I'm using Project Euler questions to help me learn clojure, and I've run into an exception I can't figure out. nillify and change-all are defined at the bottom for reference.
(loop [the-vector (vec (range 100))
queue (list 2 3 5 7)]
(if queue
(recur (nillify the-vector (first queue)) (next queue))
the-vector))
This throws a NullPointerException, and I can't figure out why. The only part of the code I can see that could throw such an exception is the call to nillify, but it doesn't seem like queue ever gets down to just one element before the exception is thrown---and even if queue were to become empty, that's what the if statement is for.
Any ideas?
"given a vector, a value, and a list of indices, return a vector w/everthing # indice=value"
(defn change-all [the-vector indices val]
(apply assoc the-vector (interleave indices (repeat (count indices) val))))
"given a vector and a val, return a vector in which all entries with indices equal to multiples of val are nilled, but leave the original untouched"
(defn nillify [coll val]
(change-all coll (range (* 2 val) (inc (last coll)) val) nil))
The problem sexpr is
(inc (last coll))
You're changing the contents of the vector, you can't use this to determine the length anymore. Instead:
(count coll)
As a matter of style, use let bindings:
(defn change-all [the-vector indices val]
(let [c (count indices)
s (interleave indices (repeat c val))]
(apply assoc the-vector s)))
(defn nillify [coll val]
(let [c (count coll)
r (range (* 2 val) c val)]
(change-all coll r nil)))
(loop [the-vector (vec (range 100))
[f & r] '(2 3 5 7)]
(if r
(recur (nillify the-vector f) r)
the-vector))
Related
I am trying to complete the Hackerrank Maximum Element challenge found here: https://www.hackerrank.com/challenges/maximum-element/problem
My solution produces the correct output, but times out on the final test cases beginning with #17.
Initially, I used a list and loop/recur to get my answer:
(defn get-query []
(map #(Integer/parseInt %) (clojure.string/split (read-line) #" ")))
(defn stack-stepper [query stack]
(condp = (first query)
1 (conj stack (second query))
2 (rest stack)
3 (do (println (apply max stack)) stack)))
(loop [stack '()
queries-left (Integer/parseInt (read-line))]
(if (> queries-left 0)
(recur (stack-stepper (get-query) stack) (dec queries-left))))
After some research and feedback from other channels, I tried a vector instead of a list, and reduce instead of loop/recur, but the results were the same.
(defn get-query []
(map #(Integer/parseInt %) (clojure.string/split (read-line) #" ")))
(defn get-queries []
(loop [queries []
queries-left (Integer/parseInt (read-line))]
(if (= queries-left 0)
queries
(recur (conj queries (get-query)) (dec queries-left)))))
(defn stack-stepper [stack query]
(condp = (first query)
1 (conj stack (second query))
2 (pop stack)
3 (do (println (apply max stack)) stack)))
(reduce stack-stepper [] (get-queries))
I am still new to FP and Clojure and I would really like to understand what I am missing. I greatly appreciate your time and help!
HackerRank problems are often very demanding from a performance point of view.
The obvious thing to try first is using a transient vector so see if that helps. I tried this:
(let [in (clojure.string/split (slurp *in*) #"\s")
tests (first in)
input-data (map #(Integer/parseInt %) (rest in))]
(loop [v (transient [])
d input-data]
(when (seq d)
(condp = (first d)
1 (recur (conj! v (second d)) (drop 2 d))
2 (recur (pop! v) (rest d))
3 (let [pv (persistent! v)] (println (apply max pv)) (recur (transient pv) (rest d)))))))
If failed at the same point as your solution. Clearly they're looking for something cleverer than that.
The obvious bottleneck is the calculation of the max value on the current stack, which gets re-calculated each time. We can instead save the previous max value on the stack, and recover it as the current max value when we pop the stack:
(defn peek! [tvec] (get tvec (dec (count tvec))))
(let [in (clojure.string/split (slurp *in*) #"\s")
tests (first in)
input-data (map #(Integer/parseInt %) (rest in))]
(loop [v (transient [])
m 0
d input-data]
(when (seq d)
(condp = (first d)
1 (let [snd (second d)
max-now (max m snd)]
(recur (conj! v {:val snd :max-prev m}) max-now (drop 2 d)))
2 (let [popped (peek! v)
max (if popped (:max-prev popped) 0)]
(recur (pop! v) max (rest d)))
3 (do
(println m)
(recur v m (rest d)))))))
Which puts me at rank 1 on the leaderboard :)
I have completed this problem on hackerrank and my solution passes most test cases but it is not fast enough for 4 out of the 11 test cases.
