I'd like to know how to create an infinite, impure sequence of unique values in Clojure.
(def generator ...) ; def, not defn
(take 4 generator) ; => (1 2 3 4)
(take 4 generator) ; => (5 6 7 8). note the generator's impurity.
I think that such a design could be more convenient than e.g. wrapping a single integer value into a reference type and increment it from its consumers, as:
The proposed approach reduces the implementation details to a single point of change: the generator. Otherwise all the consumers would have to care about both the reference type (atom), and the concrete function that provides the next value (inc)
Sequences can take advantage many clojure.core functions. 'Manually' building a list of ids out of an atom would be a bit bulky: (take 4 (repeatedly #(swap! _ inc)))
I couldn't come up with a working implementation. Is it possible at all?
You can wrap a lazy sequence around an impure class (like a java.util.concurrent.atomic.AtomicLong) to create an id sequence:
(def id-counter (java.util.concurrent.atomic.AtomicLong.))
(defn id-gen []
(cons
(.getAndIncrement id-counter)
(lazy-seq
(id-gen))))
This works, but only if you don't save the head of the sequence. If you create a var that captures the head:
(def id-seq (id-gen))
Then call it repeatedly, it will return ids from the beginning of the sequence, because you've held onto the head of the sequence:
(take 3 id-seq)
;; => (0 1 2)
(take 3 id-seq)
;; => (0 1 2)
(take 3 id-seq)
;; => (0 1 2)
If you re-create the sequence though, you'll get fresh values because of the impurity:
(take 3 (id-gen))
;; (3 4 5)
(take 3 (id-gen))
;; (6 7 8)
(take 3 (id-gen))
;; (9 10 11)
I only recommend doing the following for educational purposes (not production code), but you can create your own instance of ISeq which implements the impurity more directly:
(def custom-seq
(reify clojure.lang.ISeq
(first [this] (.getAndIncrement id-counter))
(next [this] (.getAndIncrement id-counter))
(cons [this thing]
(cons thing this))
(more [this] (cons
(.getAndIncrement id-counter)
this))
(count [this] (throw (RuntimeException. "count: not supported")))
(empty [this] (throw (RuntimeException. "empty: not supported")))
(equiv [this obj] (throw (RuntimeException. "equiv: not supported")))
(seq [this] this)))
(take 3 custom-seq)
;; (12 13 14)
(take 3 custom-seq)
;; (15 16 17)
I had a fun time discovering something during answering your question. The first thing that occured to me was that perhaps, for whatever ultimate goal you need these IDs for, the gensym function might be helpful.
Then, I thought "well hey, that seems to increment some impure counter to generate new IDs" and "well hey, what's in the source code for that?" Which led me to this:
(. clojure.lang.RT (nextID))
Which seems to do what you need. Cool! If you want to use it the way you suggest, then I would probably make it a function:
(defn generate-id []
(. clojure.lang.RT (nextID)))
Then you can do:
user> (repeatedly 5 generate-id)
=> (372 373 374 375 376)
I haven't yet tested whether this will produce always unique values "globally"--I'm not sure about terminology, but I'm talking about when you might be using this generate-id function from within different threads, but want to still be sure that it's producing unique values.
this is another solution, maybe:
user=> (defn positive-numbers
([] (positive-numbers 1))
([n] (cons n (lazy-seq (positive-numbers (inc n))))))
#'user/positive-numbers
user=> (take 4 (positive-numbers))
(1 2 3 4)
user=> (take 4 (positive-numbers 5))
(5 6 7 8)
A way that would be more idiomatic, thread-safe, and invites no weirdness over head references would be to use a closure over one of clojures built in mutable references. Here is a quick sample I worked up since I was having the same issue. It simply closes over a ref.
(def id-generator (let [counter (ref 0)]
(fn [] (dosync (let [cur-val #counter]
(do (alter counter + 1)
cur-val))))))
Every time you call (id-generator) you will get the next number in the sequence.
Here's another quick way:
user> (defn make-generator [& [ii init]]
(let [a (atom (or ii 0 ))
f #(swap! a inc)]
#(repeatedly f)))
#'user/make-generator
user> (def g (make-generator))
#'user/g
user> (take 3 (g))
(1 2 3)
user> (take 3 (g))
(4 5 6)
user> (take 3 (g))
(7 8 9)
This is hack but it works and it is extremely simple
; there be dragons !
