As a newbie to Clojure I often have difficulties to express the simplest things. For example, for replacing the last element in a vector, which would be
v[-1]=new_value
in python, I end up with the following variants in Clojure:
(assoc v (dec (count v)) new_value)
which is pretty long and inexpressive to say the least, or
(conj (vec (butlast v)) new_value)
which even worse, as it has O(n) running time.
That leaves me feeling silly, like a caveman trying to repair a Swiss watch with a club.
What is the right Clojure way to replace the last element in a vector?
To support my O(n)-claim for butlast-version (Clojure 1.8):
(def v (vec (range 1e6)))
#'user/v
user=> (time (first (conj (vec (butlast v)) 55)))
"Elapsed time: 232.686159 msecs"
0
(def v (vec (range 1e7)))
#'user/v
user=> (time (first (conj (vec (butlast v)) 55)))
"Elapsed time: 2423.828127 msecs"
0
So basically for 10 time the number of elements it is 10 times slower.
I'd use
(defn set-top [coll x]
(conj (pop coll) x))
For example,
(set-top [1 2 3] :a)
=> [1 2 :a]
But it also works on the front of lists:
(set-top '(1 2 3) :a)
=> (:a 2 3)
The Clojure stack functions - peek, pop, and conj - work on the natural open end of a sequential collection.
But there is no one right way.
How do the various solutions react to an empty vector?
Your Python v[-1]=new_value throws an exception, as does your (assoc v (dec (count v)) new_value) and my (defn set-top [coll x] (conj (pop coll) x)).
Your (conj (vec (butlast v)) new_value) returns [new_value]. The butlast has no effect.
If you insist on being "pure", your 2nd or 3rd solutions will work. I prefer to be simpler & more explicit using the helper functions from the Tupelo library:
(s/defn replace-at :- ts/List
"Replaces an element in a collection at the specified index."
[coll :- ts/List
index :- s/Int
elem :- s/Any]
...)
(is (= [9 1 2] (replace-at (range 3) 0 9)))
(is (= [0 9 2] (replace-at (range 3) 1 9)))
(is (= [0 1 9] (replace-at (range 3) 2 9)))
As with drop-at, replace-at will throw an exception for invalid values of index.
Similar helper functions exist for
insert-at
drop-at
prepend
append
Note that all of the above work equally well for either a Clojure list (eager or lazy) or a Clojure vector. The conj solution will fail unless you are careful to always coerce the input to a vector first as in your example.
What is the best way of implementing map function together with an updatable state between applications of function to each element of sequence? To illustrate the issue let's suppose that we have a following problem:
I have a vector of the numbers. I want a new sequence where each element is multiplied by 2 and then added number of 10's in the sequence up to and including the current element. For example from:
[20 30 40 10 20 10 30]
I want to generate:
[40 60 80 21 41 22 62]
Without adding the count of 10 the solution can be formulated using a high level of abstraction:
(map #(* 2 %) [20 30 40 10 20 10 30])
Having count to update forced me to "go to basic" and the solution I came up with is:
(defn my-update-state [x state]
(if (= x 10) (+ state 1) state)
)
(defn my-combine-with-state [x state]
(+ x state))
(defn map-and-update-state [vec fun state update-state combine-with-state]
(when-not (empty? vec)
(let [[f & other] vec
g (fun f)
new-state (update-state f state)]
(cons (combine-with-state g new-state) (map-and-update-state other fun new-state update-state combine-with-state))
)))
(map-and-update-state [20 30 40 50 10 20 10 30 ] #(* 2 %) 0 my-update-state my-combine-with-state )
My question: is it the appropriate/canonical way to solve the problem or I overlooked some important concepts/functions.
PS:
The original problem is walking AST (abstract syntax tree) and generating new AST together with updating symbol table, so when proposing the solution to the problem above please keep it in mind.
I do not worry about blowing up stack, so replacement with loop+recur is not
my concern here.
Is using global Vars or Refs instead of passing state as an argument a definite no-no?
