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.
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.
This is similar to Clojure get map key by value
However, there is one difference. How would you do the same thing if hm is like
{1 ["bar" "choco"]}
The idea being to get 1 (the key) where the first element if the value list is "bar"? Please feel free to close/merge this question if some other question answers it.
I tried something like this, but it doesn't work.
(def hm {:foo ["bar", "choco"]})
(keep #(when (= ((nth val 0) %) "bar")
(key %))
hm)
You can filter the map and return the first element of the first item in the resulting sequence:
(ffirst (filter (fn [[k [v & _]]] (= "bar" v)) hm))
you can destructure the vector value to access the second and/or third elements e.g.
(ffirst (filter (fn [[k [f s t & _]]] (= "choco" s))
{:foo ["bar", "choco"]}))
past the first few elements you will probably find nth more readable.
Another way to do it using some:
(some (fn [[k [v & _]]] (when (= "bar" v) k)) hm)
Your example was pretty close to working, with some minor changes:
(keep #(when (= (nth (val %) 0) "bar")
(key %))
hm)
keep and some are similar, but some only returns one result.
in addition to all the above (correct) answers, you could also want to reindex your map to desired form, especially if the search operation is called quite frequently and the the initial map is rather big, this would allow you to decrease the search complexity from linear to constant:
(defn map-invert+ [kfn vfn data]
(reduce (fn [acc entry] (assoc acc (kfn entry) (vfn entry)))
{} data))
user> (def data
{1 ["bar" "choco"]
2 ["some" "thing"]})
#'user/data
user> (def inverted (map-invert+ (comp first val) key data))
#'user/inverted
user> inverted
;;=> {"bar" 1, "some" 2}
user> (inverted "bar")
;;=> 1
This is a scenario I encountered many times, yet didn't find an idiomatic approach for it...
Suppose one would like to use a self-defined self-pred function to filter a seq. This self-pred function returns nil for unwanted elements, and useful information for wanted elements. It is desirable to keep the evaluated self-pred values for these wanted elements.
My general solution is:
;; self-pred is a pred function which returns valuable info
;; in general, they are unique and can be used as key
(let [new-seq (filter self-pred aseq)]
(zipmap (map self-pred new-seq) new-seq))
Basically, it is to call self-pred twice on all wanted elements. I feel it is so ugly...
Wonder if there is any better ways. Much appreciated for any input!
In these scenarios you can use keep, but you have to change your "predicate" function to return the full information you need, or nil, for each item.
For example:
(keep (fn [item]
(when-let [tested (some-test item)]
(assoc item :test-output tested))) aseq)
i use this kind of snippet:
(keep #(some->> % self-pred (vector %)) data)
like this:
user> (keep #(some->> % rseq (vector %)) [[1 2] [] [3 4]])
;;=> ([[1 2] (2 1)] [[3 4] (4 3)])
or if you like more verbose result:
user> (keep #(some->> % rseq (hash-map :data % :result)) [[1 2] [] [3 4]])
;;=> ({:result (2 1), :data [1 2]} {:result (4 3), :data [3 4]})
I wouldn't bother with keep, but would just use plain map & filter like so:
(def data (range 6))
(def my-pred odd?)
(defn zip [& colls] (apply map vector colls)) ; like Python zip
(defn filter-with-pred
[vals pred]
(filter #(first %)
(zip (map pred vals) vals)))
(println (filter-with-pred data my-pred))
with result:
([true 1] [true 3] [true 5])
If self-pred guarantees no duplicate key creation for differing values then I'd reach for reduce (since assoc the same key twice will override the original key value pair):
(reduce #(if-let [k (self-pred %2)]
(assoc %1 k %2)
%1)
{}
aseq)
Else we can use group-by to drive a similar result:
(dissoc (group-by self-pred aseq) nil)
Although not the same since the values will be in vectors: {k1 [v1 ..], k2 [..], ..}. but this guarantees all values are kept.
As a newbie to Clojure I often have difficulties to express the simplest things. For example, for replacing the last element in a vector, which would be
v[-1]=new_value
in python, I end up with the following variants in Clojure:
(assoc v (dec (count v)) new_value)
which is pretty long and inexpressive to say the least, or
(conj (vec (butlast v)) new_value)
which even worse, as it has O(n) running time.
That leaves me feeling silly, like a caveman trying to repair a Swiss watch with a club.
What is the right Clojure way to replace the last element in a vector?
To support my O(n)-claim for butlast-version (Clojure 1.8):
(def v (vec (range 1e6)))
#'user/v
user=> (time (first (conj (vec (butlast v)) 55)))
"Elapsed time: 232.686159 msecs"
0
(def v (vec (range 1e7)))
#'user/v
user=> (time (first (conj (vec (butlast v)) 55)))
"Elapsed time: 2423.828127 msecs"
0
So basically for 10 time the number of elements it is 10 times slower.
I'd use
(defn set-top [coll x]
(conj (pop coll) x))
For example,
(set-top [1 2 3] :a)
=> [1 2 :a]
But it also works on the front of lists:
(set-top '(1 2 3) :a)
=> (:a 2 3)
The Clojure stack functions - peek, pop, and conj - work on the natural open end of a sequential collection.
But there is no one right way.
How do the various solutions react to an empty vector?
Your Python v[-1]=new_value throws an exception, as does your (assoc v (dec (count v)) new_value) and my (defn set-top [coll x] (conj (pop coll) x)).
Your (conj (vec (butlast v)) new_value) returns [new_value]. The butlast has no effect.
If you insist on being "pure", your 2nd or 3rd solutions will work. I prefer to be simpler & more explicit using the helper functions from the Tupelo library:
(s/defn replace-at :- ts/List
"Replaces an element in a collection at the specified index."
[coll :- ts/List
index :- s/Int
elem :- s/Any]
...)
(is (= [9 1 2] (replace-at (range 3) 0 9)))
(is (= [0 9 2] (replace-at (range 3) 1 9)))
(is (= [0 1 9] (replace-at (range 3) 2 9)))
As with drop-at, replace-at will throw an exception for invalid values of index.
Similar helper functions exist for
insert-at
drop-at
prepend
append
Note that all of the above work equally well for either a Clojure list (eager or lazy) or a Clojure vector. The conj solution will fail unless you are careful to always coerce the input to a vector first as in your example.
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))))