(defn id-gen [n] (repeatedly n (fn [] (hash #()))))
(id-gen 3) ; (2133991908 877609209 1060288067 442239263 274390974)
Basically clojure creates an 'anonymous' function but since clojure itselfs needs a name for that, it uses uniques impure ids to avoid collitions. If you hash a unique name then you should get a unique number.
Hope it helps
Creating identifiers from an arbitrary collection of seed identifiers:
(defonce ^:private counter (volatile! 0))
(defn- next-int []
(vswap! counter inc))
(defn- char-range
[a b]
(mapv char
(range (int a) (int b))))
(defn- unique-id-gen
"Generates a sequence of unique identifiers seeded with ids sequence"
[ids]
;; Laziness ftw:
(apply concat
(iterate (fn [xs]
(for [x xs
y ids]
(str x y)))
(map str ids))))
(def inf-ids-seq (unique-id-gen (concat (char-range \a \z)
(char-range \A \Z)
(char-range \0 \9)
[\_ \-])))
(defn- new-class
"Returns an unused new classname"
[]
(nth inf-ids-seq (next-int)))
(repeatedly 10 new-class)
Demonstration:
(take 16 (unique-id-gen [\a 8 \c]))
;; => ("a" "8" "c" "aa" "a8" "ac" "8a" "88" "8c" "ca" "c8" "cc" "aaa" "aa8" "aac" "a8a")
Related
I need help with an assignment that uses Clojure. It is very small but the language is a bit confusing to understand. I need to create a function that behaves like count without actually using the count funtion. I know a loop can be involved with it somehow but I am at a lost because nothing I have tried even gets my code to work. I expect it to output the number of elements in list. For example:
(defn functionname []
...
...)
(println(functionname '(1 4 8)))
Output:3
Here is what I have so far:
(defn functionname [n]
(def n 0)
(def x 0)
(while (< x n)
do
()
)
)
(println(functionname '(1 4 8)))
It's not much but I think it goes something like this.
This implementation takes the first element of the list and runs a sum until it can't anymore and then returns the sum.
(defn recount [list-to-count]
(loop [xs list-to-count sum 0]
(if (first xs)
(recur (rest xs) (inc sum))
sum
)))
user=> (recount '(3 4 5 9))
4
A couple more example implementations:
(defn not-count [coll]
(reduce + (map (constantly 1) coll)))
or:
(defn not-count [coll]
(reduce (fn [a _] (inc a)) 0 coll))
or:
(defn not-count [coll]
(apply + (map (fn [_] 1) coll)))
result:
(not-count '(5 7 8 1))
=> 4
I personally like the first one with reduce and constantly.
We've been given a task to print the first ten multiples of any number for which we have written the below code. It is throwing an error. In simple words, if n is 2 then we need to create a table of 2's till 10.
(defn multiples [n]
(while ( n < 11)
(println( n * n))
(swap! n inc)))
(def n (Integer/parseInt (clojure.string/trim (read-line))))
(multiples n)
With this, we're getting the error:
Exception in thread "main" java.lang.ClassCastException: java.lang.Integer cannot be cast to clojure.lang.
(defn multiples [n]
(map #(* n %) (range 1 (+ 10 1))))
user=> (multiples 1)
;; => (1 2 3 4 5 6 7 8 9 10)
user=> (multiples 2)
;; => (2 4 6 8 10 12 14 16 18 20)
The resulting list you can loop over and println each of the elements.
(for [i (multiples 2)]
(println i))
;; or:
(map println (multiples 2)) ;; though one usually doesn't apply
;; `map` on side effect functions ...
To improve your own construct:
You, coming from an imperative language, try to work with mutations.
That is very un-idiomatic clojure.
However, by declaring a value atom, you can access using the # operator to its place. And mutate the variable's value.
(defn multiples [n]
(let [i (atom 1)] ;; i is an atom
(while (< #i 11) ;; #i is the value saved into i
(println (* #i n))
(swap! i inc)))) ;; and correctly you can increase the value
With this multiples, you can also print the values.
You can't apply swap! to normal variables, only to atoms.
while loops one should apply only if number of elements not known.
In this case, one knows very well, when to stop. So use rather
a for loop.
