hi guys how to collect data from if condition in a loop
(defn update-f [a b]
(let[data(loop [i 0]
(when (< i (count a))
;;(println i)
(if (every? (set b) (nth (for [x a] (set (list x))) i))
(nth (for [x a] (conj x "new")) i)
(nth (for [x a] (conj x "old")) i))
(recur (inc i))))])); loop i will take this value
i can print data from if condition i need take data from this function
You can collect data in a loop by adding an extra "accumulator" parameter:
(loop [i 0 acc []]
(if (< i 5)
(recur (inc i) (if (even? i)
(conj acc i)
acc))
acc))
However, it seems that what you are trying to achieve could be written much more concise (if I got what you want right):
(defn are-new? [a b]
(map (fn[i] [i (if (b i) :new :old)]) a))
We are mapping every item in a to add a :new or :old tag to it. A set also acts as a function, that returns a truthy value when the argument exists in the set. So to find if an item i is already in set b, we can simply do (b i).
And you can test it:
(let [r (are-new? [[1 2 3] [2 3 4] [1 2]] #{[1 2 3] [1 2]})]
(doseq [i r]
(println i)))
> [[1 2 3] :new]
> [[2 3 4] :old]
> [[1 2] :new]
Related
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)
I've pasted the code on this page in an IDE and it works. The problem is that when I replace the definition of target-data with this vector of pairs* it gives me this error**.
(vector [[1 2]
[2 3]
[3 4]
[4 5]] ) ; *
UnsupportedOperationException nth not supported on this type: core$vector clojure.lang.RT.nthFrom (RT.java:857) **
What should I do to use my own target-data?
UPDATED FULL CODE:
(ns evolvefn.core)
;(def target-data
; (map #(vector % (+ (* % %) % 1))
; (range -1.0 1.0 0.1)))
;; We'll use input (x) values ranging from -1.0 to 1.0 in increments
;; of 0.1, and we'll generate the target [x y] pairs algorithmically.
;; If you want to evolve a function to fit your own data then you could
;; just paste a vector of pairs into the definition of target-data instead.
(def target-data
(vec[1 2]
[2 3]
[3 4]
[4 5]))
;; An individual will be an expression made of functions +, -, *, and
;; pd (protected division), along with terminals x and randomly chosen
;; constants between -5.0 and 5.0. Note that for this problem the
;; presence of the constants actually makes it much harder, but that
;; may not be the case for other problems.
(defn random-function
[]
(rand-nth '(+ - * pd)))
(defn random-terminal
[]
(rand-nth (list 'x (- (rand 10) 5))))
(defn random-code
[depth]
(if (or (zero? depth)
(zero? (rand-int 2)))
(random-terminal)
(list (random-function)
(random-code (dec depth))
(random-code (dec depth)))))
;; And we have to define pd (protected division):
(defn pd
"Protected division; returns 0 if the denominator is zero."
[num denom]
(if (zero? denom)
0
(/ num denom)))
;; We can now evaluate the error of an individual by creating a function
;; built around the individual, calling it on all of the x values, and
;; adding up all of the differences between the results and the
;; corresponding y values.
(defn error
[individual]
(let [value-function (eval (list 'fn '[x] individual))]
(reduce + (map (fn [[x y]]
(Math/abs
(- (value-function x) y)))
target-data))))
;; We can now generate and evaluate random small programs, as with:
;; (let [i (random-code 3)] (println (error i) "from individual" i))
;; To help write mutation and crossover functions we'll write a utility
;; function that injects something into an expression and another that
;; extracts something from an expression.
(defn codesize [c]
(if (seq? c)
(count (flatten c))
1))
(defn inject
"Returns a copy of individual i with new inserted randomly somwhere within it (replacing something else)."
[new i]
(if (seq? i)
(if (zero? (rand-int (count (flatten i))))
new
(if (< (rand)
(/ (codesize (nth i 1))
(- (codesize i) 1)))
(list (nth i 0) (inject new (nth i 1)) (nth i 2))
(list (nth i 0) (nth i 1) (inject new (nth i 2)))))
new))
(defn extract
"Returns a random subexpression of individual i."
[i]
(if (seq? i)
(if (zero? (rand-int (count (flatten i))))
i
(if (< (rand) (/ (codesize (nth i 1))
(- (codesize i)) 1))
(extract (nth i 1))
(extract (nth i 2))))
i))
;; Now the mutate and crossover functions are easy to write:
(defn mutate
[i]
(inject (random-code 2) i))
(defn crossover
[i j]
(inject (extract j) i))
;; We can see some mutations with:
;; (let [i (random-code 2)] (println (mutate i) "from individual" i))
;; and crossovers with:
;; (let [i (random-code 2) j (random-code 2)]
;; (println (crossover i j) "from" i "and" j))
;; We'll also want a way to sort a populaty by error that doesn't require
;; lots of error re-computation:
(defn sort-by-error
[population]
(vec (map second
(sort (fn [[err1 ind1] [err2 ind2]] (< err1 err2))
(map #(vector (error %) %) population)))))
;; Finally, we'll define a function to select an individual from a sorted
;; population using tournaments of a given size.
