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Clojure: Semi-Flattening a nested Sequence

I have a list with embedded lists of vectors, which looks like:

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

Which I know is not ideal to work with. I'd like to flatten this to ([1 2] [3 4] [5 6] [7 8]).

flatten doesn't work: it gives me (1 2 3 4 5 6 7 8).

How do I do this? I figure I need to create a new list based on the contents of each list item, not the items, and it's this part I can't find out how to do from the docs.

like image 601
ChucK Avatar asked Mar 08 '11 12:03

ChucK


5 Answers

If you only want to flatten it one level you can use concat

(apply concat '(([1 2]) ([3 4] [5 6]) ([7 8])))
=> ([1 2] [3 4] [5 6] [7 8])
like image 76
nickik Avatar answered Nov 13 '22 04:11

nickik


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

However, there's usually a better solution: don't produce the list-of-lists to begin with! For example, let's imagine you have a function called get-names-for which takes a symbol and returns a list of all the cool things you could call that symbol:

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

If you want to get all the names for some list of symbols, you might try

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

But this leads to the problem you were having. You could glue them together with an apply concat, but better would be to use mapcat instead of map to begin with:

(mapcat get-names-for '[+ /]) 
=> (plus add cross junction slash divide stroke)
like image 25
amalloy Avatar answered Nov 13 '22 03:11

amalloy


The code for flatten is fairly short:

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

It uses tree-seq to walk through the data structure and return a sequence of the atoms. Since we want all the bottom-level sequences, we could modify it like this:

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

so we return all the sequences that don't contain sequences.

like image 9
David Minor Avatar answered Nov 13 '22 03:11

David Minor


Also you may found useful this general 1 level flatten function I found on clojuremvc:

(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)))))

Example:

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

concat exampe surely do job for you, but this flatten-1 is also allowing non seq elements inside a collection:

(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
like image 4
Nevena Avatar answered Nov 13 '22 04:11

Nevena


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))

So if your original sequence was:

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

You'd still get the same result:

([1 2] [3 4] [5 6] [7 8])
like image 3
bazeblackwood Avatar answered Nov 13 '22 05:11

bazeblackwood