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.
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])
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)
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.
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
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])
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With