You can use reduce to accumulate a pair of the number of 10s seen so far and the current vector of results.:
(defn map-update [v]
(letfn [(update [[ntens v] i]
(let [ntens (+ ntens (if (= 10 i) 1 0))]
[ntens (conj v (+ ntens (* 2 i)))]))]
(second (reduce update [0 []] v))))
To count # of 10 you can do
(defn count-10[col]
(reductions + (map #(if (= % 10) 1 0) col)))
Example:
user=> (count-10 [1 2 10 20 30 10 1])
(0 0 1 1 1 2 2)
And then a simple map for the final result
(map + col col (count-10 col)))
Reduce and reductions are good ways to traverse a sequence keeping a state. If you feel your code is not clear you can always use recursion with loop/recur or lazy-seq like this
(defn twice-plus-ntens
([coll] (twice-plus-ntens coll 0))
([coll ntens]
(lazy-seq
(when-let [s (seq coll)]
(let [item (first s)
new-ntens (if (= 10 item) (inc ntens) ntens)]
(cons (+ (* 2 item) new-ntens)
(twice-plus-ntens (rest s) new-ntens)))))))
have a look at map source code evaluating this at your repl
(source map)
I've skipped chunked optimization and multiple collection support.
You can make it a higher-order function this way
(defn map-update
([mf uf coll] (map-update mf uf (uf) coll))
([mf uf val coll]
(lazy-seq
(when-let [s (seq coll)]
(let [item (first s)
new-status (uf item val)]
(cons (mf item new-status)
(map-update mf uf new-status (rest s))))))))
(defn update-f
([] 0)
([item status]
(if (= item 10) (inc status) status)))
(defn map-f [item status]
(+ (* 2 item) status))
(map-update map-f update-f in)
The most appropriate way is to use function with state
(into
[]
(map
(let [mem (atom 0)]
(fn [val]
(when (== val 10) (swap! mem inc))
(+ #mem (* val 2)))))
[20 30 40 10 20 10 30])
also see
memoize
standard function
What if map and doseq had a baby? I'm trying to write a function or macro like Common Lisp's mapc, but in Clojure. This does essentially what map does, but only for side-effects, so it doesn't need to generate a sequence of results, and wouldn't be lazy. I know that one can iterate over a single sequence using doseq, but map can iterate over multiple sequences, applying a function to each element in turn of all of the sequences. I also know that one can wrap map in dorun. (Note: This question has been extensively edited after many comments and a very thorough answer. The original question focused on macros, but those macro issues turned out to be peripheral.)
This is fast (according to criterium):
(defn domap2
[f coll]
(dotimes [i (count coll)]
(f (nth coll i))))
but it only accepts one collection. This accepts arbitrary collections:
(defn domap3
[f & colls]
(dotimes [i (apply min (map count colls))]
(apply f (map #(nth % i) colls))))
but it's very slow by comparison. I could also write a version like the first, but with different parameter cases [f c1 c2], [f c1 c2 c3], etc., but in the end, I'll need a case that handles arbitrary numbers of collections, like the last example, which is simpler anyway. I've tried many other solutions as well.
Since the second example is very much like the first except for the use of apply and the map inside the loop, I suspect that getting rid of them would speed things up a lot. I have tried to do this by writing domap2 as a macro, but the way that the catch-all variable after & is handled keeps tripping me up, as illustrated above.
Other examples (out of 15 or 20 different versions), benchmark code, and times on a Macbook Pro that's a few years old (full source here):
(defn domap1
[f coll]
(doseq [e coll]
(f e)))
(defn domap7
[f coll]
(dorun (map f coll)))
(defn domap18
[f & colls]
(dorun (apply map f colls)))
(defn domap15
[f coll]
(when (seq coll)
(f (first coll))
(recur f (rest coll))))
(defn domap17
[f & colls]
(let [argvecs (apply (partial map vector) colls)] ; seq of ntuples of interleaved vals
(doseq [args argvecs]
(apply f args))))
I'm working on an application that uses core.matrix matrices and vectors, but feel free to substitute your own side-effecting functions below.
(ns tst
(:use criterium.core
[clojure.core.matrix :as mx]))
(def howmany 1000)
(def a-coll (vec (range howmany)))
(def maskvec (zero-vector :vectorz howmany))
(defn unmaskit!
