I'm looking for a sequential data structure which is perfect for the following operation. The lenght of the list remains constant, it will never be longer or shorter than a fixed length.
Omit the first item and add x to the end.
(0 1 2 3 4 5 6 7 8 9)
(pop-and-push "10")
(1 2 3 4 5 6 7 8 9 10)
There is only one other reading-operation that has to be done equally often:
(last coll)
pop-and-push could be implemented like this:
(defn pop-and-push [coll x]
(concat (pop coll) ["x"]))
(unfortunately this does not work with sequences produced by e.g. range, it just pops when the sequence declared by the literals '(..) is passed.)
but is this optimal?
The main issue here (once we change "x" to x) is that concat returns a lazy-seq, and lazy-seqs are invalid args to pop.
user=> (defn pop-and-push [coll x] (concat (pop coll) [x]))
#'user/pop-and-push
user=> (pop-and-push [1 2 3] 4)
(1 2 4)
user=> (pop-and-push *1 5)
ClassCastException clojure.lang.LazySeq cannot be cast to clojure.lang.IPersistentStack clojure.lang.RT.pop (RT.java:730)
My naive preference would be to use a vector. This function is easy to implement with subvec.
user=> (defn pop-and-push [v x] (conj (subvec (vec v) 1) x))
#'user/pop-and-push
user=> (pop-and-push [1 2 3] 4)
[2 3 4]
user=> (pop-and-push *1 5)
[3 4 5]
as you can see, this version can actually operate on its own return value
As suggested in the comments, PersistentQueue is made for this situation:
user=> (defn pop-and-push [v x] (conj (pop v) x))
#'user/pop-and-push
user=> (pop-and-push (into clojure.lang.PersistentQueue/EMPTY [1 2 3]) 4)
#object[clojure.lang.PersistentQueue 0x50313382 "clojure.lang.PersistentQueue#7c42"]
user=> (into [] *1)
[2 3 4]
user=> (pop-and-push *2 5)
#object[clojure.lang.PersistentQueue 0x4bd31064 "clojure.lang.PersistentQueue#8023"]
user=> (into [] *1)
[3 4 5]
The PersistentQueue data structure, though less convenient to use in some ways, is actually optimized for this usage.
Related
I would like to partition a seq, based on a seq of values
(partition-by-seq [3 5] [1 2 3 4 5 6])
((1 2 3)(4 5)(6))
The first input is a seq of split points.
The second input is a seq i would like to partition.
So, that the first list will be partitioned at the value 3 (1 2 3) and the second partition will be (4 5) where 5 is the next split point.
another example:
(partition-by-seq [3] [2 3 4 5])
result: ((2 3)(4 5))
(partition-by-seq [2 5] [2 3 5 6])
result: ((2)(3 5)(6))
given: the first seq (split points) is always a subset of the second input seq.
I came up with this solution which is lazy and quite (IMO) straightforward.
(defn part-seq [splitters coll]
(lazy-seq
(when-let [s (seq coll)]
(if-let [split-point (first splitters)]
; build seq until first splitter
(let [run (cons (first s) (take-while #(<= % split-point) (next s)))]
; build the lazy seq of partitions recursively
(cons run
(part-seq (rest splitters) (drop (count run) s))))
; just return one partition if there is no splitter
(list coll)))))
If the split points are all in the sequence:
(part-seq [3 5 8] [0 1 2 3 4 5 6 7 8 9])
;;=> ((0 1 2 3) (4 5) (6 7 8) (9))
If some split points are not in the sequence
(part-seq [3 5 8] [0 1 2 4 5 6 8 9])
;;=> ((0 1 2) (4 5) (6 8) (9))
Example with some infinite sequences for the splitters and the sequence to split.
(take 5 (part-seq (iterate (partial + 3) 5) (range)))
;;=> ((0 1 2 3 4 5) (6 7 8) (9 10 11) (12 13 14) (15 16 17))
the sequence to be partitioned is a splittee and the elements of split-points (aka. splitter) marks the last element of a partition.
from your example:
splittee: [1 2 3 4 5 6]
splitter: [3 5]
result: ((1 2 3)(4 5)(6))
Because the resulting partitions is always a increasing integer sequence and increasing integer sequence of x can be defined as start <= x < end, the splitter elements can be transformed into end of a sequence according to the definition.
so, from [3 5], we want to find subsequences ended with 4 and 6.
then by adding the start, the splitter can be transformed into sequences of [start end]. The start and end of the splittee is also used.
so, the splitter [3 5] then becomes:
[[1 4] [4 6] [6 7]]
splitter transformation could be done like this
(->> (concat [(first splittee)]
(mapcat (juxt inc inc) splitter)
[(inc (last splittee))])
(partition 2)
there is a nice symmetry between transformed splitter and the desired result.
