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I'm trying to parse a list of lines in Clojure and have the processing pause and start to accumulate if pred(line) is true and then stop accumulating when pred2(laterline) becomes true.
For example, pred will return True if value is b and then pred2 is used which returns True when the value is d
[a b c d e f] -> [a [b c d] e f]
I feel like the answer must be much simpler and idiomatic than the crazy things I'm coming up with.
(require '[flatland.useful.seq :refer [partition-between]])
(letfn [(pred [x] (= x 'b))
(pred2 [x] (= x 'd))]
(let [A '[a b c d e f g h b d x]]
(->> A
(partition-between (fn [[a b]]
(or (pred b)
(pred2 a))))
(reduce
(fn [acc v]
(if (pred (first v))
(conj acc v)
(vec (concat acc v))))
[]))))
produces:
=> [a [b c d] e f g h [b d] x]
Related
So I am trying to solve this problem, and this is the code I have come up with:
First I have a pack function, receives a list and groups same elements into a vector.
(defn pack [lst]
(def a [])
(def vect [])
(cond
(empty? lst)
lst
:else
(loop [i 0]
(def r (get lst i))
(def t (get lst (+ i 1)))
(if (= r t)
(def vect (conj vect r))
)
(if (not= r t)
(and (def vect (conj vect r)) (and (def a (conj a vect)) (def vect [])))
)
(if (= i (- (count lst) 1))
a
(recur (inc i))
)
))
)
for example if I have this vector:
(def tes '[a a a a b c c a a d e e e e])
pack function will return this:
[[a a a a] [b] [c c] [a a] [d] [e e e e]]
Then I tried doing the "encode" part of the problem with this code:
(def v1 [])
(def v2 [])
(conj v2 (conj v1 (count (get (pack tes) 0)) (get (get (pack tes) 0) 0)))
And it returned what I wanted, a vector "v2" with a vector "v1" that has the "encoded" item.
[[4 a]]
So now I try to make the function:
(defn encode [lst]
(loop [index 0 limit (count (pack lst)) v1 [] v2[]]
(if (= index limit)
lst
(conj v2 (conj v1 (count (get (pack tes) index)) (get (get (pack tes) index) index)))
)
(recur (inc index) limit v1 v2)
)
)
(encode tes)
but I get this error:
2021/03/07 00:00:21 got exception from server /usr/local/bin/lein: line 152:
28 Killed "$LEIN_JAVA_CMD" "${BOOTCLASSPATH[#]}" -Dfile.encoding=UTF-8 -Dmaven.wagon.http.ssl.easy=false -Dmaven.wagon.rto=10000 $LEIN_JVM_OPTS
-Dleiningen.original.pwd="$ORIGINAL_PWD" -Dleiningen.script="$0" -classpath "$CLASSPATH" clojure.main -m leiningen.core.main "$#"
2021/03/07 01:42:20 error reading from server EOF
Any way to fix my code or to solve the problem more efficiently but still return a vector?
juxt can be used in the pack function:
(defn pack [xs]
(map (juxt count first) (partition-by identity xs)))
(defn unpack [xs]
(mapcat #(apply repeat %) xs))
Don't use def inside function, because it creates global
variable. Use let instead.
Don't use multiple if in row, there is cond.
Format your code better- for example, put all parentheses on the end together on one line.
Here is more efficient solution:
(defn pack [lst]
(letfn [(pack-help [lst]
(if (empty? lst) '()
(let [elem (first lst)]
(cons (vec (take-while #(= % elem) lst))
(pack-help (drop-while #(= % elem) lst))))))]
(vec (pack-help lst))))
(defn pack-with-count [lst]
(mapv #(vector (count %) (first %))
(pack lst)))
(defn unpack [packed-lst]
(into [] (apply concat packed-lst)))
(pack '[a a a a b c c a a d e e e e])
(pack-with-count '[a a a a b c c a a d e e e e])
(unpack '[[a a a a] [b] [c c] [a a] [d] [e e e e]])
As a rule, whenever you reach for loop/recur, there are some pieces of the standard library which will allow you to get the desired effect using higher-order functions. You avoid needing to implement the wiring and can just concentrate on your intent.
(def tes '[a a a a b c c a a d e e e e])
(partition-by identity tes)
; => ((a a a a) (b) (c c) (a a) (d) (e e e e))
(map (juxt count first) *1)
; => ([4 a] [1 b] [2 c] [2 a] [1 d] [4 e])
(mapcat #(apply repeat %) *1)
; => (a a a a b c c a a d e e e e)
Here *1 is just the REPL shorthand for "previous result" - if you need to compose these into functions, this will be replaced with your argument.
If you really need vectors rather than sequences for the outer collection at each stage, you can wrap with vec (to convert the lazy sequence to a vector), or use mapv instead of map.
Finally - the error message you are getting from lein is a syntax error rather than a logic or code problem. Clojure generally flags an unexpected EOF if there aren't enough closing parens.
(println "because we left them open like this -"
Consider working inside a REPL within an IDE, or if that isn't possible then using a text editor that matches parens for you.