[idx]
(mx/mset! maskvec idx 1.0)) ; sets element idx of maskvec to 1.0
(defn runbench
[domapfn label]
(print (str "\n" label ":\n"))
(bench (def _ (domapfn unmaskit! a-coll))))
Mean execution times according to Criterium, in microseconds:
domap1: 12.317551 [doseq]
domap2: 19.065317 [dotimes]
domap3: 265.983779 [dotimes with apply, map]
domap7: 53.263230 [map with dorun]
domap18: 54.456801 [map with dorun, multiple collections]
domap15: 32.034993 [recur]
domap17: 95.259984 [doseq, multiple collections interleaved using map]
EDIT: It may be that dorun+map is the best way to implement domap for multiple large lazy sequence arguments, but doseq is still king when it comes to single lazy sequences. Performing the same operation as unmask! above, but running the index through (mod idx 1000), and iterating over (range 100000000), doseq is about twice as fast as dorun+map in my tests (i.e. (def domap25 (comp dorun map))).
You don't need a macro, and I don't see why a macro would be helpful here.
user> (defn do-map [f & lists] (apply mapv f lists) nil)
#'user/do-map
user> (do-map (comp println +) (range 2 6) (range 8 11) (range 22 40))
32
35
38
nil
note do-map here is eager (thanks to mapv) and only executes for side effects
Macros can use varargs lists, as the (useless!) macro version of do-map demonstrates:
user> (defmacro do-map-macro [f & lists] `(do (mapv ~f ~#lists) nil))
#'user/do-map-macro
user> (do-map-macro (comp println +) (range 2 6) (range 8 11) (range 22 40))
32
35
38
nil
user> (macroexpand-1 '(do-map-macro (comp println +) (range 2 6) (range 8 11) (range 22 40)))
(do (clojure.core/mapv (comp println +) (range 2 6) (range 8 11) (range 22 40)) nil)
Addendum:
addressing the efficiency / garbage-creation concerns:
note that below I truncate the output of the criterium bench function, for conciseness reasons:
(defn do-map-loop
[f & lists]
(loop [heads lists]
(when (every? seq heads)
(apply f (map first heads))
(recur (map rest heads)))))
user> (crit/bench (with-out-str (do-map-loop (comp println +) (range 2 6) (range 8 11) (range 22 40))))
...
Execution time mean : 11.367804 µs
...
This looks promising because it doesn't create a data structure that we aren't using anyway (unlike mapv above). But it turns out it is slower than the previous (maybe because of the two map calls?).
user> (crit/bench (with-out-str (do-map-macro (comp println +) (range 2 6) (range 8 11) (range 22 40))))
...
Execution time mean : 7.427182 µs
...
user> (crit/bench (with-out-str (do-map (comp println +) (range 2 6) (range 8 11) (range 22 40))))
...
Execution time mean : 8.355587 µs
...
Since the loop still wasn't faster, let's try a version which specializes on arity, so that we don't need to call map twice on every iteration:
(defn do-map-loop-3
[f a b c]
(loop [[a & as] a
[b & bs] b
[c & cs] c]
(when (and a b c)
(f a b c)
(recur as bs cs))))
Remarkably, though this is faster, it is still slower than the version that just used mapv:
user> (crit/bench (with-out-str (do-map-loop-3 (comp println +) (range 2 6) (range 8 11) (range 22 40))))
...
Execution time mean : 9.450108 µs
...
Next I wondered if the size of the input was a factor. With larger inputs...
user> (def test-input (repeatedly 3 #(range (rand-int 100) (rand-int 1000))))
#'user/test-input
user> (map count test-input)
(475 531 511)
user> (crit/bench (with-out-str (apply do-map-loop-3 (comp println +) test-input)))
...
Execution time mean : 1.005073 ms
...
user> (crit/bench (with-out-str (apply do-map (comp println +) test-input)))
...
Execution time mean : 756.955238 µs
...
Finally, for completeness, the timing of do-map-loop (which as expected is slightly slower than do-map-loop-3)
user> (crit/bench (with-out-str (apply do-map-loop (comp println +) test-input)))
...
Execution time mean : 1.553932 ms
As we see, even with larger input sizes, mapv is faster.
(I should note for completeness here that map is slightly faster than mapv, but not by a large degree).
I looked at maps source code which basically keeps creating lazy sequences. I would think that iterating over a collection and adding to a transient vector would be faster, but clearly it isn't. What don't I understand about clojures performance behavior?
