I need to build a seq of seqs (vec of vecs) by combining first, second, etc elements of the given seqs.
After a quick searching and looking at the cheat sheet. I haven't found one and finished with writing my own:
(defn zip
"From the sequence of sequences return a another sequence of sequenses
where first result sequense consist of first elements of input sequences
second element consist of second elements of input sequenses etc.
Example:
[[:a 0 \\a] [:b 1 \\b] [:c 2 \\c]] => ([:a :b :c] [0 1 2] [\\a \\b \\c])"
[coll]
(let [num-elems (count (first coll))
inits (for [_ (range num-elems)] [])]
(reduce (fn [cols elems] (map-indexed
(fn [idx coll] (conj coll (elems idx))) cols))
inits coll)))
I'm interested if there is a standard method for this?
(apply map vector [[:a 0 \a] [:b 1 \b] [:c 2 \c]])
;; ([:a :b :c] [0 1 2] [\a \b \c])
You can use the variable arity of map to accomplish this.
From the map docstring:
... Returns a lazy sequence consisting of the result of applying f to
the set of first items of each coll, followed by applying f to the set
of second items in each coll, until any one of the colls is exhausted.
Any remaining items in other colls are ignored....
Kyle's solution is a great one and I see no reason why not to use it, but if you want to write such a function from scratch you could write something like the following:
(defn zip
([ret s]
(let [a (map first s)]
(if (every? nil? a)
ret
(recur (conj ret a) (map rest s)))))
([s]
(reverse (zip nil s))))
Related
(defn image-of
"computes the image of the element x under R"
[R x]
(set
(for [r R]
(when (= (first r) x)
(second r)))))
Function idea: Add the second variable in R when it's first is equal to x.
So this function is supposed to compute image of a relation. This is kinda successful. When running a test I get this result:
Input: (image-of #{[1 :a] [2 :b] [1 :c] [3 :a]} 1)
Expected: #{:c :a}
Actual: #{nil :c :a}
So it includes a nil value for some reason. What in the function causes this? I guess I could filter out any nil values but would like to have the solution on a single line.
So the problem was I didn't know exactly how to use when
This solution does it:
(set (for [r R
:when (= (first r) x)]
(second r)))
Let me suggest a different approach.
The natural way to represent a relation in Clojure is as a map from keys to sets (or other collections) of values. A function to convert your collection of pairs to this form is ...
(defn pairs->map [pairs]
(reduce
(fn [acc [key value]]
(assoc acc key (conj (acc key #{}) value)))
{}
pairs))
For example, ...
(pairs->map #{[1 :a] [2 :b] [1 :c] [3 :a]})
=> {2 #{:b}, 1 #{:c :a}, 3 #{:a}}
You can use this map as a function. I you feed it a key, it returns the corresponding value:
({2 #{:b}, 1 #{:c :a}, 3 #{:a}} 1)
=> #{:c :a}
You construct this map once and or all and use it as often as you like. Looking it up as a function is effectively a constant-time operation. But you run through the entire collection of pairs every time you evaluate image-of.
I'm trying to write a function with recur that cut the sequence as soon as it encounters a repetition ([1 2 3 1 4] should return [1 2 3]), this is my function:
(defn cut-at-repetition [a-seq]
(loop[[head & tail] a-seq, coll '()]
(if (empty? head)
coll
(if (contains? coll head)
coll
(recur (rest tail) (conj coll head))))))
The first problem is with the contains? that throws an exception, I tried replacing it with some but with no success. The second problem is in the recur part which will also throw an exception
You've made several mistakes:
You've used contains? on a sequence. It only works on associative
collections. Use some instead.
You've tested the first element of the sequence (head) for empty?.
Test the whole sequence.
Use a vector to accumulate the answer. conj adds elements to the
front of a list, reversing the answer.
Correcting these, we get
(defn cut-at-repetition [a-seq]
(loop [[head & tail :as all] a-seq, coll []]
(if (empty? all)
coll
(if (some #(= head %) coll)
coll
(recur tail (conj coll head))))))
(cut-at-repetition [1 2 3 1 4])
=> [1 2 3]
The above works, but it's slow, since it scans the whole sequence for every absent element. So better use a set.
Let's call the function take-distinct, since it is similar to take-while. If we follow that precedent and make it lazy, we can do it thus:
(defn take-distinct [coll]
(letfn [(td [seen unseen]
(lazy-seq
(when-let [[x & xs] (seq unseen)]
(when-not (contains? seen x)
(cons x (td (conj seen x) xs))))))]
(td #{} coll)))
We get the expected results for finite sequences:
(map (juxt identity take-distinct) [[] (range 5) [2 3 2]]
=> ([[] nil] [(0 1 2 3 4) (0 1 2 3 4)] [[2 3 2] (2 3)])
And we can take as much as we need from an endless result:
(take 10 (take-distinct (range)))
=> (0 1 2 3 4 5 6 7 8 9)
I would call your eager version take-distinctv, on the map -> mapv precedent. And I'd do it this way:
(defn take-distinctv [coll]
(loop [seen-vec [], seen-set #{}, unseen coll]
(if-let [[x & xs] (seq unseen)]
(if (contains? seen-set x)
seen-vec
(recur (conj seen-vec x) (conj seen-set x) xs))
seen-vec)))
Notice that we carry the seen elements twice:
as a vector, to return as the solution; and
as a set, to test for membership of.