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We have to a write code in "Clojure’s" using case extension -
(defn calculator [x a b]
(case x "+" (println (+ a b))
"-" (println (- a b))
"*" (println (* a b))
"/" (println (/ a b)))
)
)
(def x (read-line))
(def a (Integer/parseInt (clojure.string/trim (read-line))))
(def b (Integer/parseInt (clojure.string/trim (read-line))))
(calculator x a b)
Please help us in above code i.e. where we went wrong.
I think you have one extra paren on calculator function.
(defn calculator [x a b]
(case x "+" (println (+ a b))
"-" (println (- a b))
"*" (println (* a b))
"/" (println (/ a b))))
(def x (read-line))
(def a (Integer/parseInt (clojure.string/trim (read-line))))
(def b (Integer/parseInt (clojure.string/trim (read-line))))
(calculator x a b)
The following expression
(reduce (fn [[c x y] [s k d]] (if (< c s) [s k d] [c x y])) [0 0 0] colls)
Is for the element [e, x, y] in colls such that e is the maximum among all in the tuples in colls.
Is there already an idiomatic expression in Clojure for that? I guess that with macro, it should be possible to express it with brevity.
I feel that Python/numpy/panda has some good example.
You can use max-key:
(apply max-key first colls)
How could I convert this:
[a b c d e]
or this:
(e d c b a) ;(rseq [a b c d e])
to this:
[a[b[c[d[e]]]]]
I've been wracking my brain and I feel like there is a simple solution! :-\
Ultimately I want to do this:
[a b c d e]
[a b c x y]
[a b c d j k]
as this:
{a {b {c {d {e}
{j {k}}
{x {y}}}}
Which I think conj will help with
(Update: added answer to the new question added in the edit below the answer to the original question.)
I've actually answered this very question in #clojure recently.
Here are two approaches: f is pretty much the spec directly transformed into code, which however creates a seq -- (next xs) -- which immediately gets poured into a new vector at each step; g is a much better version which only allocates objects which will actually occur in the output, plus a vector and the seq links to traverse it:
;; [1 2 3] -> [1 [2 [3]]]
;; naive, quadratic:
(defn f [xs]
(if (next xs)
[(first xs) (vec (f (next xs)))]
(vec xs)))
;; only allocates output + 1 vector + a linear number of seq links,
;; linear overall:
(defn g [v]
(reduce (fn [acc x]
[x acc])
[(peek v)]
(rseq (pop v))))
NB. I'm overlooking the usual logarithmic factors arising from vector operations (so this is soft-O complexity).
As for producing a nested map, the above isn't particularly useful. Here's one approach:
(defn h
([v]
(h nil v))
([m v]
(assoc-in m v nil)))
(h [1 2 3 4])
;= {1 {2 {3 {4 nil}}}}
(def data
'[[a b c d e]
[a b c x y]
[a b c d j k]])
(reduce h {} data)
;= {a {b {c {x {y nil}, d {j {k nil}, e nil}}}}}
I'm using nil as a "terminator", since {y} (as currently found in the answer text) is not a well-formed literal. true might be a more convenient choice if you plan to call these maps as functions to check for presence of keys.
Simpler solution here (using destructuring and non-tail recursion):
http://ideone.com/qchXZC
(defn wrap
([[a & as]]
(if-let [[b & cs] as]
[a (wrap as)]
[a])))
I've started learning core.logic and I'm totally lost. I am trying to write a core.logic relation which refactors an expression, renaming symbols. I want a relation that returns for a given expression, list of symbols and a list of symbols to rename those symbols:
(defn rename [exp from to]...
the expression with all the symbols in from becoming the corresponding one in to:
e.g. (rename '(defn multiply [x y] (* x y)) [x y] [a b])
returns (defn multiply [a b] (* a b))
but it needs to be aware of scope,
so (rename '(defn q [x] ((fn [x] (* x 5)) x)) [x] [a])
would return (defn q [a] ((fn [x] (* x 5)) a))
I don't know where to start solving this - any hints would be greatly appreciated!
This problem is more suitable for FP as it is just a tree traversal and replace operation, where as LP is more about specifying constrains and asking all possible solution around those constrains for a specific input. But if you really want to do this logical way, I tried something that does try to do it LP way but it doesn't handle a lot of cases and is just a starting point.
(defrel replace-with a b)
(fact replace-with 'x 'a)
(fact replace-with 'y 'b)
(defn replace [a b]
(conde
[(replace-with a b)]
[(== a b)]))
(defn replace-list [from to]
(conde
[(== from []) (== to [])]
[(fresh [f t f-rest t-rest]
(resto from f-rest)
(resto to t-rest)
(firsto from f) (firsto to t)
(conda [(replace-list f t)]
[(replace f t)])
(replace-list f-rest t-rest))]))
(first (run 1 [q]
(fresh [from]
(== from '(defn multiply [x y] (* x y)))
(replace-list from q))))
==> (defn multiply (a b) (* a b))