;=> (time (do-with / (range 1 1000) (range 1 1000)))
;"Elapsed time: 23.1808 msecs"
;
; vs
;=> (time (doall (map #(/ %1 %2) (range 1 1000) (range 1 1000))))
;"Elapsed time: 2.604174 msecs"
(defn do-with
[fn coll1 coll2]
(let [end (count coll1)]
(loop [i 0
res (transient [])]
(if
(= i end)
(persistent! res)
(let [x (nth coll1 i)
y (nth coll2 i)
r (fn x y)]
(recur (inc i) (conj! res r)))
))))
In order of conjectured impact on relative results:
Your do-with function uses nth to access the individual items in the input collections. nth operates in linear time on ranges, making do-with quadratic. Needless to say, this will kill performance on large collections.
range produces chunked seqs and map handles those extremely efficiently. (Essentially it produces chunks of up to 32 elements -- here it will in fact be exactly 32 -- by running a tight loop over the internal array of each input chunk in turn, placing results in internal arrays of output chunks.)
Benchmarking with time doesn't give you steady state performance. (Which is why one should really use a proper benchmarking library; in the case of Clojure, Criterium is the standard solution.)
Incidentally, (map #(/ %1 %2) xs ys) can simply be written as (map / xs ys).
Update:
I've benchmarked the map version, the original do-with and a new do-with version with Criterium, using (range 1 1000) as both inputs in each case (as in the question text), obtaining the following mean execution times:
;;; (range 1 1000)
new do-with 170.383334 µs
(doall (map ...)) 230.756753 µs
original do-with 15.624444 ms
Additionally, I've repeated the benchmark using a vector stored in a Var as input rather than ranges (that is, with (def r (vec (range 1 1000))) at the start and using r as both collection arguments in each benchmark). Unsurprisingly, the original do-with came in first -- nth is very fast on vectors (plus using nth with a vector avoids all the intermediate allocations involved in seq traversal).
;;; (vec (range 1 1000))
original do-with 73.975419 µs
new do-with 87.399952 µs
(doall (map ...)) 153.493128 µs
Here's the new do-with with linear time complexity:
(defn do-with [f xs ys]
(loop [xs (seq xs)
ys (seq ys)
ret (transient [])]
(if (and xs ys)
(recur (next xs)
(next ys)
(conj! ret (f (first xs) (first ys))))
(persistent! ret))))
I've been working on solving Project Euler problems in Clojure to get better, and I've already run into prime number generation a couple of times. My problem is that it is just taking way too long. I was hoping someone could help me find an efficient way to do this in a Clojure-y way.
When I fist did this, I brute-forced it. That was easy to do. But calculating 10001 prime numbers took 2 minutes this way on a Xeon 2.33GHz, too long for the rules, and too long in general. Here was the algorithm:
(defn next-prime-slow
"Find the next prime number, checking against our already existing list"
([sofar guess]
(if (not-any? #(zero? (mod guess %)) sofar)
guess ; Then we have a prime
(recur sofar (+ guess 2))))) ; Try again
(defn find-primes-slow
"Finds prime numbers, slowly"
([]
(find-primes-slow 10001 [2 3])) ; How many we need, initial prime seeds
([needed sofar]
(if (<= needed (count sofar))
sofar ; Found enough, we're done
(recur needed (concat sofar [(next-prime-slow sofar (last sofar))])))))
By replacing next-prime-slow with a newer routine that took some additional rules into account (like the 6n +/- 1 property) I was able to speed things up to about 70 seconds.
Next I tried making a sieve of Eratosthenes in pure Clojure. I don't think I got all the bugs out, but I gave up because it was simply way too slow (even worse than the above, I think).