Two of the three mistakes were commented on by #cfrick.
There is a tradeoff between saving a line or two and making the logic as simple & explicit as possible. To make it as obvious as possible, I would do it something like this:
(defn cut-at-repetition
[values]
(loop [remaining-values values
result []]
(if (empty? remaining-values)
result
(let [found-values (into #{} result)
new-value (first remaining-values)]
(if (contains? found-values new-value)
result
(recur
(rest remaining-values)
(conj result new-value)))))))
(cut-at-repetition [1 2 3 1 4]) => [1 2 3]
Also, be sure to bookmark The Clojure Cheatsheet and always keep a browser tab open to it.
I'd like to hear feedback on this utility function which I wrote for myself (uses filter with stateful pred instead of a loop):
(defn my-distinct
"Returns distinct values from a seq, as defined by id-getter."
[id-getter coll]
(let [seen-ids (volatile! #{})
seen? (fn [id] (if-not (contains? #seen-ids id)
(vswap! seen-ids conj id)))]
(filter (comp seen? id-getter) coll)))
(my-distinct identity "abracadabra")
; (\a \b \r \c \d)
(->> (for [i (range 50)] {:id (mod (* i i) 21) :value i})
(my-distinct :id)
pprint)
; ({:id 0, :value 0}
; {:id 1, :value 1}
; {:id 4, :value 2}
; {:id 9, :value 3}
; {:id 16, :value 4}
; {:id 15, :value 6}
; {:id 7, :value 7}
; {:id 18, :value 9})
Docs of filter says "pred must be free of side-effects" but I'm not sure if it is ok in this case. Is filter guaranteed to iterate over the sequence in order and not for example take skips forward?
When doing
(map f [0 1 2] [:0 :1])
f will get called twice, with the arguments being
0 :0
1 :1
Is there a simple yet efficient way, i.e. without producing more intermediate sequences etc., to make f get called for every value of the first collection, with the following arguments?
0 :0
1 :1
2 nil
Edit Addressing question by #fl00r in the comments.
The actual use case that triggered this question needed map to always work exactly (count first-coll) times, regardless if the second (or third, or ...) collection was longer.
It's a bit late in the game now and somewhat unfair after having accepted an answer, but if a good answer gets added that only does what I specifically asked for - mapping (count first-coll) times - I would accept that.
You could do:
(map f [0 1 2] (concat [:0 :1] (repeat nil)))
Basically, pad the second coll with an infinite sequence of nils. map stops when it reaches the end of the first collection.
An (eager) loop/recur form that walks to end of longest:
(loop [c1 [0 1 2] c2 [:0 :1] o []]
(if (or (seq c1) (seq c2))
(recur (rest c1) (rest c2) (conj o (f (first c1) (first c2))))
o))
Or you could write a lazy version of map that did something similar.
A general lazy version, as suggested by Alex Miller's answer, is
(defn map-all [f & colls]
(lazy-seq
(when-not (not-any? seq colls)
(cons
(apply f (map first colls))
(apply map-all f (map rest colls))))))
For example,
(map-all vector [0 1 2] [:0 :1])
;([0 :0] [1 :1] [2 nil])
You would probably want to specialise map-all for one and two collections.
just for fun
this could easily be done with common lisp's do macro. We could implement it in clojure and do this (and much more fun things) with it:
(defmacro cl-do [clauses [end-check result] & body]
(let [clauses (map #(if (coll? %) % (list %)) clauses)
bindings (mapcat (juxt first second) clauses)
nexts (map #(nth % 2 (first %)) clauses)]
`(loop [~#bindings]
(if ~end-check
~result
(do
~#body
(recur ~#nexts))))))
and then just use it for mapping (notice it can operate on more than 2 colls):
(defn map-all [f & colls]
(cl-do ((colls colls (map next colls))
(res [] (conj res (apply f (map first colls)))))
((every? empty? colls) res)))
in repl:
user> (map-all vector [1 2 3] [:a :s] '[z x c v])
;;=> [[1 :a z] [2 :s x] [3 nil c] [nil nil v]]
I have the following variable
(def a [[1 2] [3 4] [5 6]])
and want to return
[[1 3 5][2 4 6]]
and if input is
[[1 2] [3 4] [5 6] [7 8 9]] then the required result is
[[1 3 5 7] [2 4 6 8] [9]]
How to do it in clojure?
(persistent!
