If I am returning a lazy-seq from a function like this:
(letfn [(permutations [s]
(lazy-seq
(if (seq (rest s))
(apply concat (for [x s]
(map #(cons x %) (permutations (remove #{x} s)))))
[s])))])
If I use loop recur like below, will the list be eagerly evaluated?
(loop [perms (permutations chain)]
(if (empty? perms)
(prn "finised")
(recur (rest perms))))
If it is eagerly evaluated, can I use loop..recur to lazily loop over what is returned from the permutations function?
The list is lazily evaluated by your loop-recur code.
You can try it out yourself. Let's make permutations print something every time it returns a value by adding a println call.
(defn permutations [s]
(lazy-seq
(if (seq (rest s))
(apply concat (for [x s]
(map #(cons x %) (permutations (remove #{x} s)))))
(do
(println "returning a value")
[s]))))
When using loop, let's also print the values with we're looping over with (prn (first perms).
(loop [perms (permutations [1 2 3])]
(if (empty? perms)
(prn "finised")
(do
(prn (first perms))
(recur (rest perms)))))
Here's what it prints:
returning a value
(1 2 3)
returning a value
(1 3 2)
returning a value
(2 1 3)
returning a value
(2 3 1)
returning a value
(3 1 2)
returning a value
(3 2 1)
"finished"
As you can see, "returning a value" and the value lines are interlaced. The evaluation of the lazy seq can be forced with doall. If you loop over (doall (permutations [1 2 3])), first it prints all the "returning a value" lines and only then the values.
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.
I am new to Clojure, and doing my best to forget all my previous experience with more procedural languages (java, ruby, swift) and embrace Clojure for what it is. I am actually really enjoying the way it makes me think differently -- however, I have come up against a pattern that I just can't seem to figure out. The easiest way to illustrate, is with some code:
(defn char-to-int [c] (Integer/valueOf (str c)))
(defn digits-dont-decrease? [str]
(let [digits (map char-to-int (seq str)) i 0]
(when (< i 5)
(if (> (nth digits i) (nth digits (+ i 1)))
false
(recur (inc i))))))
(def result (digits-dont-decrease? "112233"))
(if (= true result)
(println "fit rules")
(println "doesn't fit rules"))
The input is a 6 digit number as a string, and I am simply attempting to make sure that each digit from left to right is >= the previous digit. I want to return false if it doesn't, and true if it does. The false situation works great -- however, given that recur needs to be the last thing in the function (as far as I can tell), how do I return true. As it is, when the condition is satisfied, I get an illegal argument exception:
Execution error (IllegalArgumentException) at clojure.exercise.two/digits-dont-decrease? (four:20).
Don't know how to create ISeq from: java.lang.Long
How should I be thinking about this? I assume my past training is getting in my mental way.
This is not answering your question, but also shows an alternative. While the (apply < ...) approach over the whole string is very elegant for small strings (it is eager), you can use every? for an short-circuiting approach. E.g.:
user=> (defn nr-seq [s] (map #(Integer/parseInt (str %)) s))
#'user/nr-seq
user=> (every? (partial apply <=) (partition 2 1 (nr-seq "123")))
true
You need nothing but
(apply <= "112233")
Reason: string is a sequence of character and comparison operator works on character.