(defn clean-sieve
"Clean the sieve of what we know isn't prime based"
[seeds-left sieve]
(if (zero? (count seeds-left))
sieve ; Nothing left to filter the list against
(recur
(rest seeds-left) ; The numbers we haven't checked against
(filter #(> (mod % (first seeds-left)) 0) sieve)))) ; Filter out multiples
(defn self-clean-sieve ; This seems to be REALLY slow
"Remove the stuff in the sieve that isn't prime based on it's self"
([sieve]
(self-clean-sieve (rest sieve) (take 1 sieve)))
([sieve clean]
(if (zero? (count sieve))
clean
(let [cleaned (filter #(> (mod % (last clean)) 0) sieve)]
(recur (rest cleaned) (into clean [(first cleaned)]))))))
(defn find-primes
"Finds prime numbers, hopefully faster"
([]
(find-primes 10001 [2]))
([needed seeds]
(if (>= (count seeds) needed)
seeds ; We have enough
(recur ; Recalculate
needed
(into
seeds ; Stuff we've already found
(let [start (last seeds)
end-range (+ start 150000)] ; NOTE HERE
(reverse
(self-clean-sieve
(clean-sieve seeds (range (inc start) end-range))))))))))
This is bad. It also causes stack overflows if the number 150000 is smaller. This despite the fact I'm using recur. That may be my fault.
Next I tried a sieve, using Java methods on a Java ArrayList. That took quite a bit of time, and memory.
My latest attempt is a sieve using a Clojure hash-map, inserting all the numbers in the sieve then dissoc'ing numbers that aren't prime. At the end, it takes the key list, which are the prime numbers it found. It takes about 10-12 seconds to find 10000 prime numbers. I'm not sure it's fully debugged yet. It's recursive too (using recur and loop), since I'm trying to be Lispy.
So with these kind of problems, problem 10 (sum up all primes under 2000000) is killing me. My fastest code came up with the right answer, but it took 105 seconds to do it, and needed quite a bit of memory (I gave it 512 MB just so I wouldn't have to fuss with it). My other algorithms take so long I always ended up stopping them first.
I could use a sieve to calculate that many primes in Java or C quite fast and without using so much memory. I know I must be missing something in my Clojure/Lisp style that's causing the problem.
Is there something I'm doing really wrong? Is Clojure just kinda slow with large sequences? Reading some of the project Euler discussions people have calculated the first 10000 primes in other Lisps in under 100 miliseconds. I realize the JVM may slow things down and Clojure is relatively young, but I wouldn't expect a 100x difference.
Can someone enlighten me on a fast way to calculate prime numbers in Clojure?
Here's another approach that celebrates Clojure's Java interop. This takes 374ms on a 2.4 Ghz Core 2 Duo (running single-threaded). I let the efficient Miller-Rabin implementation in Java's BigInteger#isProbablePrime deal with the primality check.
(def certainty 5)
(defn prime? [n]
(.isProbablePrime (BigInteger/valueOf n) certainty))
(concat [2] (take 10001
(filter prime?
(take-nth 2
(range 1 Integer/MAX_VALUE)))))
The Miller-Rabin certainty of 5 is probably not very good for numbers much larger than this. That certainty is equal to 96.875% certain it's prime (1 - .5^certainty)
I realize this is a very old question, but I recently ended up looking for the same and the links here weren't what I'm looking for (restricted to functional types as much as possible, lazily generating ~every~ prime I want).
I stumbled upon a nice F# implementation, so all credits are his. I merely ported it to Clojure:
(defn gen-primes "Generates an infinite, lazy sequence of prime numbers"
[]
(letfn [(reinsert [table x prime]
(update-in table [(+ prime x)] conj prime))
(primes-step [table d]
(if-let [factors (get table d)]
(recur (reduce #(reinsert %1 d %2) (dissoc table d) factors)
(inc d))
(lazy-seq (cons d (primes-step (assoc table (* d d) (list d))
(inc d))))))]
(primes-step {} 2)))
Usage is simply
(take 5 (gen-primes))
Very late to the party, but I'll throw in an example, using Java BitSets:
(defn sieve [n]
"Returns a BitSet with bits set for each prime up to n"
(let [bs (new java.util.BitSet n)]
(.flip bs 2 n)
(doseq [i (range 4 n 2)] (.clear bs i))
(doseq [p (range 3 (Math/sqrt n))]
(if (.get bs p)
(doseq [q (range (* p p) n (* 2 p))] (.clear bs q))))
bs))
Running this on a 2014 Macbook Pro (2.3GHz Core i7), I get:
user=> (time (do (sieve 1e6) nil))
"Elapsed time: 64.936 msecs"
See the last example here:
http://clojuredocs.org/clojure_core/clojure.core/lazy-seq
;; An example combining lazy sequences with higher order functions
;; Generate prime numbers using Eratosthenes Sieve
;; See http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
;; Note that the starting set of sieved numbers should be
;; the set of integers starting with 2 i.e., (iterate inc 2)
(defn sieve [s]
(cons (first s)
(lazy-seq (sieve (filter #(not= 0 (mod % (first s)))
(rest s))))))
user=> (take 20 (sieve (iterate inc 2)))
(2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71)
Here's a nice and simple implementation:
http://clj-me.blogspot.com/2008/06/primes.html
... but it is written for some pre-1.0 version of Clojure. See lazy_seqs in Clojure Contrib for one that works with the current version of the language.
