Reputation: 9206
I have a recursion using loop
recur
with some fairly complex logic. It turns out I should do identical things in two of the branches, both which should lead to a recursion. I don't believe I can abstract this away using a function because of the limitations of recur, so how can I do it and not have duplicate code. Must I use a macro?
The code is below, and repeated code is highlighted with REPEATED CODE HERE
(defn overlapping-sampler [n-samples]
(let [...]
(loop [samples [] cache (zipmap boxes []) n-samples n-samples]
(cond
(zero? n-samples)
samples
:else
(let [box (categorical boxes volumes)
cache-item (peek (cache box))]
(if (nil? cache-item)
; REPEATED CODE HERE
(let [sample (interval-sample (:internals box))
all-boxes (map #(apply (:formula %) sample) boxes)
pos-dominant (max-pred-index true? all-boxes)
pos-box (max-pred-index #(= box %) boxes)]
(if (= pos-dominant pos-box)
(recur (conj samples sample) cache (dec n-samples))
(recur samples
(update-in cache [(nth boxes pos-dominant)]
#(conj % {:from box :sample sample}))
n-samples)))
; Otherwise with prob p=ratio of overlapping region/box, take sample
(if (flip (/ (volume (overlap box (:from cache-item))) (volume box)))
(recur (conj samples (:sample cache-item)) ; I should take the sample
(update-in cache [box]
#(pop %))
(dec n-samples))
(let [sample (gen-until #(interval-sample (:internals box))
#(and (apply (:formula box) %)
(not (apply (:formula (:from cache-item)) %))))
;REPEATED CODE HERE
all-boxes (map #(apply (:formula %) sample) boxes)
pos-dominant (max-pred-index true? all-boxes)
pos-box (max-pred-index #(= box %) boxes)]
(if (= pos-dominant pos-box)
(recur (conj samples sample) cache (dec n-samples))
(recur samples
(update-in cache [(nth boxes pos-dominant)]
#(conj % {:from box :sample sample}))
n-samples))))))))))
Upvotes: 0
Views: 114
Reputation: 84379
Perhaps rewrite to
(let [cache-item ...]
(if (and (nil? cache-item) (flip ...))
(recur ...)
(let [sample (if (nil? cache-item)
(... calculate sample ... )
(... use cache-item ... ))
; REPEATED CODE HERE
...]
...)))
This tests (nil? cache-item)
twice in the "else" branch of the if
. You could introduce a cache-hit?
local to store the value of (nil? cache-item)
to avoid repeatedly typing in the nil?
call; not for performance reasons, however, as there should be virtually no difference.
Upvotes: 0
Reputation: 33657
Using a local function:
(defn overlapping-sampler [n-samples]
(let [fun (fn [sample samples]
(let [all-boxes (map #(apply (:formula %) sample) boxes)
pos-dominant (max-pred-index true? all-boxes)
pos-box (max-pred-index #(= box %) boxes)]
(if (= pos-dominant pos-box)
[(conj samples sample) cache (dec n-samples)]
[samples
(update-in cache [(nth boxes pos-dominant)]
#(conj % {:from box :sample sample}))
n-samples])))]
(loop [[samples cache n-samples] [[] (zipmap boxes []) n-samples]]
(cond
(zero? n-samples)
samples
:else
(let [box (categorical boxes volumes)
cache-item (peek (cache box))]
(if (nil? cache-item)
; REPEATED CODE HERE
(let [sample (interval-sample (:internals box))]
(recur (fun sample) samples))
; Otherwise with prob p=ratio of overlapping region/box, take sample
(if (flip (/ (volume (overlap box (:from cache-item))) (volume box)))
(recur [(conj samples (:sample cache-item)) ; I should take the sample
(update-in cache [box]
#(pop %))
(dec n-samples)])
(let [sample (gen-until #(interval-sample (:internals box))
#(and (apply (:formula box) %)
(not (apply (:formula (:from cache-item)) %))))]
(recur (fun sample) samples)))))))))
Upvotes: 1