(defn multiples [n]
(for [i (range 1 11)]
(println (* i n))))
Look at what iterate function does here
(defn multiples-of [n]
(iterate (partial * n) n))
(def ten-multiples-of-ten
(take 10 (multiples-of 10)))
EDIT: I misread the author of the question, I believe he wants to just generate a sequence of squares. Here is one way using transducers, cause why not ;)
(def xf
(comp
(map inc)
(map #(* % %))))
(defn first-n-squares [n]
(into [] xf (take n (range))))
You can use recur in a loop:
(defn multiples [n]
(if (< n 11)
(do ; then
(println (* n n))
(recur (inc n)))
nil)) ; else return nil
Running this by invoking
(multiples 1)
in a REPL will produce
1
4
9
16
25
36
49
64
81
100
nil
A reduce call has its f argument first. Visually speaking, this is often the biggest part of the form.
e.g.
(reduce
(fn [[longest current] x]
(let [tail (last current)
next-seq (if (or (not tail) (> x tail))
(conj current x)
[x])
new-longest (if (> (count next-seq) (count longest))
next-seq
longest)]
[new-longest next-seq]))
[[][]]
col))
The problem is, the val argument (in this case [[][]]) and col argument come afterward, below, and it's a long way for your eyes to travel to match those with the parameters of f.
It would look more readable to me if it were in this order instead:
(reduceb val col
(fn [x y]
...))
Should I implement this macro, or am I approaching this entirely wrong in the first place?
You certainly shouldn't write that macro, since it is easily written as a function instead. I'm not super keen on writing it as a function, either, though; if you really want to pair the reduce with its last two args, you could write:
(-> (fn [x y]
...)
(reduce init coll))
Personally when I need a large function like this, I find that a comma actually serves as a good visual anchor, and makes it easier to tell that two forms are on that last line:
(reduce (fn [x y]
...)
init, coll)
Better still is usually to not write such a large reduce in the first place. Here you're combining at least two steps into one rather large and difficult step, by trying to find all at once the longest decreasing subsequence. Instead, try splitting the collection up into decreasing subsequences, and then take the largest one.
(defn decreasing-subsequences [xs]
(lazy-seq
(cond (empty? xs) []
(not (next xs)) (list xs)
:else (let [[x & [y :as more]] xs
remainder (decreasing-subsequences more)]
(if (> y x)
(cons [x] remainder)
(cons (cons x (first remainder)) (rest remainder)))))))
Then you can replace your reduce with:
(apply max-key count (decreasing-subsequences xs))
Now, the lazy function is not particularly shorter than your reduce, but it is doing one single thing, which means it can be understood more easily; also, it has a name (giving you a hint as to what it's supposed to do), and it can be reused in contexts where you're looking for some other property based on decreasing subsequences, not just the longest. You can even reuse it more often than that, if you replace the > in (> y x) with a function parameter, allowing you to split up into subsequences based on any predicate. Plus, as mentioned it is lazy, so you can use it in situations where a reduce of any sort would be impossible.
Speaking of ease of understanding, as you can see I misunderstood what your function is supposed to do when reading it. I'll leave as an exercise for you the task of converting this to strictly-increasing subsequences, where it looked to me like you were computing decreasing subsequences.
You don't have to use reduce or recursion to get the descending (or ascending) sequences. Here we are returning all the descending sequences in order from longest to shortest:
(def in [3 2 1 0 -1 2 7 6 7 6 5 4 3 2])
(defn descending-sequences [xs]
(->> xs
(partition 2 1)
(map (juxt (fn [[x y]] (> x y)) identity))
(partition-by first)
(filter ffirst)
(map #(let [xs' (mapcat second %)]
(take-nth 2 (cons (first xs') xs'))))
(sort-by (comp - count))))
(descending-sequences in)
;;=> ((7 6 5 4 3 2) (3 2 1 0 -1) (7 6))
(partition 2 1) gives every possible comparison and partition-by allows you to mark out the runs of continuous decreases. At this point you can already see the answer and the rest of the code is removing the baggage that is no longer needed.
If you want the ascending sequences instead then you only need to change the < to a >:
;;=> ((-1 2 7) (6 7))
If, as in the question, you only want the longest sequence then put a first as the last function call in the thread last macro. Alternatively replace the sort-by with:
(apply max-key count)
For maximum readability you can name the operations:
(defn greatest-continuous [op xs]
(let [op-pair? (fn [[x y]] (op x y))
take-every-second #(take-nth 2 (cons (first %) %))
make-canonical #(take-every-second (apply concat %))]
(->> xs
(partition 2 1)
(partition-by op-pair?)
(filter (comp op-pair? first))
(map make-canonical)
(apply max-key count))))
I feel your pain...they can be hard to read.