My solution looks like this:
(ns scratch.core
(require [clojure.string :as str :only (split-lines join split)]))
(defn ascii [char]
(int (.charAt (str char) 0)))
(defn process [text]
(let [parts (split-at (int (Math/floor (/ (count text) 2))) text)
left (first parts)
right (if (> (count (last parts)) (count (first parts)))
(rest (last parts))
(last parts))]
(reduce (fn [acc i]
(let [a (ascii (nth left i))
b (ascii (nth (reverse right) i))]
(if (> a b)
(+ acc (- a b))
(+ acc (- b a))))
) 0 (range (count left)))))
(defn print-result [[x & xs]]
(prn x)
(if (seq xs)
(recur xs)))
(let [input (slurp "/Users/paulcowan/Downloads/input10.txt")
inputs (str/split-lines input)
length (read-string (first inputs))
texts (rest inputs)]
(time (print-result (map process texts))))
Can anyone give me any advice about what I should look at to make this faster?
Would using recursion instead of reduce be faster or maybe this line is expensive:
right (if (> (count (last parts)) (count (first parts)))
(rest (last parts))
(last parts))
Because I am getting a count twice.
You are redundantly calling reverse on every iteration of the reduce:
user=> (let [c [1 2 3]
noisey-reverse #(doto (reverse %) println)]
(reduce (fn [acc e] (conj acc (noisey-reverse c) e))
[]
[:a :b :c]))
(3 2 1)
(3 2 1)
(3 2 1)
[(3 2 1) :a (3 2 1) :b (3 2 1) :c]
The reversed value could be calculated inside the containing let, and would then only need to be calculated once.
Also, due to the way your parts is defined, you are doing linear time lookups with each call to nth. It would be better to put parts in a vector and do indexed lookup. In fact you wouldn't need a reversed parts, and could do arithmetic based on the count of the vector to find the item to look up.
I have two columns (vectors) of different length and want to create a new vector of rows (if the column has enough elements). I'm trying to create a new vector (see failed attempt below). In Java this would involve the steps: iterate vector, check condition, append to vector, return vector. Do I need recursion here? I'm sure this is not difficult to solve, but it's very different than procedural code.
(defn rowmaker [colA colB]
"create a row of two columns of possibly different length"
(let [mia (map-indexed vector colA)
rows []]
(doseq [[i elA] mia]
;append if col has enough elements
(if (< i (count colA)) (vec (concat rows elA))) ; ! can't append to rows
(if (< i (count colB)) (vec (concat rows (nth colB i)))
;return rows
rows)))
Expected example input/output
(rowMaker ["A1"] ["B1" "B2"])
; => [["A1" "B1“] [“" "B2"]]
(defn rowMaker [colA colB]
"create a row from two columns"
(let [ca (count colA) cb (count colB)
c (max ca cb)
colA (concat colA (repeat (- c ca) ""))
colB (concat colB (repeat (- c cb) ""))]
(map vector colA colB)))
(defn rowmaker
[cols]
(->> cols
(map #(concat % (repeat "")))
(apply map vector)
(take (->> cols
(map count)
(apply max)))))
I prefer recursion to counting the number of items in collections. Here is my solution.
(defn row-maker
[col-a col-b]
(loop [acc []
as (seq col-a)
bs (seq col-b)]
(if (or as bs)
(recur (conj acc [(or (first as) "") (or (first bs) "")])
(next as)
(next bs))
acc)))
The following does the trick with the given example:
(defn rowMaker [v1 v2]
(mapv vector (concat v1 (repeat "")) v2))
(rowMaker ["A1"] ["B1" "B2"])
;[["A1" "B1"] ["" "B2"]]
However, it doesn't work the other way round:
(rowMaker ["B1" "B2"] ["A1"])
;[["B1" "A1"]]
To make it work both ways, we are going to have to write a version of mapv that fills in for sterile sequences so long as any sequence is fertile. Here is a corresponding lazy version for map, which will work for infinite sequences too:
(defn map-filler [filler f & colls]
(let [filler (vec filler)
colls (vec colls)
live-coll-map (->> colls
(map-indexed vector)
(filter (comp seq second))
(into {}))
split (fn [lcm] (reduce
(fn [[x xm] [i coll]]
(let [[c & cs] coll]
[(assoc x i c) (if cs (assoc xm i cs) xm)]))
[filler {}]
lcm))]
((fn expostulate [lcm]
(lazy-seq
(when (seq lcm)
(let [[this thoses] (split lcm)]
(cons (apply f this) (expostulate thoses))))))
live-coll-map)))
The idea is that you supply a filler sequence with one entry for each of the collections that follow. So we can now define your required rowmaker function thus:
(defn rowmaker [& colls]
(apply map-filler (repeat (count colls) "") vector colls))
This will take any number of collections, and will fill in blank strings for exhausted collections.
(rowmaker ["A1"] ["B1" "B2"])
;(["A1" "B1"] ["" "B2"])
(rowmaker ["B1" "B2"] ["A1"])
;(["B1" "A1"] ["B2" ""])
It works!
