I'm looking for an elegant way to generate a sequence of the rolling average of a sequence of numbers. Hopefully something more elegant than using lazy-seq
Without any consideration of efficiency:
(defn average [lst] (/ (reduce + lst) (count lst)))
(defn moving-average [window lst] (map average (partition window 1 lst)))
user> (moving-average 5 '(1 2 3 4 5 6 7 8))
(3 4 5 6)
If you need it to be fast, there are some fairly obvious improvements to be made!
But it will get less elegant.
There's a very similar question on SO: Calculating the Moving Average of a List. It's more general -- a number of FP-friendly languages are represented, with the accepted answer using Scala -- but there are a few nice Clojure solutions.
I've posted my own solution over there. Note that it does use lazy-seq, but that's because I wanted it to perform well for large periods (which means adjusting the average at each step rather than calculating a separate average for each window of size = period into the input list). Look around that Q for nice solutions which made the other tradeoff, resulting in shorter code with a somewhat more declarative feel, which actually performs better for very short periods (although suffers significant slowdowns for longer periods, as is to be expected).
This version is a bit faster, especially for long windows, since it keeps a rolling sum and avoids repeatedly adding the same things.
Because of the lazy-seq, it's also perfectly general and won't blow stack
(defn partialsums [start lst]
(lazy-seq
(if-let [lst (seq lst)]
(cons start (partialsums (+ start (first lst)) (rest lst)))
(list start))))
(defn sliding-window-moving-average [window lst]
(map #(/ % window)
(let [start (apply + (take window lst))
diffseq (map - (drop window lst) lst)]
(partialsums start diffseq))))
;; To help see what it's doing:
(sliding-window-moving-average 5 '(1 2 3 4 5 6 7 8 9 10 11))
start = (+ 1 2 3 4 5) = 15
diffseq = - (6 7 8 9 10 11)
(1 2 3 4 5 6 7 8 9 10 11)
= (5 5 5 5 5 5)
(partialsums 15 '(5 5 5 5 5 5) ) = (15 20 25 30 35 40 45)
(map #(/ % 5) (20 25 30 35 40 45)) = (3 4 5 6 7 8 9)
;; Example
(take 20 (sliding-window-moving-average 5 (iterate inc 0)))
Instead of the partialsums fn (which is helpful to see what's going on), you can use reductions in clojure.core:
(defn sliding-window-moving-average [window lst]
(map #(/ % window)
(let [start (apply + (take window lst))
diffseq (map - (drop window lst) lst)]
(reductions + start diffseq))))
Related
I'm currently learning Clojure, and I'm trying to learn how to do things the best way. Today I'm looking at the basic concept of doing things on a sequence, I know the basics of map, filter and reduce. Now I want to try to do a thing to pairs of elements in a sequence, and I found two ways of doing it. The function I apply is println. The output is simply 12 34 56 7
(def xs [1 2 3 4 5 6 7])
(defn work_on_pairs [xs]
(loop [data xs]
(if (empty? data)
data
(do
(println (str (first data) (second data)))
(recur (drop 2 data))))))
(work_on_pairs xs)
I mean, I could do like this
(map println (zipmap (take-nth 2 xs) (take-nth 2 (drop 1 xs))))
;; prints [1 2] [3 4] [5 6], and we loose the last element because zip.
But it is not really nice.. My background is in Python, where I could just say zip(xs[::2], xs[1::2]) But I guess this is not the Clojure way to do it.
So I'm looking for suggestions on how to do this same thing, in the best Clojure way.
I realize I'm so new to Clojure I don't even know what this kind of operation is called.
Thanks for any input
This can be done with partition-all:
(def xs [1 2 3 4 5 6 7])
(->> xs
(partition-all 2) ; Gives ((1 2) (3 4) (5 6) (7))
(map (partial apply str)) ; or use (map #(apply str %))
(apply println))
12 34 56 7
The map line is just to join the pairs so the "()" don't end up in the output.
If you want each pair printed on its own line, change (apply println) to (run! println). Your expected output seems to disagree with your code, so that's unclear.
If you want to dip into transducers, you can do something similar to the threading (->>) form of the accepted answer, but in a single pass over the data.
Assuming
(def xs [1 2 3 4 5 6 7])
has been evaluated already,
(transduce
(comp
(partition-all 2)
(map #(apply str %)))
conj
[]
xs)
should give you the same output if you wrap it in
(apply println ...)