I see 2 possible improvements. The simplest is to write a wrapper similar to the Plumatic Plumbing defnk style:
(fnk-reduce { :fn (fn [state val] ... <new state value>)
:init []
:coll some-collection } )
so the function call has a single map arg, where each of the 3 pieces is labelled & can come in any order in the map literal.
Another possibility is to just extract the reducing fn and give it a name. This can be either internal or external to the code expression containing the reduce:
(let [glommer (fn [state value] (into state value)) ]
(reduce glommer #{} some-coll))
or possibly
(defn glommer [state value] (into state value))
(reduce glommer #{} some-coll))
As always, anything that increases clarity is preferred. If you haven't noticed already, I'm a big fan of Martin Fowler's idea of Introduce Explaining Variable refactoring. :)
I will apologize in advance for posting a longer solution to something where you wanted more brevity/clarity.
We are in the new age of clojure transducers and it appears a bit that your solution was passing the "longest" and "current" forward for record-keeping. Rather than passing that state forward, a stateful transducer would do the trick.
(def longest-decreasing
(fn [rf]
(let [longest (volatile! [])
current (volatile! [])
tail (volatile! nil)]
(fn
([] (rf))
([result] (transduce identity rf result))
([result x] (do (if (or (nil? #tail) (< x #tail))
(if (> (count (vswap! current conj (vreset! tail x)))
(count #longest))
(vreset! longest #current))
(vreset! current [(vreset! tail x)]))
#longest)))))))
Before you dismiss this approach, realize that it just gives you the right answer and you can do some different things with it:
(def coll [2 1 10 9 8 40])
(transduce longest-decreasing conj coll) ;; => [10 9 8]
(transduce longest-decreasing + coll) ;; => 27
(reductions (longest-decreasing conj) [] coll) ;; => ([] [2] [2 1] [2 1] [2 1] [10 9 8] [10 9 8])
Again, I know that this may appear longer but the potential to compose this with other transducers might be worth the effort (not sure if my airity 1 breaks that??)
I believe that iterate can be a more readable substitute for reduce. For example here is the iteratee function that iterate will use to solve this problem:
(defn step-state-hof [op]
(fn [{:keys [unprocessed current answer]}]
(let [[x y & more] unprocessed]
(let [next-current (if (op x y)
(conj current y)
[y])
next-answer (if (> (count next-current) (count answer))
next-current
answer)]
{:unprocessed (cons y more)
:current next-current
:answer next-answer}))))
current is built up until it becomes longer than answer, in which case a new answer is created. Whenever the condition op is not satisfied we start again building up a new current.
iterate itself returns an infinite sequence, so needs to be stopped when the iteratee has been called the right number of times:
(def in [3 2 1 0 -1 2 7 6 7 6 5 4 3 2])
(->> (iterate (step-state-hof >) {:unprocessed (rest in)
:current (vec (take 1 in))})
(drop (- (count in) 2))
first
:answer)
;;=> [7 6 5 4 3 2]
Often you would use a drop-while or take-while to short circuit just when the answer has been obtained. We could so that here however there is no short circuiting required as we know in advance that the inner function of step-state-hof needs to be called (- (count in) 1) times. That is one less than the count because it is processing two elements at a time. Note that first is forcing the final call.
I wanted this order for the form:
reduce
val, col
f
I was able to figure out that this technically satisfies my requirements:
> (apply reduce
(->>
[0 [1 2 3 4]]
(cons
(fn [acc x]
(+ acc x)))))
10
But it's not the easiest thing to read.
This looks much simpler:
> (defn reduce< [val col f]
(reduce f val col))
nil
> (reduce< 0 [1 2 3 4]
(fn [acc x]
(+ acc x)))
10
(< is shorthand for "parameters are rotated left"). Using reduce<, I can see what's being passed to f by the time my eyes get to the f argument, so I can just focus on reading the f implementation (which may get pretty long). Additionally, if f does get long, I no longer have to visually check the indentation of the val and col arguments to determine that they belong to the reduce symbol way farther up. I personally think this is more readable than binding f to a symbol before calling reduce, especially since fn can still accept a name for clarity.
This is a general solution, but the other answers here provide many good alternative ways to solve the specific problem I gave as an example.
Consider a query function q that returns, with a delay, some (let say ten) results.
Delay function:
(defn dlay [x]
(do
(Thread/sleep 1500)
x))
Query function:
(defn q [pg]
(lazy-seq
(let [a [0 1 2 3 4 5 6 7 8 9 ]]
(println "q")
(map #(+ (* pg 10) %) (dlay a)))))
Wanted behaviour:
I would like to produce an infinite lazy sequence such that when I take a value only needed computations are evaluated
Wrong but explicative example:
(drop 29 (take 30 (mapcat q (range))))
If I'm not wrong, it needs to evaluate every sequence because it really doesn't now how long the sequences will be.
