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)
Related
I have to display the index of the odd items in a given list of numbers.
I tried getting the remainder but I have to divide the given list by [2 3 5 10] in order to know which element is odd.
(defn odd_one_out [y]
(println (map #(rem % 2) y)))
(odd_one_out [2 8 9 200 56])
I expect the output 9 or index of 9 since it is the only element which cannot be divided by 2.
The output i am getting is 0 0 1 0 0
If I understand correctly, you want to find the number which is uniquely indivisible for given divisors. You could use group-by to group the numbers by their divisibility, then find the one(s) that are indivisible by exactly one divisor.
(defn odd-one-out [nums divs]
(->> nums
(group-by #(map (fn [d] (zero? (mod % d))) divs))
(some (fn [[div-flags nums']]
(and (= 1 (count nums'))
(= 1 (count (filter true? div-flags)))
(first nums'))))))
(odd-one-out [2 8 9 200 56] [2 3 5 10]) ;; => 9
(odd-one-out [2 10 20 60 90] [2 3 5 10]) ;; => 2
If you just want to extend your current function, you could use map-indexed,which will give you this list ([0 0] [1 0] [2 1] [3 0] [4 0]), which you can then filter to keep only the vectors that have 1 in the second position. This will return the index of the odd character.
(defn odd-one-out [y]
(->> y
(map #(rem % 2))
(map-indexed vector)
(filter #(= 1 (second %)))
(map first)))
(odd-one-out [2 8 9 200 56])
(2)
Even better would be to use the function odd? from Clojure's standard library.
(->> [2 8 9 200 56]
(map odd?)
(map-indexed vector)
(filter #(second %))
(map first))
Another version using keep.
(to return the index)
(->> [2 8 9 200 56]
(map-indexed vector)
(keep #(when (odd? (second %))
(first %))))
(2)
(to return the value)
(->> [2 8 9 200 56]
(map-indexed vector)
(keep #(when (odd? (second %))
(second %))))
(9)
I'd like to have a function, such that,
(f '([1 4 7] [2 5 9] [3 6]))
would give
([1 2 3] [4 5 6] [7 9])
I tried
(apply map vector '([1 4 7] [2 5 9] [3 6]))
would only produce:
([1 2 3] [4 5 6])
I find it hard to describe my requirements that it's difficult for me to search for a ready solution.
Please help me either to improve my description, or pointer to a solution.
Thanks in advance!
I'd solve a more general problem which means you might reuse that function in the future. I'd change map so that it keeps going past the smallest map.
(defn map-all
"Like map but if given multiple collections will call the function f
with as many arguments as there are elements still left."
([f] (map f))
([f coll] (map f coll))
([f c1 & colls]
(let [step (fn step [cs]
(lazy-seq
(let [ss (keep seq cs)]
(when (seq ss)
(cons (map first ss)
(step (map rest ss)))))))]
(map #(apply f %) (step (conj colls c1))))))
(apply map-all vector '([1 4 7] [2 5 9] [3 6]))
(apply map-all vector '([1 false 7] [nil 5 9] [3 6] [8]))
Note, that as opposed to many other solutions, this one works fine even if any of the sequences contain nil or false.
or this way with loop/recur:
user> (defn transpose-all-2 [colls]
(loop [colls colls res []]
(if-let [colls (seq (filter seq colls))]
(recur (doall (map next colls))
(conj res (mapv first colls)))
res)))
#'user/transpose-all-2
user> (transpose-all-2 x)
[[1 2 3] [4 5 6] [7 9]]
user> (transpose-all-2 '((0 1 2 3) (4 5 6 7) (8 9)))
[[0 4 8] [1 5 9] [2 6] [3 7]]
If you know the maximum length of the vectors ahead of time, you could define
(defn tx [colls]
(lazy-seq
(cons (filterv identity (map first colls))
(tx (map rest colls)))))
then
(take 3 (tx '([1 4 7] [2 5 9] [3 6])))
A simple solution is
(defn transpose-all
[colls]
(lazy-seq
(let [ss (keep seq colls)]
(when (seq ss)
(cons (map first ss) (transpose-all (map rest ss)))))))
For example,
(transpose-all '([1 4 7] [2 5 9] [3 6] [11 12 13 14]))
;((1 2 3 11) (4 5 6 12) (7 9 13) (14))
Here is my own attempt:
(defn f [l]
(let [max-count (apply max (map count l))
l-patched (map (fn [e] (if (< (count e) max-count)
(concat e (take (- max-count (count e)) (repeat nil)))
e)) l)]
(map (fn [x] (filter identity x)) (apply map vector l-patched))
))
Another simple solution:
(->> jagged-list
(map #(concat % (repeat nil)))
(apply map vector)
(take-while (partial some identity)))
A jagged-list like this
'([1 4 7 ]
[2 5 9 ]
[3 6 ]
[11 12 13 14])
will produce:
'([1 2 3 11]
[4 5 6 12]
[7 9 nil 13]
[nil nil nil 14])
Here is another go that doesn't require you to know the vector length in advance:
(defn padzip [& [colls]]
(loop [acc [] colls colls]
(if (every? empty? colls) acc
(recur (conj acc (filterv some?
(map first colls))) (map rest colls)))))
What is the best way to remove n instances of matched elements of collection-2 from collection-1?
(let [coll-1 [8 2]
coll-2 [8 8 8 2]
Here's what I first came up with to solve original problem:
...
;; (remove (set coll-1) coll-2))
;; --> ()
But realised I must achieve:
...
;; (some-magic coll-1 coll-2))
;; --> (8 8)
Clarification:
(some-magic {8 2} [8 8 8 2]) ;;Removes 1x8 and 1x2 from vector.
(some-magic {8 8 2} [8 8 8 2]) ;;Removes 2x8 and 1x2 from vector.
Edit:
Preserving the order is desired.
Here is a lazy solution, written in the style of distinct:
(defn some-magic [count-map coll]
(let [step (fn step [xs count-map]
(lazy-seq
((fn [[f :as xs] count-map]
(when-let [s (seq xs)]
(if (pos? (get count-map f 0))
(recur (rest s) (update-in count-map [f] dec))
(cons f (step (rest s) count-map)))))
xs count-map)))]
(step coll count-map)))
The first argument needs to be a map indicating how many of each value to remove:
(some-magic {8 1, 2 1} [8 8 8 2]) ;; Removes 1x8 and 1x2
;=> (8 8)
(some-magic {8 2, 2 1} [8 8 8 2]) ;; Removes 2x8 and 1x2
;=> (8)
Here is an example dealing with falsey values and infinite input:
(take 10 (some-magic {3 4, 2 2, nil 1} (concat [3 nil 3 false nil 3 2] (range))))
;=> (false nil 0 1 4 5 6 7 8 9)
I don't see any of the built in sequence manipulation functions quite solving this, though a straitforward loop can build the result nicely:
user> (loop [coll-1 (set coll-1) coll-2 coll-2 result []]
(if-let [[f & r] coll-2]
(if (coll-1 f)
(recur (disj coll-1 f) r result)
(recur coll-1 r (conj result f)))
result))
[8 8]
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 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.