We supply conj (reducing fn) and [] (initial data structure) to specify how the reduce process inside transduce should build up the result.
I wouldn't use a transducer for a list that small, or a process that simple, but it's good to know what's possible!
I am learning clojure. While solving one of the problem, I had to use first + filter. I noted that the filter is running unnecessarily for all the inputs.
How can I make the filter to run lazily so that it need not apply the predicate for the whole input.
The below is an example showing that it is not lazy,
(defn filter-even
[n]
(println n)
(= (mod n 2) 0))
(first (filter filter-even (range 1 4)))
The above code prints
1
2
3
Whereas it need not go beyond 2. How can we make it lazy?
This happens because range is a chunked sequence:
(chunked-seq? (range 1))
=> true
And it will actually take the first 32 elements if available:
(first (filter filter-even (range 1 100)))
1
2
. . .
30
31
32
=> 2
This overview shows an unchunk function that prevents this from happening. Unfortunately, it isn't standard:
(defn unchunk [s]
(when (seq s)
(lazy-seq
(cons (first s)
(unchunk (next s))))))
(first (filter filter-even (unchunk (range 1 100))))
2
=> 2
Or, you could apply list to it since lists aren't chunked:
(first (filter filter-even (apply list (range 1 100))))
2
=> 2
But then obviously, the entire collection needs to be realized pre-filtering.
This honestly isn't something that I've ever been too concerned about though. The filtering function usually isn't too expensive, and 32 element chunks aren't that big in the grand scheme of things.
I have the following bit of code that produces the correct results:
(ns scratch.core
(require [clojure.string :as str :only (split-lines join split)]))
(defn numberify [str]
(vec (map read-string (str/split str #" "))))
(defn process [acc sticks]
(let [smallest (apply min sticks)
cuts (filter #(> % 0) (map #(- % smallest) sticks))]
(if (empty? cuts)
acc
(process (conj acc (count cuts)) cuts))))
(defn print-result [[x & xs]]
(prn x)
(if (seq xs)
(recur xs)))
(let [input "8\n1 2 3 4 3 3 2 1"
lines (str/split-lines input)
length (read-string (first lines))
inputs (first (rest lines))]
(print-result (process [length] (numberify inputs))))
The process function above recursively calls itself until the sequence sticks is empty?.
I am curious to know if I could have used something like take-while or some other technique to make the code more succinct?
If ever I need to do some work on a sequence until it is empty then I use recursion but I can't help thinking there is a better way.
Your core problem can be described as
stop if count of sticks is zero
accumulate count of sticks
subtract the smallest stick from each of sticks
filter positive sticks
go back to 1.
Identify the smallest sub-problem as steps 3 and 4 and put a box around it
(defn cuts [sticks]
(let [smallest (apply min sticks)]
(filter pos? (map #(- % smallest) sticks))))
Notice that sticks don't change between steps 5 and 3, that cuts is a fn sticks->sticks, so use iterate to put a box around that:
(defn process [sticks]
(->> (iterate cuts sticks)
;; ----- 8< -------------------
This gives an infinite seq of sticks, (cuts sticks), (cuts (cuts sticks)) and so on
Incorporate step 1 and 2
(defn process [sticks]
(->> (iterate cuts sticks)
(map count) ;; count each sticks
(take-while pos?))) ;; accumulate while counts are positive
(process [1 2 3 4 3 3 2 1])
;-> (8 6 4 1)
Behind the scene this algorithm hardly differs from the one you posted, since lazy seqs are a delayed implementation of recursion. It is more idiomatic though, more modular, uses take-while for cancellation which adds to its expressiveness. Also it doesn't require one to pass the initial count and does the right thing if sticks is empty. I hope it is what you were looking for.
I think the way your code is written is a very lispy way of doing it. Certainly there are many many examples in The Little Schema that follow this format of reduction/recursion.
To replace recursion, I usually look for a solution that involves using higher order functions, in this case reduce. It replaces the min calls each iteration with a single sort at the start.
(defn process [sticks]
(drop-last (reduce (fn [a i]
(let [n (- (last a) (count i))]
(conj a n)))
[(count sticks)]
(partition-by identity (sort sticks)))))
(process [1 2 3 4 3 3 2 1])
=> (8 6 4 1)
I've changed the algorithm to fit reduce by grouping the same numbers after sorting, and then counting each group and reducing the count size.
