我最近开始学习Clojure并决定练习Euler问题，以获取可用数据结构并练习递归和循环。
我尝试了Problem 50的各种方法，但不管我做了什么，找到1000000的解决方案都没有完成。在我查看其他人所做的事情之后，我猜想我所做的事情也不应该永远消失，所以我在Python中输入了等效的算法，看看问题在于我是否缺乏对某些Clojure事物或Java设置的理解。 Python在10秒内完成。对于100000以下的素数，Python版本在0.5秒内完成，Clojure在5秒内完成。
我发布了专门为匹配Python代码而创建的Clojure版本。你能帮我理解为什么性能会有这么大差异吗？我应该使用unchecked-add，输入提示，原语（但在哪里？）或什么？
所以这是Clojure：
(defn prime? [n]
(let [r (int (Math/sqrt n))]
(loop [d 2]
(cond
(= n 1) false
(> d r) true
(zero? (rem n d)) false
:other (recur (inc d))))))
(defn primes []
(filter prime? (iterate inc 2)))
(defn cumulative-sum [s]
(reduce
(fn [v, x] (conj v (+ (last v) x)))
[(first s)]
(rest s)))
(defn longest-seq-under [n]
"Longest prime seq with sum under n"
(let [ps (vec (take-while #(< % n) (primes))) ; prime numbers up to n
prime-set (set ps) ; set for testing of inclusion
cs (cumulative-sum ps)
cnt (count ps)
max-len (count (take-while #(< % n) cs)) ; cannot have longer sequences
sub-sum (fn [i j] ; sum of primes between the i-th and j-th
(- (cs j) (get cs (dec i) 0)))
seq-with-len (fn [m] ; try m length prime sequences and return the first where the sum is prime
(loop [i 0] ; try with the lowest sum
(if (> i (- cnt m)) ; there are no more elements for and m length sequence
nil ; could not find any
(let [j (+ i (dec m)) ; fix length
s (sub-sum i j)]
(if (>= s n) ; overshoot
nil
(if (prime-set s) ; sum is prime
[i (inc j)] ; we just looked for the first
(recur (inc i))))))))] ; shift window
(loop [m max-len] ; try with the longest sequence
(if (not (zero? m))
(let [[i j] (seq-with-len m) ]
(if j
(subvec ps i j)
(recur (dec m))))))))
(assert (= [2 3 5 7 11 13] (longest-seq-under 100)))
(let [s1000 (longest-seq-under 1000)]
(assert (= 21 (count s1000)))
(assert (= 953 (reduce + s1000))))
; (time (reduce + (longest-seq-under 100000))) ; "Elapsed time: 5707.784369 msecs"
在Python中也是如此：
from math import sqrt
from itertools import takewhile
def is_prime(n) :
for i in xrange(2, int(sqrt(n))+1) :
if n % i == 0 :
return False
return True
def next_prime(n):
while not is_prime(n) :
n += 1
return n
def primes() :
i = 1
while True :
i = next_prime(i+1)
yield i
def cumulative_sum(s):
cs = []
css = 0
for si in s :
css += si
cs.append( css )
return cs
def longest_seq_under(n) :
ps = list(takewhile( lambda p : p < n, primes()))
pss = set(ps)
cs = cumulative_sum(ps)
cnt = len(ps)
max_len = len(list(takewhile(lambda s : s < n, cs)))
def subsum(i, j):
return cs[j] - (cs[i-1] if i > 0 else 0)
def interval_with_length(m) :
for i in xrange(0, cnt-m+1) :
j = i + m - 1
sij = subsum(i,j)
if sij >= n :
return None, None
if sij in pss : # prime
return i, j+1
return None, None
for m in xrange(max_len, 0, -1) :
f, t = interval_with_length(m)
if t :
return ps[f:t]
assert longest_seq_under(100) == [2, 3, 5, 7, 11, 13]
assert sum(longest_seq_under(1000)) == 953
# import timeit
# timeit.Timer("sum(longest_seq_under(100000))", "from __main__ import longest_seq_under").timeit(1) # 0.51235757617223499
由于
答案 0 :(得分：15)
我认为减速来自你在longest-seq-under中迭代序列的次数;每次迭代都会造成损失。这是一个吸烟快速版本，基于您的代码和发布的答案here的组合。请注意，primes是惰性的，因此我们可以将其与def vs defn绑定：
(defn prime? [n]
(let [r (int (Math/sqrt n))]
(loop [d 2]
(cond (= n 1) false
(> d r) true
(zero? (rem n d)) false
:else (recur (inc d))))))
(def primes (filter prime? (iterate inc 2)))
(defn make-seq-accumulator
[[x & xs]]
(map first (iterate
(fn [[sum [s & more]]]
[(+ sum s) more])
[x xs])))
(def prime-sums
(conj (make-seq-accumulator primes) 0))
(defn euler-50 [goal]
(loop [c 1]
(let [bots (reverse (take c prime-sums))
tops (take c (reverse (take-while #(> goal (- % (last bots)))
(rest prime-sums))))]
(or (some #(when (prime? %) %)
(map - tops bots))
(recur (inc c))))))
在我的机器上大约6毫秒完成：
user> (time (euler-50 1000000))
"Elapsed time: 6.29 msecs"
997651
答案 1 :(得分：4)
我将接受我自己的评论作为Python工作原因和Clojure没有的问题的答案：使用向量的last是一个线性操作，阻止累积和按照我的预期方式计算
更新函数以使用这样的瞬态向量：
(defn cumulative-sum-2 [s]
(loop [[x & xs] s
ss 0
acc (transient [])]
(if x
(let [ssx (+ ss x)]
(recur xs ssx (conj! acc ssx)))
(persistent! acc))))
导致Clojure版本的运行速度只是Python的两倍。我希望Clojure可以比Python更快地进行同样的操作，想知道我是否还会遗漏一些东西。我顺便使用1.2。
由于