(defn make-row
[cola colb r]
(let [pad ""]
(cond
(and (not (empty? cola))
(not (empty? colb))) (recur (rest cola)
(rest colb)
(conj r [(first cola) (first colb)]))
(and (not (empty? cola))
(empty? colb)) (recur (rest cola)
(rest colb)
(conj r [(first cola) pad]))
(and (empty? cola)
(not (empty? colb))) (recur (rest cola)
(rest colb)
(conj r [pad (first colb)]))
:else r)))
I have function
(defn goneSeq [inseq uptil]
(loop [counter 0 newSeq [] orginSeq inseq]
(if (== counter uptil)
newSeq
(recur (inc counter) (conj newSeq (first orginSeq)) (rest orginSeq)))))
(defn insert [sorted-seq n]
(loop [currentSeq sorted-seq counter 0]
(cond (empty? currentSeq) (concat sorted-seq (vector n))
(<= n (first currentSeq)) (concat (goneSeq sorted-seq counter) (vector n) currentSeq)
:else (recur (rest currentSeq) (inc counter)))))
that takes in a sorted-sequence and insert the number n at its appropriate position for example: (insert [1 3 4] 2) returns [1 2 3 4].
Now I want to use this function with reduce to sort a given sequence so something like:
(reduce (insert seq n) givenSeq)
What is thr correct way to achieve this?
If the function works for inserting a single value, then this would work:
(reduce insert [] givenSeq)
for example:
user> (reduce insert [] [0 1 2 30.5 0.88 2.2])
(0 0.88 1 2 2.2 30.5)
Also, it should be noted that sort and sort-by are built in and are better than most hand-rolled solutions.
May I suggest some simpler ways to do insert:
A slowish lazy way is
(defn insert [s x]
(let [[fore aft] (split-with #(> x %) s)]
(concat fore (cons x aft))))
A faster eager way is
(defn insert [coll x]
(loop [fore [], coll coll]
(if (and (seq coll) (> x (first coll)))
(recur (conj fore x) (rest coll))
(concat fore (cons x coll)))))
By the way, you had better put your defns in bottom-up order, if possible. Use declare if there is mutual recursion. You had me thinking your solution did not compile.
I'm working on a project to learn Clojure in practice. I'm doing well, but sometimes I get stuck. This time I need to transform sequence of the form:
[":keyword0" "word0" "word1" ":keyword1" "word2" "word3"]
into:
[[:keyword0 "word0" "word1"] [:keyword1 "word2" "word3"]]
I'm trying for at least two hours, but I know not so many Clojure functions to compose something useful to solve the problem in functional manner.
I think that this transformation should include some partition, here is my attempt:
(partition-by (fn [x] (.startsWith x ":")) *1)
But the result looks like this:
((":keyword0") ("word1" "word2") (":keyword1") ("word3" "word4"))
Now I should group it again... I doubt that I'm doing right things here... Also, I need to convert strings (only those that begin with :) into keywords. I think this combination should work:
(keyword (subs ":keyword0" 1))
How to write a function which performs the transformation in most idiomatic way?
Here is a high performance version, using reduce
(reduce (fn [acc next]
(if (.startsWith next ":")
(conj acc [(-> next (subs 1) keyword)])
(conj (pop acc) (conj (peek acc)
next))))
[] data)
Alternatively, you could extend your code like this
(->> data
(partition-by #(.startsWith % ":"))
(partition 2)
(map (fn [[[kw-str] strs]]
(cons (-> kw-str
(subs 1)
keyword)
strs))))
what about that:
(defn group-that [ arg ]
(if (not-empty arg)
(loop [list arg, acc [], result []]
(if (not-empty list)
(if (.startsWith (first list) ":")
(if (not-empty acc)
(recur (rest list) (vector (first list)) (conj result acc))
(recur (rest list) (vector (first list)) result))
(recur (rest list) (conj acc (first list)) result))
(conj result acc)
))))
Just 1x iteration over the Seq and without any need of macros.
Since the question is already here... This is my best effort:
(def data [":keyword0" "word0" "word1" ":keyword1" "word2" "word3"])
(->> data
(partition-by (fn [x] (.startsWith x ":")))
(partition 2)
(map (fn [[[k] w]] (apply conj [(keyword (subs k 1))] w))))
I'm still looking for a better solution or criticism of this one.
First, let's construct a function that breaks vector v into sub-vectors, the breaks occurring everywhere property pred holds.
(defn breakv-by [pred v]
(let [break-points (filter identity (map-indexed (fn [n x] (when (pred x) n)) v))
starts (cons 0 break-points)
finishes (concat break-points [(count v)])]
(mapv (partial subvec v) starts finishes)))
For our case, given
(def data [":keyword0" "word0" "word1" ":keyword1" "word2" "word3"])
then
(breakv-by #(= (first %) \:) data)
produces
[[] [":keyword0" "word0" "word1"] [":keyword1" "word2" "word3"]]
Notice that the initial sub-vector is different:
It has no element for which the predicate holds.
It can be of length zero.
All the others
start with their only element for which the predicate holds and
are at least of length 1.
So breakv-by behaves properly with data that
doesn't start with a breaking element or
has a succession of breaking elements.
For the purposes of the question, we need to muck about with what breakv-by produces somewhat:
(let [pieces (breakv-by #(= (first %) \:) data)]
(mapv
#(update-in % [0] (fn [s] (keyword (subs s 1))))
(rest pieces)))
;[[:keyword0 "word0" "word1"] [:keyword1 "word2" "word3"]]