Question

我有一份包含病媒清单的清单,其中涉及:

([1 2])([3 4] [5 6])([7 8])



我知道的是,与我合作并不理想。 我愿把这段话打上<代码>[1 2][3 4] [5 6] [7 8]>。

flatten doesn t work: it gives me (1 2 3 4 5 6 7 8).

我如何这样做? 图一

Answer 1

如果你只想平整一个等级,你可以使用<条码>。

(apply concat  (([1 2]) ([3 4] [5 6]) ([7 8])))
=> ([1 2] [3 4] [5 6] [7 8])

Answer 2

To turn a list-of-lists into a single list containing the elements of every sub-list, you want apply concat as nickik suggests.

然而,通常的解决办法是:不要提出从名单上开始的名单! 例如,请想一下,你有一份名为的植被名称-for的功能,该功能具有象征意义,并列出了你可以称之为这一标志的所有冷却物品:

(get-names-for  +) => (plus add cross junction)

如果你想获得某些标志清单的所有名字,你可能会尝试尝试。

(map get-names-for  [+ /]) 
=> ((plus add cross junction) (slash divide stroke))

但这导致了你面临的问题。您可以将其与<条码>应用于目录/代码”一起浏览,但最好使用<条码>,而不是<条码><>>>>。

(mapcat get-names-for  [+ /]) 
=> (plus add cross junction slash divide stroke)

Answer 3


The code for flatten is fairly short:

(defn flatten
  [x]
  (filter (complement sequential?)
    (rest (tree-seq sequential? seq x))))

它使用<条码>树-seq,通过数据结构行走,并回归原子序列。由于我们想到底线上的所有顺序,我们可以修改如下:

(defn almost-flatten
  [x]
  (filter #(and (sequential? %) (not-any? sequential? %))
    (rest (tree-seq #(and (sequential? %) (some sequential? %)) seq x))))

因此,我们回去了没有顺序的所有顺序。

Answer 4

也可发现,在:

(defn flatten-1 
  "Flattens only the first level of a given sequence, e.g. [[1 2][3]] becomes
   [1 2 3], but [[1 [2]] [3]] becomes [1 [2] 3]."
  [seq]
  (if (or (not (seqable? seq)) (nil? seq))
    seq ; if seq is nil or not a sequence, don t do anything
    (loop [acc [] [elt & others] seq]
      (if (nil? elt) acc
        (recur
          (if (seqable? elt)
            (apply conj acc elt) ; if elt is a sequence, add each element of elt
            (conj acc elt))      ; if elt is not a sequence, add elt itself 
       others)))))

例:

(flatten-1 (([1 2]) ([3 4] [5 6]) ([7 8])))
=>[[1 2] [3 4] [5 6] [7 8]]

考试当然确实是你的工作,但code>> 也是allowing non seq elements。在收集过程中:

(flatten-1  (1 2 ([3 4] [5 6]) ([7 8])))
=>[1 2 [3 4] [5 6] [7 8]]
;whereas 
(apply concat  (1 2 ([3 4] [5 6]) ([7 8])))
=> java.lang.IllegalArgumentException: 
   Don t know how to create ISeq from: java.lang.Integer

Answer 5

Here s a function that will flatten down to the sequence level, regardless of uneven nesting:

(fn flt [s] (mapcat #(if (every? coll? %) (flt %) (list %)) s))

因此,如果你最初的顺序是:

 (([1 2]) (([3 4]) ((([5 6])))) ([7 8]))

你也取得了同样的结果:

([1 2] [3 4] [5 6] [7 8])

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