[[1 4] [4 6] [6 7]]
((1 2 3) (4 5) (6))
then the problem becomes how to extract subsequences inside splittee that is ranged by [start end] inside transformed splitter
clojure has subseq function that can be used to find a subsequence inside ordered sequence by start and end criteria. I can just map the subseq of splittee for each elements of transformed-splitter
(map (fn [[x y]]
(subseq (apply sorted-set splittee) <= x < y))
transformed-splitter)
by combining the steps above, my answer is:
(defn partition-by-seq
[splitter splittee]
(->> (concat [(first splittee)]
(mapcat (juxt inc inc) splitter)
[(inc (last splittee))])
(partition 2)
(map (fn [[x y]]
(subseq (apply sorted-set splittee) <= x < y)))))
This is the solution i came up with.
(def a [1 2 3 4 5 6])
(def p [2 4 5])
(defn partition-by-seq [s input]
(loop [i 0
t input
v (transient [])]
(if (< i (count s))
(let [x (split-with #(<= % (nth s i)) t)]
(recur (inc i) (first (rest x)) (conj! v (first x))))
(do
(conj! v t)
(filter #(not= (count %) 0) (persistent! v))))))
(partition-by-seq p a)
If I use the reductions function like so:
(reductions + [1 2 3 4 5])
Then I get
(1 3 6 10 15)
Which is great - but I'd like to apply a binary function in the same way without the state being carried forward - something like
(magic-hof + [1 2 3 4 5])
leads to
(1 3 5 7 9)
ie it returns the operation applied to the first pair, then steps 1 to the next pair.
Can someone tell me the higher-order function I'm looking for? (Something like reductions)
This is my (non-working) go at it:
(defn thisfunc [a b] [(+ a b) b])
(reduce thisfunc [1 2 3 4 5])
You can do it with map:
(map f coll (rest coll))
And if you want a function:
(defn map-pairwise [f coll]
(map f coll (rest coll)))
And if you really need the first element to remain untouched (thanx to juan.facorro's comment):
(defn magic-hof [f [x & xs :as s]]
(cons x (map f s xs)))
partition will group your seq:
user> (->> [1 2 3 4 5] (partition 2 1) (map #(apply + %)) (cons 1))
(1 3 5 7 9)
So, you want to apply a function to subsequent pairs of elements?
(defn pairwise-apply
[f sq]
(when (seq sq)
(->> (map f sq (next sq))
(cons (first sq)))))
Let's try it:
(pairwise-apply + (range 1 6))
;; => (1 3 5 7 9)
This is sufficient:
(#(map + (cons 0 %) %) [1 2 3 4 5])
;; => (1 3 5 7 9)
I'm trying to write a function adjacents that returns a vector of a sequence's adjacent pairs. So (adjacents [1 2 3]) would return [[1 2] [2 3]].
(defn adjacents [s]
(loop [[a b :as remaining] s
acc []]
(if (empty? b)
acc
(recur (rest remaining) (conj acc (vector a b))))))
My current implementation works for sequences of strings but with integers or characters the REPL outputs this error:
IllegalArgumentException Don't know how to create ISeq from: java.lang.Long clojure.lang.RT.seqFrom (RT.java:494)
The problem here is in the first evaluation loop of (adjacents [1 2 3]), a is bound to 1 and b to 2. Then you ask if b is empty?. But empty? works on sequences and b is not a sequence, it is a Long, namely 2. The predicate you could use for this case here is nil?:
user=> (defn adjacents [s]
#_=> (loop [[a b :as remaining] s acc []]
#_=> (if (nil? b)
#_=> acc
#_=> (recur (rest remaining) (conj acc (vector a b))))))
#'user/adjacents
user=> (adjacents [1 2 3 4 5])
[[1 2] [2 3] [3 4] [4 5]]
But, as #amalloy points out, this may fail to give the desired result if you have legitimate nils in your data:
user=> (adjacents [1 2 nil 4 5])
[[1 2]]
See his comment for suggested implementation using lists.
Note that Clojure's partition can be used to do this work without the perils of defining your own:
user=> (partition 2 1 [1 2 3 4 5])
((1 2) (2 3) (3 4) (4 5))
user=> (partition 2 1 [1 2 nil 4 5])
((1 2) (2 nil) (nil 4) (4 5))
Here is my short answer. Everything becomes a vector, but it works for all sequences.
(defn adjacent-pairs [s]
{:pre [(sequential? s)]}
(map vector (butlast s) (rest s)))
Testing:
user=> (defn adjacent-pairs [s] (map vector (butlast s) (rest s)))
#'user/adjacent-pairs
user=> (adjacent-pairs '(1 2 3 4 5 6))
([1 2] [2 3] [3 4] [4 5] [5 6])
user=> (adjacent-pairs [1 2 3 4 5 6])
([1 2] [2 3] [3 4] [4 5] [5 6])
user=>
This answer is probably less efficient than the one using partition above, however.
I need to modify map function behavior to provide mapping not with minimum collection size but with maximum and use zero for missing elements.
Standard behavior:
(map + [1 2 3] [4 5 6 7 8]) => [5 7 9]
Needed behavior:
(map + [1 2 3] [4 5 6 7 8]) => [5 7 9 7 8]
I wrote function to do this, but it seems not very extensible with varargs.