(->> "0123456789" (mapcat #(repeat 1000 %)) (apply str) (def loooong))
(count loooong)
10000
(time (apply <= loooong))
"Elapsed time: 21.006625 msecs"
true
(->> "9123456789" (mapcat #(repeat 1000 %)) (apply str) (def bad-loooong))
(count bad-loooong)
10000
(time (apply <= bad-loooong))
"Elapsed time: 2.581750 msecs"
false
(above runs on my iPhone)
In this case, you don't really need loop/recur. Just use the built-in nature of <= like so:
(ns tst.demo.core
(:use demo.core tupelo.core tupelo.test))
(def true-samples
["123"
"112233"
"13"])
(def false-samples
["10"
"12324"])
(defn char->int
[char-or-str]
(let [str-val (str char-or-str)] ; coerce any chars to len-1 strings
(assert (= 1 (count str-val)))
(Integer/parseInt str-val)))
(dotest
(is= 5 (char->int "5"))
(is= 5 (char->int \5))
(is= [1 2 3] (mapv char->int "123"))
; this shows what we are going for
(is (<= 1 1 2 2 3 3))
(isnt (<= 1 1 2 1 3 3))
and now test the char sequences:
;-----------------------------------------------------------------------------
; using built-in `<=` function
(doseq [true-samp true-samples]
(let [digit-vals (mapv char->int true-samp)]
(is (apply <= digit-vals))))
(doseq [false-samp false-samples]
(let [digit-vals (mapv char->int false-samp)]
(isnt (apply <= digit-vals))))
if you want to write your own, you can like so:
(defn increasing-equal-seq?
"Returns true iff sequence is non-decreasing"
[coll]
(when (< (count coll) 2)
(throw (ex-info "coll must have at least 2 vals" {:coll coll})))
(loop [prev (first coll)
remaining (rest coll)]
(if (empty? remaining)
true
(let [curr (first remaining)
prev-next curr
remaining-next (rest remaining)]
(if (<= prev curr)
(recur prev-next remaining-next)
false)))))
;-----------------------------------------------------------------------------
; using home-grown loop/recur
(doseq [true-samp true-samples]
(let [digit-vals (mapv char->int true-samp)]
(is (increasing-equal-seq? digit-vals))))
(doseq [false-samp false-samples]
(let [digit-vals (mapv char->int false-samp)]
(isnt (increasing-equal-seq? digit-vals))))
)
with result
-------------------------------
Clojure 1.10.1 Java 13
-------------------------------
Testing tst.demo.core
Ran 2 tests containing 15 assertions.
0 failures, 0 errors.
Passed all tests
Finished at 23:36:17.096 (run time: 0.028s)
You an use loop with recur.
Assuming you require following input v/s output -
"543221" => false
"54321" => false
"12345" => true
"123345" => true
Following function can help
;; Assuming char-to-int is defined by you before as per the question
(defn digits-dont-decrease?
[strng]
(let [digits (map char-to-int (seq strng))]
(loop [;;the bindings in loop act as initial state
decreases true
i (- (count digits) 2)]
(let [decreases (and decreases (>= (nth digits (+ i 1)) (nth digits i)))]
(if (or (< i 1) (not decreases))
decreases
(recur decreases (dec i)))))))
This should work for numeric string of any length.
Hope this helps. Please let me know if you were looking for something else :).
(defn non-decreasing? [str]
(every?
identity
(map
(fn [a b]
(<= (int a) (int b)))
(seq str)
(rest str))))
(defn non-decreasing-loop? [str]
(loop [a (seq str) b (rest str)]
(if-not (seq b)
true
(if (<= (int (first a)) (int (first b)))
(recur (rest a) (rest b))
false))))
(non-decreasing? "112334589")
(non-decreasing? "112324589")
(non-decreasing-loop? "112334589")
(non-decreasing-loop? "112324589")
I have completed this problem on hackerrank and my solution passes most test cases but it is not fast enough for 4 out of the 11 test cases.
My solution looks like this:
(ns scratch.core
(require [clojure.string :as str :only (split-lines join split)]))
(defn ascii [char]
(int (.charAt (str char) 0)))
(defn process [text]
(let [parts (split-at (int (Math/floor (/ (count text) 2))) text)
left (first parts)
right (if (> (count (last parts)) (count (first parts)))
(rest (last parts))
(last parts))]
(reduce (fn [acc i]
(let [a (ascii (nth left i))
b (ascii (nth (reverse right) i))]
(if (> a b)
(+ acc (- a b))
(+ acc (- b a))))
) 0 (range (count left)))))
(defn print-result [[x & xs]]
(prn x)
(if (seq xs)
(recur xs)))
(let [input (slurp "/Users/paulcowan/Downloads/input10.txt")
inputs (str/split-lines input)
length (read-string (first inputs))
texts (rest inputs)]
(time (print-result (map process texts))))
Can anyone give me any advice about what I should look at to make this faster?