How would you obtain the correct behaviour?
My attempt to correct this behaviour:
(defn getq [coll n]
(nth
(nth coll (quot n 10))
(mod n 10)))
(defn results-seq []
(let [a (map q (range))]
(map (partial getq a)
(iterate inc 0)))) ; using iterate instead of range, this way i don't have a chunked sequence
But
(drop 43 (take 44 (results-seq)))
still realizes the "unneeded" q sequences.
Now, I verified that a is lazy, iterate and map should produce lazy sequences, so the problem must be with getq. But I can't understand really how it breaks my laziness...perhaps does nth realize things while walking through a sequence? If this would be true, is there a viable alternative in this case or my solution suffers from bad design?
Here is the function I'm trying to run...
(defn mongean [cards times]
(let [_cards (transient cards)]
(loop [i 0 c (get cards i) _count (count cards) _current (/ _count 2)]
(assoc! _cards _current c)
(if ((rem i 2) = 0)
(def _newcur (- _current (inc i)))
(def _newcur (+ _current (inc i))))
(if (<= i _count)
(recur (inc i) (get cards i) _count _newcur )))
(persistent! _cards)))
It's resulting in this Exception...
Exception in thread "main" java.lang.ClassCastException: clojure.lang.PersistentHashSet$TransientHashSet cannot be cast to clojure.lang.ITransientAssociative
Being new to clojure, I'd also appreciate any constructive criticism of my approach above. The goal is to take a List, and return a re-ordered list.
I assume that you are trying to implement the Mongean shuffle. Your approach is very imperative and you should try to use a more functional approach.
This would be a possible implementation, were we calculate the final order of the cards (as per Wikipedia formula) and then we use the built-in replace function to do the mapping:
(defn mongean [cards]
(let [num-cards (count cards)
final-order (concat (reverse (range 1 num-cards 2)) (range 0 num-cards 2))]
(replace cards final-order)))
user> (mongean [1 2 3 4 5 6 7 8])
(8 6 4 2 1 3 5 7)
How do you call that function? It looks like you're passing a set, so that its transient version will also be a set and hence can't be used with any of the assoc functions, as they work on associative data structures and vectors:
user=> (assoc #{} :a 1)
ClassCastException clojure.lang.PersistentHashSet cannot be cast to clojure.lang.Associative clojure.lang.RT.assoc (RT.java:691)
user=> (assoc! (transient #{}) :a 1)
ClassCastException clojure.lang.PersistentHashSet$TransientHashSet cannot be cast to clojure.lang.ITransientAssociative clojure.core/assoc! (core.clj:2959)
; the following works as it uses maps and vectors
user=> (assoc {} :a 1)
{:a 1}
user=> (assoc! (transient {}) :a 1)
#<TransientArrayMap clojure.lang.PersistentArrayMap$TransientArrayMap#65cd1dff>
user=> (assoc [] 0 :a)
[:a]
Now, let's try to discuss the code itself. It's a bit hard to follow your code and try to understand what the goal really is without some more hints on what you want to achieve, but as general comments:
you have a times input parameter you don't use at all
you are supposed to use the result of a transient mutation, not assume that the transient will mutate in place
avoid transients if you can, they're only meant as a performance optimization
the binding _current (/ _count 2) is probably not what you want, as (/ 5 2) really returns 5/2 and it seems that you want to use it as a position in the result
constants like _count don't need to be part of the loop binding, you can use the outer let so that you don't have to pass them at each and every iteration
use let instead of def for naming things inside a function
(if ((rem 1 2) = 0)) is definitely not what you want
Now, leaving aside the shuffling algorithm, if you need to rearrange a sequence you might just produce a sequence of new positions, map them with the original cards to produce pairs of [position card] and finally reduce them by placing the card at the new position, using the original sequence as the seed:
(defn generate [coll] ; counts down from (count coll) to 0, change to
; implement your shuffling algorithm
(range (dec (count coll)) -1 -1))
(defn mongean [cards times]
(let [positions (generate cards) ; get the new positions
assemble (fn [dest [pos card]] ; assoc the card at the wanted position
(assoc dest pos card))]
(reduce assemble cards (map vector positions cards))))
If you simply want to shuffle:
(defn mongean [cards times] (shuffle cards))