" "foo 3" "foo 4"]
I was trying to implement selection sort O(n2) in clojure. Yes the underlying sort uses the very efficient java's array sort. However this is a learning exercise.
The code below works, however I was wondering if there is a more idiomatic way of rewriting it, as the below seems clunky -
(defn mins [coll]
(reduce (fn[[min-coll rest-coll] val]
(case (compare val (first min-coll))
-1 [[val] (apply conj rest-coll min-coll)]
0 [(conj min-coll val) rest-coll]
1 [min-coll (conj rest-coll val)]))
[[(first coll)] []]
(rest coll)))
;; (mins [3 1 1 2]) => [[1 1] [3 2]]
(defn selection-sort [coll]
(loop [[sorted coll] [[] coll]]
(let [[s c] (mins coll)]
(if-not (seq coll)
sorted
(recur [(concat sorted s) c])))))
(selection-sort [3 1 1 2 5 7 8 8 4 6])
A functional solution could be:
(defn selection-sort
[input]
(let [ixs (vec (range (count input)))
min-key-from (fn [acc ix] (apply min-key acc (subvec ixs ix)))
swap (fn [coll i j] (assoc coll i (coll j) j (coll i)))]
(reduce
(fn [acc ix] (swap acc ix (min-key-from acc ix))) input ixs)))
You can use the following:
(defn remove-first [coll e]
(if-let [pos (and (seq coll) (.indexOf coll e))]
(vec (concat
(subvec coll 0 pos)
(subvec coll (inc pos))))
coll))
(defn best [coll f]
(reduce f
(first coll)
(rest coll)))
(defn select-sort
([coll] (select-sort coll min))
([coll fmin]
(loop [sorted (transient []) c (vec coll)]
(if (seq c)
(let [n (best c fmin)]
(recur (conj! sorted n) (remove-first c n)))
(persistent! sorted)))))
=> (select-sort [3 5 2])
=> [2 3 5]
=> (select-sort [3 5 2] max)
=> [5 3 2]
I need to take some amount of elements from a sequence based on some quantity rule. Here is a solution I came up with:
(defn take-while-not-enough
[p len xs]
(loop [ac 0
r []
s xs]
(if (empty? s)
r
(let [new-ac (p ac (first s))]
(if (>= new-ac len)
r
(recur new-ac (conj r (first s)) (rest s)))))))
(take-while-not-enough + 10 [2 5 7 8 2 1]) ; [2 5]
(take-while-not-enough #(+ %1 (%2 1)) 7 [[2 5] [7 8] [2 1]]) ; [[2 5]]
Is there any better way to achieve the same?
Thanks.
UPDATE:
Somebody posted that solution, but then removed it. It does the same is the answer that I accepted, but is more readable. Thank you, anonymous well-wisher!
(defn take-while-not-enough [reducer-fn limit data]
(->> (reductions reducer-fn 0 data) ; 1. the sequence of accumulated values
(map vector data) ; 2. paired with the original sequence
(take-while #(< (second %) limit)) ; 3. until a certain accumulated value
(map first))) ; 4. then extract the original values
My first thought is to view this problem as a variation on reduce and thus to break the problem into two steps:
count the number of items in the result
take that many from the input
I also took some liberties with the argument names:
user> (defn take-while-not-enough [reducer-fn limit data]
(take (dec (count (take-while #(< % limit) (reductions reducer-fn 0 data))))
data))
#'user/take-while-not-enough
user> (take-while-not-enough #(+ %1 (%2 1)) 7 [[2 5] [7 8] [2 1]])
([2 5])
user> (take-while-not-enough + 10 [2 5 7 8 2 1])
(2 5)
This returns a sequence and your examples return a vector, if this is important then you can add a call to vec
Something that would traverse the input sequence only once:
(defn take-while-not-enough [r v data]
(->> (rest (reductions (fn [s i] [(r (s 0) i) i]) [0 []] data))
(take-while (comp #(< % v) first))
(map second)))
Well, if you want to use flatland/useful, this is a kinda-okay way to use glue:
(defn take-while-not-enough [p len xs]
(first (glue conj []
(constantly true)
#(>= (reduce p 0 %) len)
xs)))
But it's rebuilding the accumulator for the entire "processed so far" chunk every time it decides whether to grow the chunk more, so it's O(n^2), which will be unacceptable for larger inputs.
The most obvious improvement to your implementation is to make it lazy instead of tail-recursive:
(defn take-while-not-enough [p len xs]
((fn step [acc coll]
(lazy-seq
(when-let [xs (seq coll)]
(let [x (first xs)
acc (p acc x)]
(when-not (>= acc len)
(cons x (step acc xs)))))))
0 xs))
Sometimes lazy-seq is straight-forward and self-explaining.
(defn take-while-not-enough
([f limit coll] (take-while-not-enough f limit (f) coll))
([f limit acc coll]
(lazy-seq
(when-let [s (seq coll)]
(let [fst (first s)
nacc (f acc fst)]
(when (< nxt-sd limit)
(cons fst (take-while-not-enough f limit nacc (rest s)))))))))
Note: f is expected to follow the rules of reduce.