Would using recursion instead of reduce be faster or maybe this line is expensive:
right (if (> (count (last parts)) (count (first parts)))
(rest (last parts))
(last parts))
Because I am getting a count twice.
You are redundantly calling reverse on every iteration of the reduce:
user=> (let [c [1 2 3]
noisey-reverse #(doto (reverse %) println)]
(reduce (fn [acc e] (conj acc (noisey-reverse c) e))
[]
[:a :b :c]))
(3 2 1)
(3 2 1)
(3 2 1)
[(3 2 1) :a (3 2 1) :b (3 2 1) :c]
The reversed value could be calculated inside the containing let, and would then only need to be calculated once.
Also, due to the way your parts is defined, you are doing linear time lookups with each call to nth. It would be better to put parts in a vector and do indexed lookup. In fact you wouldn't need a reversed parts, and could do arithmetic based on the count of the vector to find the item to look up.
I have a function that produces lazy-sequences called a-function.
If I run the code:
(map a-function a-sequence-of-values)
it returns a lazy sequence as expected.
But when I run the code:
(mapcat a-function a-sequence-of-values)
it breaks the lazyness of my function. In fact it turns that code into
(apply concat (map a-function a-sequence-of-values))
So it needs to realize all the values from the map before concatenating those values.
What I need is a function that concatenates the result of a map function on demand without realizing all the map beforehand.
I can hack a function for this:
(defn my-mapcat
[f coll]
(lazy-seq
(if (not-empty coll)
(concat
(f (first coll))
(my-mapcat f (rest coll))))))
But I can't believe that clojure doesn't have something already done. Do you know if clojure has such feature? Only a few people and I have the same problem?
I also found a blog that deals with the same issue: http://clojurian.blogspot.com.br/2012/11/beware-of-mapcat.html
Lazy-sequence production and consumption is different than lazy evaluation.
Clojure functions do strict/eager evaluation of their arguments. Evaluation of an argument that is or that yields a lazy sequence does not force realization of the yielded lazy sequence in and of itself. However, any side effects caused by evaluation of the argument will occur.
The ordinary use case for mapcat is to concatenate sequences yielded without side effects. Therefore, it hardly matters that some of the arguments are eagerly evaluated because no side effects are expected.
Your function my-mapcat imposes additional laziness on the evaluation of its arguments by wrapping them in thunks (other lazy-seqs). This can be useful when significant side effects - IO, significant memory consumption, state updates - are expected. However, the warning bells should probably be going off in your head if your function is doing side effects and producing a sequence to be concatenated that your code probably needs refactoring.
Here is similar from algo.monads
(defn- flatten*
"Like #(apply concat %), but fully lazy: it evaluates each sublist
only when it is needed."
[ss]
(lazy-seq
(when-let [s (seq ss)]
(concat (first s) (flatten* (rest s))))))
Another way to write my-mapcat:
(defn my-mapcat [f coll] (for [x coll, fx (f x)] fx))
Applying a function to a lazy sequence will force realization of a portion of that lazy sequence necessary to satisfy the arguments of the function. If that function itself produces lazy sequences as a result, those are not realized as a matter of course.
Consider this function to count the realized portion of a sequence
(defn count-realized [s]
(loop [s s, n 0]
(if (instance? clojure.lang.IPending s)
(if (and (realized? s) (seq s))
(recur (rest s) (inc n))
n)
(if (seq s)
(recur (rest s) (inc n))
n))))
Now let's see what's being realized
(let [seq-of-seqs (map range (list 1 2 3 4 5 6))
concat-seq (apply concat seq-of-seqs)]
(println "seq-of-seqs: " (count-realized seq-of-seqs))
(println "concat-seq: " (count-realized concat-seq))
(println "seqs-in-seq: " (mapv count-realized seq-of-seqs)))
;=> seq-of-seqs: 4
; concat-seq: 0
; seqs-in-seq: [0 0 0 0 0 0]
So, 4 elements of the seq-of-seqs got realized, but none of its component sequences were realized nor was there any realization in the concatenated sequence.