(defn map-ext [f coll1 coll2]
(let [mx (max (count coll1) (count coll2))]
(map f
(concat coll1 (repeat (- mx (count coll1)) 0))
(concat coll2 (repeat (- mx (count coll2)) 0)))))
Is there a better way to do this?
Your method is concise, but inefficient (it calls count). A more efficient solution, which does not require the entirety of its input sequences to be stored in memory follows:
(defn map-pad [f pad & colls]
(lazy-seq
(let [seqs (map seq colls)]
(when (some identity seqs)
(cons (apply f (map #(or (first %) pad) seqs))
(apply map-pad f pad (map rest seqs)))))))
Used like this:
user=> (map-pad + 0 [] [1] [1 1] (range 1 10))
(3 3 3 4 5 6 7 8 9)
Edit: Generalized map-pad to arbitrary arity.
Another lazy variant, usable with an arbitrary number of input sequences:
(defn map-ext [f ext & seqs]
(lazy-seq
(if (some seq seqs)
(cons (apply f (map #(if (seq %) (first %) ext) seqs))
(apply map-ext f ext (map rest seqs)))
())))
Usage:
user> (map-ext + 0 [1 2 3] [4 5 6 7 8])
(5 7 9 7 8)
user> (map-ext + 0 [1 2 3] [4 5 6 7 8] [3 4])
(8 11 9 7 8)
If you just want it to work for any number of collections, try:
(defn map-ext [f & colls]
(let [mx (apply max (map count colls))]
(apply map f (map #(concat % (repeat (- mx (count %)) 0)) colls))))
Clojure> (map-ext + [1 2] [1 2 3] [1 2 3 4])
(3 6 6 4)
I suspect there may be better solutions though (as Trevor Caira suggests, this solution isn't lazy due to the calls to count).
How about that:
(defn map-ext [f x & xs]
(let [colls (cons x xs)
res (apply map f colls)
next (filter not-empty (map #(drop (count res) %) colls))]
(if (empty? next) res
(lazy-seq (concat res (apply map-ext f next))))))
user> (map-ext + [1 2 3] [4] [5 6] [7 8 9 10])
(17 16 12 10)
Along the lines of #LeNsTR's solution, but simpler and faster:
(defn map-ext [f & colls]
(lazy-seq
(let [colls (filter seq colls)
firsts (map first colls)
rests (map rest colls)]
(when (seq colls)
(cons (apply f firsts) (apply map-ext f rests))))))
(map-ext + [1 2 3] [4] [5 6] [7 8 9 10])
;(17 16 12 10)
I've just noticed Michał Marczyk's accepted solution, which is superior: it deals properly with asymmetric mapping functions such as -.
We can make Michał Marczyk's answer neater by using the convention - which many core functions follow - that you get a default or identity value by calling the function with no arguments. For examples:
(+) ;=> 0
(concat) ;=> ()
The code becomes
(defn map-ext [f & seqs]
(lazy-seq
(when (some seq seqs)
(cons (apply f (map #(if (seq %) (first %) (f)) seqs))
(apply map-ext f (map rest seqs)))
)))
(map-ext + [1 2 3] [4 5 6 7 8] [3 4])
;(8 11 9 7 8)
I've made the minimum changes. It could be speeded up a bit.
We may need a function that will inject such a default value into a function that lacks it:
(defn with-default [f default]
(fn
([] default)
([& args] (apply f args))))
((with-default + 6)) ;=> 6
((with-default + 6) 7 8) ;=> 15
This could be speeded up or even turned into a macro.
I have a list [2 3 5] which I want to use to remove items from another list like [1 2 3 4 5], so that I get [1 4].
thanks
Try this:
(let [a [1 2 3 4 5]
b [2 3 5]]
(remove (set b) a))
which returns (1 4).
The remove function, by the way, takes a predicate and a collection, and returns a sequence of the elements that don't satisfy the predicate (a set, in this example).
user=> (use 'clojure.set)
nil
user=> (difference (set [1 2 3 4 5]) (set [2 3 5]))
#{1 4}
Reference:
http://clojure.org/data_structures#toc22
http://clojure.org/api#difference
You can do this yourself with something like:
(def a [2 3 5])
(def b [1 2 3 4 5])
(defn seq-contains?
[coll target] (some #(= target %) coll))
(filter #(not (seq-contains? a %)) b)
; (3 4 5)
A version based on the reducers library could be:
(require '[clojure.core.reducers :as r])
(defn seq-contains?
[coll target]
(some #(= target %) coll))
(defn my-remove
"remove values from seq b that are present in seq a"
[a b]
(into [] (r/filter #(not (seq-contains? b %)) a)))
(my-remove [1 2 3 4 5] [2 3 5] )
; [1 4]
EDIT Added seq-contains? code
Here is my take without using sets;
(defn my-diff-func [X Y]
(reduce #(remove (fn [x] (= x %2)) %1) X Y ))