(defn select
[population tournament-size]
(let [size (count population)]
(nth population
(apply min (repeatedly tournament-size #(rand-int size))))))
;; Now we can evolve a solution by starting with a random population and
;; repeatedly sorting, checking for a solution, and producing a new
;; population.
(defn evolve
[popsize]
(println "Starting evolution...")
(loop [generation 0
population (sort-by-error (repeatedly popsize #(random-code 2)))]
(let [best (first population)
best-error (error best)]
(println "======================")
(println "Generation:" generation)
(println "Best error:" best-error)
(println "Best program:" best)
(println " Median error:" (error (nth population
(int (/ popsize 2)))))
(println " Average program size:"
(float (/ (reduce + (map count (map flatten population)))
(count population))))
(if (< best-error 0.1) ;; good enough to count as success
(println "Success:" best)
(recur
(inc generation)
(sort-by-error
(concat
(repeatedly (* 1/2 popsize) #(mutate (select population 7)))
(repeatedly (* 1/4 popsize) #(crossover (select population 7)
(select population 7)))
(repeatedly (* 1/4 popsize) #(select population 7)))))))))
;; Run it with a population of 1000:
(evolve 1000)
And the error is:
(evolve 1000)
Starting evolution...
IllegalArgumentException No matching method found: abs clojure.lang.Reflector.invokeMatchingMethod (Reflector.java:80)
evolvefn.core=>
I have the following snippet:
(defn explode [e]
(seq [e e e e]))
(defn f [coll]
(when-first [e coll]
(cons e
(lazy-seq (f (lazy-cat (next coll)
(explode e)))))))
When I try to access an element, I get a StackOverflow error:
user=> (nth (f (seq [1 2 3])) 1000)
3
user=> (nth (f (seq [1 2 3])) 10000)
StackOverflowError clojure.core/concat/fn--3923 (core.clj:678)
How can I structure this code in a way that doesn't blow the stack?
You'll have to keep track of the remaining work explicitly, perhaps like so:
(defn f [coll]
(letfn [(go [xs q]
(lazy-seq
(cond
(seq xs)
(cons (first xs)
(go (next xs) (conj q (explode (first xs)))))
(seq q)
(go (peek q) (pop q)))))]
(go coll clojure.lang.PersistentQueue/EMPTY)))
From the REPL:
(nth (f [1 2 3]) 1000)
;= 3
(nth (f [1 2 3]) 10000)
;= 2
;; f-orig is f as found in the question text
(= (take 1000 (f-orig [1 2 3])) (take 1000 (f [1 2 3])))
;= true
I need to take some amount of elements from a sequence based on some quantity rule. Here is a solution I came up with:
(defn take-while-not-enough
[p len xs]
(loop [ac 0
r []
s xs]
(if (empty? s)
r
(let [new-ac (p ac (first s))]
(if (>= new-ac len)
r
(recur new-ac (conj r (first s)) (rest s)))))))
(take-while-not-enough + 10 [2 5 7 8 2 1]) ; [2 5]
(take-while-not-enough #(+ %1 (%2 1)) 7 [[2 5] [7 8] [2 1]]) ; [[2 5]]
Is there any better way to achieve the same?
Thanks.
UPDATE:
Somebody posted that solution, but then removed it. It does the same is the answer that I accepted, but is more readable. Thank you, anonymous well-wisher!
(defn take-while-not-enough [reducer-fn limit data]
(->> (reductions reducer-fn 0 data) ; 1. the sequence of accumulated values
(map vector data) ; 2. paired with the original sequence
(take-while #(< (second %) limit)) ; 3. until a certain accumulated value
(map first))) ; 4. then extract the original values
My first thought is to view this problem as a variation on reduce and thus to break the problem into two steps:
count the number of items in the result
take that many from the input
I also took some liberties with the argument names:
user> (defn take-while-not-enough [reducer-fn limit data]
(take (dec (count (take-while #(< % limit) (reductions reducer-fn 0 data))))
data))
#'user/take-while-not-enough
user> (take-while-not-enough #(+ %1 (%2 1)) 7 [[2 5] [7 8] [2 1]])
([2 5])
user> (take-while-not-enough + 10 [2 5 7 8 2 1])
(2 5)
This returns a sequence and your examples return a vector, if this is important then you can add a call to vec
Something that would traverse the input sequence only once:
(defn take-while-not-enough [r v data]
(->> (rest (reductions (fn [s i] [(r (s 0) i) i]) [0 []] data))
(take-while (comp #(< % v) first))
(map second)))
Well, if you want to use flatland/useful, this is a kinda-okay way to use glue:
(defn take-while-not-enough [p len xs]
(first (glue conj []
(constantly true)
#(>= (reduce p 0 %) len)
xs)))
But it's rebuilding the accumulator for the entire "processed so far" chunk every time it decides whether to grow the chunk more, so it's O(n^2), which will be unacceptable for larger inputs.