(defn sieve
[[p & rst]]
;; make sure the stack size is sufficiently large!
(lazy-seq (cons p (sieve (remove #(= 0 (mod % p)) rst)))))
(def primes (sieve (iterate inc 2)))
with a 10M stack size, I get the 1001th prime in ~ 33 seconds on a 2.1Gz macbook.
So I've just started with Clojure, and yeah, this comes up a lot on Project Euler doesn't it? I wrote a pretty fast trial division prime algorithm, but it doesn't really scale too far before each run of divisions becomes prohibitively slow.
So I started again, this time using the sieve method:
(defn clense
"Walks through the sieve and nils out multiples of step"
[primes step i]
(if (<= i (count primes))
(recur
(assoc! primes i nil)
step
(+ i step))
primes))
(defn sieve-step
"Only works if i is >= 3"
[primes i]
(if (< i (count primes))
(recur
(if (nil? (primes i)) primes (clense primes (* 2 i) (* i i)))
(+ 2 i))
primes))
(defn prime-sieve
"Returns a lazy list of all primes smaller than x"
[x]
(drop 2
(filter (complement nil?)
(persistent! (sieve-step
(clense (transient (vec (range x))) 2 4) 3)))))
Usage and speed:
user=> (time (do (prime-sieve 1E6) nil))
"Elapsed time: 930.881 msecs
I'm pretty happy with the speed: it's running out of a REPL running on a 2009 MBP. It's mostly fast because I completely eschew idiomatic Clojure and instead loop around like a monkey. It's also 4X faster because I'm using a transient vector to work on the sieve instead of staying completely immutable.
Edit: After a couple of suggestions / bug fixes from Will Ness it now runs a whole lot faster.
Here's a simple sieve in Scheme:
http://telegraphics.com.au/svn/puzzles/trunk/programming-in-scheme/primes-up-to.scm
Here's a run for primes up to 10,000:
#;1> (include "primes-up-to.scm")
; including primes-up-to.scm ...
#;2> ,t (primes-up-to 10000)
0.238s CPU time, 0.062s GC time (major), 180013 mutations, 130/4758 GCs (major/minor)
(2 3 5 7 11 13...
Here is a Clojure solution. i is the current number being considered and p is a list of all prime numbers found so far. If division by some prime numbers has a remainder of zero, the number i is not a prime number and recursion occurs with the next number. Otherwise the prime number is added to p in the next recursion (as well as continuing with the next number).
(defn primes [i p]
(if (some #(zero? (mod i %)) p)
(recur (inc i) p)
(cons i (lazy-seq (primes (inc i) (conj p i))))))
(time (do (doall (take 5001 (primes 2 []))) nil))
; Elapsed time: 2004.75587 msecs
(time (do (doall (take 10001 (primes 2 []))) nil))
; Elapsed time: 7700.675118 msecs
Update:
Here is a much slicker solution based on this answer above.
Basically the list of integers starting with two is filtered lazily. Filtering is performed by only accepting a number i if there is no prime number dividing the number with remainder of zero. All prime numbers are tried where the square of the prime number is less or equal to i.
Note that primes is used recursively but Clojure manages to prevent endless recursion. Also note that the lazy sequence primes caches results (that's why the performance results are a bit counter intuitive at first sight).