(reduce
(fn [acc e]
(reduce-kv
(fn [acc2 i2 e2]
(assoc! acc2 i2 ((fnil conj []) (get acc2 i2) e2)))
acc
e))
(transient [])
[[1 2 3] [:a :b] [\a] [111] [200 300 400 500]]))
;;=> [[1 :a \a 111 200] [2 :b 300] [3 400] [500]]
An empty vector can be updated via the update-in fn at the 0th index, a non-empty vector can be, additionally, updated at the index immediately following the last value.
The reduction here is about passing the outer accumulator to the inner reducing function, updating it accordingly, and then returning it back to the outer reducing function, which in turn will pass again to the inner rf for processing the next element.
EDIT: Updated to fastest version.
I like ifett's implementation, though it seems weird to use reduce-kv to build a vector that could be easily build with map/mapv.
So, here is how I would've done it:
(defn transpose [v]
(mapv (fn [ind]
(mapv #(get % ind)
(filter #(contains? % ind) v)))
(->> (map count v)
(apply max)
range)))
(->> (range)
(map (fn [i]
(->> a
(filter #(contains? % i))
(map #(nth % i)))))
(take-while seq))
Notice that this algorithm creates a lazy seq of lazy seqs so you that you will only pay for the transformations you really consume. If you insist on creating vectors instead, wrap the forms in vec at the necessary places - or if you are using Clojurescript or don't mind a Clojure 1.7 alpha use transducers to create vectors eagerly without paying for laziness or immutability:
(into []
(comp
(map (fn [i]
(into [] (comp (filter #(contains? % i))
(map #(nth % i)))
a)))
(take-while seq))
(range))
I find this easy to understand:
(defn nth-column [matrix n]
(for [row matrix] (nth row n)))
(defn transpose [matrix]
(for [column (range (count (first matrix)))]
(nth-column matrix column)))
(transpose a)
=> ((1 3 5) (2 4 6))
nth-column is a list comprehension generating a sequence from the nth element of each sequence (of rows).
Then transpose-matrix is simply iterating over the columns creating a sequence element for each, consisting of (nth-column matrix column) i.e. the sequence of elements for that column.
(map
(partial filter identity) ;;remove nil in each sub-list
(take-while
#(some identity %) ;;stop on all nil sub-list
(for [i (range)]
(map #(get % i) a)))) ;; get returns nil on missing values
Use get to have nil on missing values, iterate (for) on an infinite range, stop on all nil sub-list, remove nil from sub-lists. Add vector constructor before first map and in it's function (first argument) if you really need vectors.
EDIT: please leave a comment if you think this is not useful. We can all learn from mistakes.
I am trying to find a Clojure-idiomatic way to "compress" a vector:
(shift-nils-left [:a :b :c :a nil :d nil])
;=> (true [nil nil :a :b :c :a :d])
(shift-nils-left [nil :a])
;=> (false [nil :a])
(shift-nils-left [:a nil])
;=> (true [nil :a])
(shift-nils-left [:a :b])
;=> (false [:a :b])
In other words, I want to move all of the nil values to the left end of the vector, without changing the length. The boolean indicates whether any shifting occurred. The "outside" structure can be any seq, but the inside result should be a vector.
I suspect that the function will involve filter (on the nil values) and into to add to a vector of nils of the same length as the original, but I'm not sure how to reduce the result back to the original length. I know how to this "long-hand", but I suspect that Clojure will be able to do it in a single line.
I am toying with the idea of writing a Bejeweled player as an exercise to learn Clojure.
Thanks.
I would write it like this:
(ns ...
(:require [clojure.contrib.seq-utils :as seq-utils]))
(defn compress-vec
"Returns a list containing a boolean value indicating whether the
vector was changed, and a vector with all the nils in the given
vector shifted to the beginning."
([v]
(let [shifted (vec (apply concat (seq-utils/separate nil? v)))]
(list (not= v shifted)
shifted))))
Edit: so, the same as what Thomas beat me to posting, but I wouldn't use flatten just in case you end up using some sort of seqable object to represent the jewels.
Maybe this way:
(defn shift-nils-left
"separate nil values"
[s]
(let [s1 (vec (flatten (clojure.contrib.seq/separate nil? s)))]
(list (not (= s s1)) s1)))
A little more low-level approach. It traverses the input seq just once as well as the vector of non-nils once. The two more highlevel approaches traverse the input sequence two times (for nil? and (complenent nil?)). The not= traverses the input a third time in the worst-case of no shift.
(defn compress-vec
[v]
(let [[shift? nils non-nils]
(reduce (fn [[shift? nils non-nils] x]
(if (nil? x)
[(pos? (count non-nils)) (conj nils nil) non-nils]
[shift? nils (conj non-nils x)]))
[false [] []] v)]
[shift? (into nils non-nils)]))
(def v [1 2 nil 4 5 nil 7 8] )
(apply vector (take 8 (concat (filter identity v) (repeat nil))))
This creates a sequence of the non- nil values in the vector using filter and then appends nils to the end of the sequence. This gives the values you want as a sequence and then converts them into a vector. The take 8 ensures that the vector is right size.