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.
Related
I am trying to extract two elements of a map with the largest distance. For that, I defined the function for calculating the distance and can obtain the distance between the first element (p1) and other elements of the map. But I need to calculate distances between the second item (p2) and the next ones (p3, p4, p5), the third item (p3) and (p4, p5), the fourth item (p4) and fifth item (p5). Then I need to identify the maximum amount between all distances and return the 2 items with the largest distance and the distance itself. Any help is highly appreciated.
Here is my code:
(defn eclid-dist
[u v]
(Math/sqrt (apply + (map #(* % %) (mapv - u v)))))
(def error
{:p1 [1 2 3]
:p2 [4 5 6]
:p3 [7 8 9]
:p4 [1 2 3]
:p5 [6 5 4]})
(dotimes [i (dec (count error))]
(let [dis (eclid-dist (second (nth (seq error) 0))
(second (nth (seq error) (+ i 1))))
max-error (max dis)]
(println [':dis' dis ':max-error' max-error])))
I tried to save each calculated distance as a vector element separately to prevent overwriting but it was not successful.
You could use the for macro for this. It let's you combine two nested loops to test for all pairs. Then you can use max-key to pick the pair with largest distance:
(defn find-largest-dist-pair [vec-map]
(apply max-key second
(for [[[k0 v0] & r] (iterate rest vec-map)
:while r
[k1 v1] r]
[[k0 k1] (eclid-dist v0 v1)])))
(find-largest-dist-pair error)
;; => [[:p3 :p4] 10.392304845413264]
There is nothing wrong with eclid-dist, you could just use the dedicated Clojure library clojure.math (and ->> thread-last macro for better readability) and rewrite it like this:
(:require [clojure.math :as m])
(defn distance [u v]
(->> (mapv - u v)
(mapv #(m/pow % 2))
(reduce +)
m/sqrt))
Your main problem is, how to create unique pairs of points from your data. You could write a recursive function for this:
(defn unique-pairs [point-seq]
(let [[f & r] point-seq]
(when (seq r)
(concat (map #(vector f %) r)
(unique-pairs r)))))
(def error {:p1 [1 2 3]
:p2 [4 5 6]
:p3 [7 8 9]
:p4 [1 2 3]
:p5 [6 5 4]})
(unique-pairs (vals error))
or use library clojure.math.combinatorics:
Dependency: [org.clojure/math.combinatorics "0.1.6"]
(:require [clojure.math.combinatorics :as combi])
(combi/combinations (vals error) 2)
Note that these functions have slightly different results- it doesn't affect the final result, but if you can, you should use combinations.
Now, you have to compute distance for all these pairs and return the pair with the largest one:
(defn max-distance [point-map]
(->> (combi/combinations (vals point-map) 2)
(map (fn [[u v]] {:u u :v v :distance (distance u v)}))
(apply max-key :distance)))
(max-distance error)
=> {:u [1 2 3], :v [7 8 9], :distance 10.392304845413264}
I am trying to get into Lisps and FP by trying out the 99 problems.
Here is the problem statement (Problem 15)
Replicate the elements of a list a given number of times.
I have come up with the following code which simply returns an empty list []
I am unable to figure out why my code doesn't work and would really appreciate some help.
(defn replicateList "Replicates each element of the list n times" [l n]
(loop [initList l returnList []]
(if (empty? initList)
returnList
(let [[head & rest] initList]
(loop [x 0]
(when (< x n)
(conj returnList head)
(recur (inc x))))
(recur rest returnList)))))
(defn -main
"Main" []
(test/is (=
(replicateList [1 2] 2)
[1 1 2 2])
"Failed basic test")
)
copying my comment to answer:
this line: (conj returnList head) doesn't modify returnlist, rather it just drops the result in your case. You should restructure your program to pass the accumulated list further to the next iteration. But there are better ways to do this in clojure. Like (defn replicate-list [data times] (apply concat (repeat times data)))
If you still need the loop/recur version for educational reasons, i would go with this:
(defn replicate-list [data times]
(loop [[h & t :as input] data times times result []]
(if-not (pos? times)
result
(if (empty? input)
(recur data (dec times) result)
(recur t times (conj result h))))))
user> (replicate-list [1 2 3] 3)
;;=> [1 2 3 1 2 3 1 2 3]
user> (replicate-list [ ] 2)
;;=> []
user> (replicate-list [1 2 3] -1)
;;=> []
update
based on the clarified question, the simplest way to do this is
(defn replicate-list [data times]
(mapcat (partial repeat times) data))
user> (replicate-list [1 2 3] 3)
;;=> (1 1 1 2 2 2 3 3 3)
and the loop/recur variant:
(defn replicate-list [data times]
(loop [[h & t :as data] data n 0 res []]
(cond (empty? data) res
(>= n times) (recur t 0 res)
:else (recur data (inc n) (conj res h)))))
user> (replicate-list [1 2 3] 3)
;;=> [1 1 1 2 2 2 3 3 3]
user> (replicate-list [1 2 3] 0)
;;=> []
user> (replicate-list [] 10)
;;=> []
Here is a version based on the original post, with minimal modifications:
;; Based on the original version posted
(defn replicateList "Replicates each element of the list n times" [l n]
(loop [initList l returnList []]
(if (empty? initList)
returnList
(let [[head & rest] initList]
(recur
rest
(loop [inner-returnList returnList
x 0]
(if (< x n)
(recur (conj inner-returnList head) (inc x))
inner-returnList)))))))
Please keep in mind that Clojure is mainly a functional language, meaning that most functions produce their results as a new return value instead of updating in place. So, as pointed out in the comment, the line (conj returnList head) will not have an effect, because it's return value is ignored.