(def primes
(lazy-seq
(filter (fn [i] (not-any? #(zero? (rem i %))
(take-while #(<= (* % %) i) primes)))
(drop 2 (range)))))
(time (first (drop 10000 primes)))
; Elapsed time: 542.204211 msecs
(time (first (drop 20000 primes)))
; Elapsed time: 786.667644 msecs
(time (first (drop 40000 primes)))
; Elapsed time: 1780.15807 msecs
(time (first (drop 40000 primes)))
; Elapsed time: 8.415643 msecs
Based on Will's comment, here is my take on postponed-primes:
(defn postponed-primes-recursive
([]
(concat (list 2 3 5 7)
(lazy-seq (postponed-primes-recursive
{}
3
9
(rest (rest (postponed-primes-recursive)))
9))))
([D p q ps c]
(letfn [(add-composites
[D x s]
(loop [a x]
(if (contains? D a)
(recur (+ a s))
(persistent! (assoc! (transient D) a s)))))]
(loop [D D
p p
q q
ps ps
c c]
(if (not (contains? D c))
(if (< c q)
(cons c (lazy-seq (postponed-primes-recursive D p q ps (+ 2 c))))
(recur (add-composites D
(+ c (* 2 p))
(* 2 p))
(first ps)
(* (first ps) (first ps))
(rest ps)
(+ c 2)))
(let [s (get D c)]
(recur (add-composites
(persistent! (dissoc! (transient D) c))
(+ c s)
s)
p
q
ps
(+ c 2))))))))
Initial submission for comparison:
Here is my attempt to port this prime number generator from Python to Clojure. The below returns an infinite lazy sequence.
(defn primes
[]
(letfn [(prime-help
[foo bar]
(loop [D foo
q bar]
(if (nil? (get D q))
(cons q (lazy-seq
(prime-help
(persistent! (assoc! (transient D) (* q q) (list q)))
(inc q))))
(let [factors-of-q (get D q)
key-val (interleave
(map #(+ % q) factors-of-q)
(map #(cons % (get D (+ % q) (list)))
factors-of-q))]
(recur (persistent!
(dissoc!
(apply assoc! (transient D) key-val)
q))
(inc q))))))]
(prime-help {} 2)))
Usage:
user=> (first (primes))
2
user=> (second (primes))
3
user=> (nth (primes) 100)
547
user=> (take 5 (primes))
(2 3 5 7 11)
user=> (time (nth (primes) 10000))
"Elapsed time: 409.052221 msecs"
104743
edit:
Performance comparison, where postponed-primes uses a queue of primes seen so far rather than a recursive call to postponed-primes:
user=> (def counts (list 200000 400000 600000 800000))
#'user/counts
user=> (map #(time (nth (postponed-primes) %)) counts)
("Elapsed time: 1822.882 msecs"
"Elapsed time: 3985.299 msecs"
"Elapsed time: 6916.98 msecs"
"Elapsed time: 8710.791 msecs"
2750161 5800139 8960467 12195263)
user=> (map #(time (nth (postponed-primes-recursive) %)) counts)
("Elapsed time: 1776.843 msecs"
"Elapsed time: 3874.125 msecs"
"Elapsed time: 6092.79 msecs"
"Elapsed time: 8453.017 msecs"
2750161 5800139 8960467 12195263)
Idiomatic, and not too bad
(def primes
(cons 1 (lazy-seq
(filter (fn [i]
(not-any? (fn [p] (zero? (rem i p)))
(take-while #(<= % (Math/sqrt i))
(rest primes))))
(drop 2 (range))))))
=> #'user/primes
(first (time (drop 10000 primes)))
"Elapsed time: 0.023135 msecs"
=> 104729
From: http://steloflute.tistory.com/entry/Clojure-%ED%94%84%EB%A1%9C%EA%B7%B8%EB%9E%A8-%EC%B5%9C%EC%A0%81%ED%99%94
Using Java array
(defmacro loopwhile [init-symbol init whilep step & body]
`(loop [~init-symbol ~init]
(when ~whilep ~#body (recur (+ ~init-symbol ~step)))))
(defn primesUnderb [limit]
(let [p (boolean-array limit true)]
(loopwhile i 2 (< i (Math/sqrt limit)) 1
(when (aget p i)
(loopwhile j (* i 2) (< j limit) i (aset p j false))))
(filter #(aget p %) (range 2 limit))))
Usage and speed:
user=> (time (def p (primesUnderb 1e6)))
"Elapsed time: 104.065891 msecs"
After coming to this thread and searching for a faster alternative to those already here, I am surprised nobody linked to the following article by Christophe Grand :