I have a higher order predicate
(defn not-factor-of-x? [x]
(fn [n]
(cond
(= n x) true
(zero? (rem n x)) false
:else true)))
which returns a predicate that checks if the given argument n is not a factor of x.
Now I want to filter a list of numbers and find which are not factors of say '(2 3). One way to do this would be :
(filter (not-factor-of-x? 3) (filter (not-factor-of-x? 2) (range 2 100)))
But one can only type so much. In order to do this dynamically I tried function composition :
(comp (partial filter (not-factor-of-x? 2)) (partial filter (not-factor-of-x? 3)))
And it works. So I tried reducing the filters, like this:
(defn compose-filters [fn1 fn2]
(comp (partial filter fn1) (partial filter fn2)))
(def composed-filter (reduce compose-filters (map not-factor-of-x? '(2 3 5 7 11))))
(composed-filter (range 2 122)) ; returns (2 3 4 5 6 7 8 9 10 .......)
So, why the filter composition is not working as intended ?
There are many ways to compose functions and/or improve your code. Here's one:
(defn factor? [n x]
(and (not= n x) (zero? (rem n x))))
(->> (range 2 100)
(remove #(factor? % 2))
(remove #(factor? % 3)))
;; the same as the above
(->> (range 2 100)
(remove (fn [n] (some #(factor? n %) [2 3]))))
To see your problem with (reduce compose-filters ... let's look a bit at what that actually does. First, it uses filter on the first two predicates and composes them.. The result of that is a new function from sequences to sequences. The next iteration then calls filter on that function, when filter expects a predicate. Every sequence is a truthy value, so that new filter will now never remove any values because it's using a "predicate" which always returns truthy values. So in the end, only the very last filter actually does any filtering - in my REPL your code removes the numbers 22, 33, 44 and so on because 11 is a factor in them. I think the reduce you want to do here is more like
(reduce comp (map (comp (partial partial filter) not-factor-of-x?) '(2 3 5 7 11)))
Note how because we only want to call (partial filter) once per number, you can move that into the mapping step of the mapreduce. As to how I'd do this, considering that you produce all your predicates together:
(map not-factor-of-x? '(2 3 5 7 11))
it seems more natural to me to just combine the predicates at that point using every-pred
(apply every-pred (map not-factor-of-x? '(2 3 5 7 11)))
and use one filter on that predicate. It seems to communicate the intent a little more clearly ("I want values satisfying every one of these preds") and unlike composition of (partial filter ...) it avoids making an intermediate sequence for each predicate.
(In Clojure 1.7+ you can also avoid this by composing the transducer version of filter).
I am working on some Lisp exercises using Clojure. I am trying to work these exercises without taking advantage of vectors and some Clojure functions.
This function
(defn rev-seq
[s1]
(concat (pop s1) (list (peek s1))))
puts the first element of a list at the end. I want to call this function as many times as it takes to reverse the list (without calling Clojure's reverse function).
I am not sure what to use in its place. I have experimented with map, apply, and repeat with no success. I would rather have a way to think differently about this than a straight answer, but I am not asking for a discussion.
Firstly, I think you'll need to convert rev-seq to use first/rest rather than peek/pop if you want to work on general sequences - at leas in Clojure 1.4 peek/pop seems to require a PersistentStack:
(defn rev-seq
[s1]
(concat (rest s1) (list (first s1))))
Then you should probably note that applying this function repeatedly will "cycle" a list rather than reversing it. You can see that if you look at the result of a small number of applications using iterate:
(def s '(1 2 3 4 5 6 7 8 9))
(nth (iterate rev-seq s) 3)
=> (4 5 6 7 8 9 1 2 3)
An option that would work is to reverse with a recursive function:
(defn reverse-seq [s]
(concat (reverse (next s)) (list (first s))))
(reverse-seq s)
=> (9 8 7 6 5 4 3 2 1)
Or alternatively you can do a reverse using the technique in clojure.core:
(defn reverse-seq [s]
(reduce conj () s))
(reverse-seq s)
=> (9 8 7 6 5 4 3 2 1)
Hope this gives you some ideas!
Recursion is powerful!
I translated
the solution
into Clojure.
(defn- inverte-aux
[lista resto]
(if lista
(recur (next lista) (cons (first lista) resto))
resto))
(defn inverte
[lista]
(inverte-aux lista nil))
user> (inverte [4 3 2 1 3])
(3 1 2 3 4)