The most obvious improvement to your implementation is to make it lazy instead of tail-recursive:
(defn take-while-not-enough [p len xs]
((fn step [acc coll]
(lazy-seq
(when-let [xs (seq coll)]
(let [x (first xs)
acc (p acc x)]
(when-not (>= acc len)
(cons x (step acc xs)))))))
0 xs))
Sometimes lazy-seq is straight-forward and self-explaining.
(defn take-while-not-enough
([f limit coll] (take-while-not-enough f limit (f) coll))
([f limit acc coll]
(lazy-seq
(when-let [s (seq coll)]
(let [fst (first s)
nacc (f acc fst)]
(when (< nxt-sd limit)
(cons fst (take-while-not-enough f limit nacc (rest s)))))))))
Note: f is expected to follow the rules of reduce.
I have a sequence s and a list of indexes into this sequence indexes. How do I retain only the items given via the indexes?
Simple example:
(filter-by-index '(a b c d e f g) '(0 2 3 4)) ; => (a c d e)
My usecase:
(filter-by-index '(c c# d d# e f f# g g# a a# b) '(0 2 4 5 7 9 11)) ; => (c d e f g a b)
You can use keep-indexed:
(defn filter-by-index [coll idxs]
(keep-indexed #(when ((set idxs) %1) %2)
coll))
Another version using explicit recur and lazy-seq:
(defn filter-by-index [coll idxs]
(lazy-seq
(when-let [idx (first idxs)]
(if (zero? idx)
(cons (first coll)
(filter-by-index (rest coll) (rest (map dec idxs))))
(filter-by-index (drop idx coll)
(map #(- % idx) idxs))))))
make a list of vectors containing the items combined with the indexes,
(def with-indexes (map #(vector %1 %2 ) ['a 'b 'c 'd 'e 'f] (range)))
#'clojure.core/with-indexes
with-indexes
([a 0] [b 1] [c 2] [d 3] [e 4] [f 5])
filter this list
lojure.core=> (def filtered (filter #(#{1 3 5 7} (second % )) with-indexes))
#'clojure.core/filtered
clojure.core=> filtered
([b 1] [d 3] [f 5])
then remove the indexes.
clojure.core=> (map first filtered)
(b d f)
then we thread it together with the "thread last" macro
(defn filter-by-index [coll idxs]
(->> coll
(map #(vector %1 %2)(range))
(filter #(idxs (first %)))
(map second)))
clojure.core=> (filter-by-index ['a 'b 'c 'd 'e 'f 'g] #{2 3 1 6})
(b c d g)
The moral of the story is, break it into small independent parts, test them, then compose them into a working function.
The easiest solution is to use map:
(defn filter-by-index [coll idx]
(map (partial nth coll) idx))
I like Jonas's answer, but neither version will work well for an infinite sequence of indices: the first tries to create an infinite set, and the latter runs into a stack overflow by layering too many unrealized lazy sequences on top of each other. To avoid both problems you have to do slightly more manual work:
(defn filter-by-index [coll idxs]
((fn helper [coll idxs offset]
(lazy-seq
(when-let [idx (first idxs)]
(if (= idx offset)
(cons (first coll)
(helper (rest coll) (rest idxs) (inc offset)))
(helper (rest coll) idxs (inc offset))))))
coll idxs 0))
With this version, both coll and idxs can be infinite and you will still have no problems:
user> (nth (filter-by-index (range) (iterate #(+ 2 %) 0)) 1e6)
2000000
Edit: not trying to single out Jonas's answer: none of the other solutions work for infinite index sequences, which is why I felt a solution that does is needed.
I had a similar use case and came up with another easy solution. This one expects vectors.
I've changed the function name to match other similar clojure functions.
(defn select-indices [coll indices]
(reverse (vals (select-keys coll indices))))
(defn filter-by-index [seq idxs]
(let [idxs (into #{} idxs)]
(reduce (fn [h [char idx]]
(if (contains? idxs idx)
(conj h char) h))
[] (partition 2 (interleave seq (iterate inc 0))))))
(filter-by-index [\a \b \c \d \e \f \g] [0 2 3 4])
=>[\a \c \d \e]
=> (defn filter-by-index [src indexes]
(reduce (fn [a i] (conj a (nth src i))) [] indexes))
=> (filter-by-index '(a b c d e f g) '(0 2 3 4))
[a c d e]
I know this is not what was asked, but after reading these answers, I realized in my own personal use case, what I actually wanted was basically filtering by a mask.
So here was my take. Hopefully this will help someone else.
(defn filter-by-mask [coll mask]
(filter some? (map #(if %1 %2) mask coll)))
(defn make-errors-mask [coll]
(map #(nil? (:error %)) coll))
Usage
(let [v [{} {:error 3} {:ok 2} {:error 4 :yea 7}]
data ["one" "two" "three" "four"]
mask (make-errors-mask v)]
(filter-by-mask data mask))
; ==> ("one" "three")