The above version works, but does not really take advantage of Clojure's sequence processing facilities. So here are two other suggestions for solving your problem:
;; Using lazy seqs and reduce
(defn replicateList2 [l n]
(reduce into [] (map #(take n (repeat %)) l)))
;; Yet another way using transducers
(defn replicateList3 [l n]
(transduce
(comp (map #(take n (repeat %)))
cat
)
conj
[]
l))
One thing is not clear about your question though: From your implementation, it looks like you want to create a new list where each element is repeated n times, e.g.
playground.replicate> (replicateList [1 2 3] 4)
[1 1 1 1 2 2 2 2 3 3 3 3]
But if you would instead like this result
playground.replicate> (replicateList [1 2 3] 4)
[1 2 3 1 2 3 1 2 3 1 2 3]
the answer to your question will be different.
If you want to learn idiomatic Clojure you should try to find a solution without such low level facilities as loop. Rather try to combine higher level functions like take, repeat, repeatedly. If you're feeling adventurous you might throw in laziness as well. Clojure's sequences are lazy, that is they get evaluated only when needed.
One example I came up with would be
(defn repeat-list-items [l n]
(lazy-seq
(when-let [s (seq l)]
(concat (repeat n (first l))
(repeat-list-items (next l) n)))))
Please also note the common naming with kebab-case
This seems to do what you want pretty well and works for an unlimited input (see the call (range) below), too:
experi.core> (def l [:a :b :c])
#'experi.core/
experi.core> (repeat-list-items l 2)
(:a :a :b :b :c :c)
experi.core> (repeat-list-items l 0)
()
experi.core> (repeat-list-items l 1)
(:a :b :c)
experi.core> (take 10 (drop 10000 (repeat-list-items (range) 4)))
(2500 2500 2500 2500 2501 2501 2501 2501 2502 2502)
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.
I would like to compute the weighted mean of vectors in an idiomatic way.
To illustrate what I want, imagine I have this data :
data 1 = [2 1] , weight 1 = 1
data 2 = [3 4], weight 2 = 2
Then mean = [(2*1 + 3*2)/(1+2) (1*1 + 2*4)/(1+2)] = [2.67 3.0]
Here is my code :
(defn meanv
"Returns the vector that is the mean of input ones.
You can also pass weights just like apache-maths.stats/mean"
([data]
(let [n (count (first data))]
(->> (for [i (range 0 n)]
(vec (map (i-partial nth i) data)))
(mapv stats/mean))))
([data weights]
(let [n (count (first data))]
(->> (for [i (range 0 n)]
(vec (map (i-partial nth i) data)))
(mapv (i-partial stats/mean weights))))))
Then
(meanv [[2 1] [3 4]] [1 2]) = [2.67 3.0]
Few notes :
stats/means takes 1 or 2 inputs.
One input version has weights = 1 by default.
Two inputs is the weighted version.
i-partial is like partial but the fn has reversed args
Ex : ((partial / 2) 1) = 2
((i-partial / 2) 1 = 1/2
So my function works, no problem.
But in a way I would like to implement it in a more idiomatic Clojure.
I tried many combinations with things like (map (fn [&xs ... but it does not work.
Is it possible to take all nth elements of undefined number of vectors and directly apply stats/mean ? I mean a one-liner
Thanks
EDIT (birdspider answer)
(defn meanv
([data]
(->> (apply mapv vector data)
(mapv stats/mean)))
([data weights]
(->> (apply mapv vector data)
(mapv (i-partial stats/mean weights)))))
And with
(defn transpose [m]
(apply mapv vector m))
(defn meanv
([data]
(->> (transpose data)
(mapv stats/mean)))
([data weights]
(->> (transpose data)
(mapv (i-partial stats/mean weights)))))
(def mult-v (partial mapv *))
(def sum-v (partial reduce +))
(def transpose (partial apply mapv vector))
(defn meanv [data weights]
(->> data
transpose
(map (partial mult-v weights))
(map sum-v)
(map #(/ % (sum-v weights)))))
First thing you want to do is to transpose the matrix (get the firsts, seconds, thirds, etc.)
See this SO page.
; https://stackoverflow.com/a/10347404/2645347
(defn transpose [m]
(apply mapv vector m))
Then I would do it as follows, input checks are utterly absent.
(defn meanv
([data]
; no weigths default to (1 1 1 ...
(meanv data (repeat (count data) 1))))
([data weigths]
(let [wf (mapv #(partial * %) weigths) ; vector of weight mult fns
wsum (reduce + weigths)]
(map-indexed
(fn [i datum]
(/
; map over datum apply corresponding weight-fn - then sum
(apply + (map-indexed #((wf %1) %2) datum))
wsum))
(transpose data)))))
(meanv [[2 1] [3 4]] [1 2]) => (8/3 3) ; (2.6666 3.0)
Profit!
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))))