Why 4? Because the applicable arity overloaded version of concat takes 4 arguments [x y & xs] (count the &).
Compare to
(let [seq-of-seqs (map range (list 1 2 3 4 5 6))
foo-seq (apply (fn foo [& more] more) seq-of-seqs)]
(println "seq-of-seqs: " (count-realized seq-of-seqs))
(println "seqs-in-seq: " (mapv count-realized seq-of-seqs)))
;=> seq-of-seqs: 2
; seqs-in-seq: [0 0 0 0 0 0]
(let [seq-of-seqs (map range (list 1 2 3 4 5 6))
foo-seq (apply (fn foo [a b c & more] more) seq-of-seqs)]
(println "seq-of-seqs: " (count-realized seq-of-seqs))
(println "seqs-in-seq: " (mapv count-realized seq-of-seqs)))
;=> seq-of-seqs: 5
; seqs-in-seq: [0 0 0 0 0 0]
Clojure has two solutions to making the evaluation of arguments lazy.
One is macros. Unlike functions, macros do not evaluate their arguments.
Here's a function with a side effect
(defn f [n] (println "foo!") (repeat n n))
Side effects are produced even though the sequence is not realized
user=> (def x (concat (f 1) (f 2)))
foo!
foo!
#'user/x
user=> (count-realized x)
0
Clojure has a lazy-cat macro to prevent this
user=> (def y (lazy-cat (f 1) (f 2)))
#'user/y
user=> (count-realized y)
0
user=> (dorun y)
foo!
foo!
nil
user=> (count-realized y)
3
user=> y
(1 2 2)
Unfortunately, you cannot apply a macro.
The other solution to delay evaluation is wrap in thunks, which is exactly what you've done.
Your premise is wrong. Concat is lazy, apply is lazy if its first argument is, and mapcat is lazy.
user> (class (mapcat (fn [x y] (println x y) (list x y)) (range) (range)))
0 0
1 1
2 2
3 3
clojure.lang.LazySeq
note that some of the initial values are evaluated (more on this below), but clearly the whole thing is still lazy (or the call would never have returned, (range) returns an endless sequence, and will not return when used eagerly).
The blog you link to is about the danger of recursively using mapcat on a lazy tree, because it is eager on the first few elements (which can add up in a recursive application).
I'm reading Programming Clojure now and I find out next example
(defn by-pairs [coll]
(let
[take-pair (fn [c] (when (next c) (take 2 c)))]
(lazy-seq
(when-let [pair (seq (take-pair coll))] ;seq calls here
(cons pair (by-pairs (rest coll)))))))
it breaks list into pairs, like
(println (by-pairs [1 2 1])) ((1 2) (2 1))
(println (by-pairs [1 2 1 3])) ((1 2) (2 1) (1 3))
(println (by-pairs [])) ()
(println (by-pairs [1])) ()
What I can not get is why we should invoke seq on take-pair result? So why we can not just write
(defn by-pairs [coll]
(let
[take-pair (fn [c] (when (next c) (take 2 c)))]
(lazy-seq
(when-let [pair (take-pair coll)]
(cons pair (by-pairs (rest coll)))))))
In witch cases there are will be different results or are there are any performance reasons?
Both the code are same and there will be no difference because next and take functions that are being applied to coll in take-pair function, do call seq on the passed parameter i.e next c will first call seq on c or try to check if it is an object which implements ISeq and same in being doing by the take function. So basically in this case if you don't call seq yourself, the next and take will call seq on it.