This is sweet. Thanks a lot Linus and Alex. So there is no function to do
it, you just have to build it yourself. I was not sure of that.
I like Alex's solution. This is something I can read and understand. Also,
it would be nice to compare and split in one loop. This is what I came up
with, based on Alex's work:
(defn partition-by-seq
[sub-seq coll]
(letfn [(take-sub-seq [coll]
(loop [[fcoll & rcoll] (seq coll)
[fsub & rsub] (seq sub-seq)]
(if (= fcoll fsub)
(if (nil? rsub)
[sub-seq rcoll]
(if rcoll
(recur rcoll rsub)
[nil coll]))
[nil coll])))
(step [coll]
(loop [run [] coll (seq coll)]
(let [[match more] (take-sub-seq coll)]
(if match
(if (seq run)
(cons run (cons match (lazy-seq (step more))))
(cons match (lazy-seq (step more))))
(if (seq coll)
(recur (conj run (first coll)) (drop 1 coll))
(when (seq run)
[run])
)))))]
(if (seq sub-seq)
(step coll)
[coll])))
;; Case: empty sub-seq
(= (partition-by-seq [] [:a]) '([:a]))
;; Case: empty coll
(= (partition-by-seq [:b :c] []) nil)
;; Case: no match
(= (partition-by-seq [:b :c] [:a :d]) '([:a :d]))
;; Case: example input
(= (partition-by-seq [:b :c] [:a :b :c :d :a :b :c :d :a :b :c])
'((:a) (:b :c) (:d :a) (:b :c) (:d :a) (:b :c)))
;; Case: false match :b, followed by true match :b :c
(= (partition-by-seq [:b :c] [:a :b :b :c :e :d :a :b :c :d :a :b :c])
'((:a :b) (:b :c) (:e :d :a) (:b :c) (:d :a) (:b :c)))
;; Case: non-matched remainder
(= (partition-by-seq [:b :c] [:a :b :c :b]) '([:a] [:b :c] [:b]))
;; Case: handle nil gracefully
(= (partition-by-seq [:b nil nil] [:a :b nil nil :c :b]) '([:a] [:b nil
nil] [:c :b]))
On Sun, Jun 8, 2014 at 6:27 PM, Alex Walker <alex.wal...@answers.com> wrote:
> https://gist.github.com/alexpw/f20c7b3ac858003e07e2
>
> This version supports the missing case:
>
> ;; Case: false match :b, followed by true match :b :c
> (= (partition-by-seq [:b :c] [:a :b :b :c :e :d :a :b :c :d :a :b :c])
> '((:a :b) (:b :c) (:e :d :a) (:b :c) (:d :a) (:b :c)))
>
>
> On Sunday, June 8, 2014 8:52:09 AM UTC-5, Alex Walker wrote:
>>
>> Err, this was bugging me all day when I went afk. I wrote it too quickly
>> and am missing the case where the sub-seq starts right after seeing the
>> first val, [:b :b :c].
>>
>> Will fix it if I have time today, but may need to take a slightly
>> different approach. Fun problem, btw. :)
>>
>> On Saturday, June 7, 2014 2:31:04 PM UTC-5, Alex Walker wrote:
>>>
>>> Here's a solution based on your description that behaves like core
>>> partition fns, taking liberty to presume your example should've matched the
>>> output below.
>>>
>>> (defn partition-by-seq
>>> [sub-seq coll]
>>> (letfn [(step [coll]
>>> (when-let [coll (seq coll)]
>>> (let [[run more] (split-with (partial not= (first sub-seq))
>>> coll)
>>> [possible-match more] (split-at (count sub-seq) more)]
>>> (if (= possible-match sub-seq)
>>> (if (seq run)
>>> (cons run (cons possible-match (lazy-seq (step
>>> more))))
>>> (cons possible-match (lazy-seq (step
>>> more))))
>>> (cons (concat run possible-match) (lazy-seq (step
>>> more)))))))]
>>> (or (step coll) ())))
>>>
>>> => (partition-by-seq [:b :c] [])
>>>
>>> ()
>>>
>>> => (partition-by-seq [:b :c] [:a :d])
>>>
>>> ((:a :d))
>>>
>>> => (partition-by-seq [:b :c] [:a :b :c :d :a :b :c :d :a :b :c])
>>>
>>> ((:a) (:b :c) (:d :a) (:b :c) (:d :a) (:b :c))
>>>
>>>
>>>
>>> On Tuesday, June 3, 2014 4:05:16 AM UTC-5, Ulrich Küttler wrote:
>>>>
>>>> Hi,
>>>>
>>>> what is the preferred way to find sub-seqs in a seq? I am trying to
>>>> convert
>>>>
>>>> [:a :b :c :d :a :b :c :d :a :b :c]
>>>>
>>>> into
>>>>
>>>> ((:a) (:b :c) (:a :d) (:b :c) (:a))
>>>>
>>>> using the sub-seq (:b :c) instead of positions.
>>>>
>>>> partition, partition-by and the like all look at one element at a time.
>>>> What I need is a search based on seqs. Are there functions that support
>>>> such a search / split?
>>>>
>>>> Uli
>>>>
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