(defn primes3 [max]
(let [enqueue (fn [sieve n factor]
(let [m (+ n (+ factor factor))]
(if (sieve m)
(recur sieve m factor)
(assoc sieve m factor))))
next-sieve (fn [sieve candidate]
(if-let [factor (sieve candidate)]
(-> sieve
(dissoc candidate)
(enqueue candidate factor))
(enqueue sieve candidate candidate)))]
(cons 2 (vals (reduce next-sieve {} (range 3 max 2))))))
As well as a lazy version :
(defn lazy-primes3 []
(letfn [(enqueue [sieve n step]
(let [m (+ n step)]
(if (sieve m)
(recur sieve m step)
(assoc sieve m step))))
(next-sieve [sieve candidate]
(if-let [step (sieve candidate)]
(-> sieve
(dissoc candidate)
(enqueue candidate step))
(enqueue sieve candidate (+ candidate candidate))))
(next-primes [sieve candidate]
(if (sieve candidate)
(recur (next-sieve sieve candidate) (+ candidate 2))
(cons candidate
(lazy-seq (next-primes (next-sieve sieve candidate)
(+ candidate 2))))))]
(cons 2 (lazy-seq (next-primes {} 3)))))
Plenty of answers already, but I have an alternative solution which generates an infinite sequence of primes. I was also interested on bechmarking a few solutions.
First some Java interop. for reference:
(defn prime-fn-1 [accuracy]
(cons 2
(for [i (range)
:let [prime-candidate (-> i (* 2) (+ 3))]
:when (.isProbablePrime (BigInteger/valueOf prime-candidate) accuracy)]
prime-candidate)))
Benjamin # https://stackoverflow.com/a/7625207/3731823 is primes-fn-2
nha # https://stackoverflow.com/a/36432061/3731823 is primes-fn-3
My implementations is primes-fn-4:
(defn primes-fn-4 []
(let [primes-with-duplicates
(->> (for [i (range)] (-> i (* 2) (+ 5))) ; 5, 7, 9, 11, ...
(reductions
(fn [known-primes candidate]
(if (->> known-primes
(take-while #(<= (* % %) candidate))
(not-any? #(-> candidate (mod %) zero?)))
(conj known-primes candidate)
known-primes))
[3]) ; Our initial list of known odd primes
(cons [2]) ; Put in the non-odd one
(map (comp first rseq)))] ; O(1) lookup of the last element of the vec "known-primes"
; Ugh, ugly de-duplication :(
(->> (map #(when (not= % %2) %) primes-with-duplicates (rest primes-with-duplicates))
(remove nil?))))
Reported numbers (time in milliseconds to count first N primes) are the fastest from the run of 5, no JVM restarts between experiments so your mileage may vary:
1e6 3e6
(primes-fn-1 5) 808 2664
(primes-fn-1 10) 952 3198
(primes-fn-1 20) 1440 4742
(primes-fn-1 30) 1881 6030
(primes-fn-2) 1868 5922
(primes-fn-3) 489 1755 <-- WOW!
(primes-fn-4) 2024 8185
If you don't need a lazy solution and you just want a sequence of primes below a certain limit, the straight forward implementation of the Sieve of Eratosthenes is pretty fast. Here's my version using transients:
(defn classic-sieve
"Returns sequence of primes less than N"
[n]
(loop [nums (transient (vec (range n))) i 2]
(cond
(> (* i i) n) (remove nil? (nnext (persistent! nums)))
(nums i) (recur (loop [nums nums j (* i i)]
(if (< j n)
(recur (assoc! nums j nil) (+ j i))
nums))
(inc i))
:else (recur nums (inc i)))))
I just started using Clojure so I don't know if it's good but here is my solution:
(defn divides? [x i]
(zero? (mod x i)))
(defn factors [x]
(flatten (map #(list % (/ x %))
(filter #(divides? x %)
(range 1 (inc (Math/floor (Math/sqrt x))))))))
(defn prime? [x]
(empty? (filter #(and divides? (not= x %) (not= 1 %))
(factors x))))
(def primes
(filter prime? (range 2 java.lang.Integer/MAX_VALUE)))
(defn sum-of-primes-below [n]
(reduce + (take-while #(< % n) primes)))