I've written a probability function in Clojure that takes an optional hash-map of options:
(defn roll-lte
([n d] (/ n d))
([n d options]
(let [p (/ n d)
roll-type (:type options :normal)]
(cond
(= roll-type :advantage) (- (* p 2) (* p p))
(= roll-type :disadvantage) (* p p)
(= roll-type :normal) p
:else (throw (IllegalArgumentException. "Invalid roll type."))))))
This works as intended, but the idea is to write other functions that build off of this one -- for example:
(defn roll-gte
([n d] (roll-lte (- d n -1) d))
([n d options] (roll-lte (- d n -1) d options)))
The two arities in roll-lte make building off of the function awkward and repetitive, especially in cases like the above where options is simply being forwarded to roll-lte. Is there a more concise and less repetitive way to achieve this?
When I have functions with multiple arities, I usually try to have the lower-arity versions call the higher-arity versions with safe default arguments. The "main" implementation of the function usually ends up being the highest-arity body:
(defn roll-lte
([n d] (roll-lte n d nil))
([n d {:keys [type]
:or {type :normal}}]
(let [p (/ n d)]
(case type ;; used case instead of cond here
:advantage (- (* p 2) (* p p))
:disadvantage (* p p)
:normal p
(throw (IllegalArgumentException. "Invalid roll type."))))))
I also used :or in the options map destructuring above to set the default value for type, which allows the lower-arity functions to just pass a nil options map.
(defn roll-gte
([n d] (roll-gte n d nil))
([n d options] (roll-lte (- d n -1) d options)))
(roll-gte 3 4) ;=> 1/2
(roll-gte 3 4 {:type :advantage}) ;=> 3/4
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.
One of my favorite ways to test the power of a language I'm learning is to try and implement various fixed-point combinators. Since I'm learning Clojure (though I'm not new to lisps), I did the same for it.
First, a little "testable" code, factorial:
(def !'
"un-fixed factorial function"
(fn [f]
(fn [n]
(if (zero? n)
1
(* n (f (dec n)))))))
(defn !
"factorial"
[n]
(if (zero? n)
1
(apply * (map inc (range n)))))
For any combinator c I implement, I want to verify that ((c !') n) is equal to (! n).
We start with the traditional Y:
(defn Y
"pure lazy Y combinator => stack overflow"
[f]
(let [A (fn [x] (f (x x)))]
(A A)))
But of course Clojure is not nearly so lazy as that, so we pivot to Z:
(defn Z
"strict fixed-point combinator"
[f]
(let [A (fn [x] (f (fn [v] ((x x) v))))]
(A A)))
And indeed, it holds that (= ((Z !') n) (! n)).
Now comes my issue: I cannot get either of U or the Turing combinator (theta-v) to work correctly. I suspect with U it's a language limit, while with theta-v I'm more inclined to believe it's a misread of Wikipedia's notation:
(defn U
"the U combinator => broken???"
[f]
(f f))
(defn theta-v
"Turing fixed-point combinator by-value"
[f]
(let [A (fn [x] (fn [y] (y (fn [z] ((x x) y) z))))]
(A A)))
A sample REPL experience:
((U !') 5)
;=> Execution error (ClassCastException) at fix/!'$fn (fix.clj:55).
;=> fix$_BANG__SINGLEQUOTE_$fn__180 cannot be cast to java.lang.Number
((theta-v !') 5)
;=> Execution error (ClassCastException) at fix/theta-v$A$fn (fix.clj:36).
;=> java.lang.Long cannot be cast to clojure.lang.IFn
Can anyone explain
Why these implementations of U and theta-v are not working; and
How to fix them?
Your definition of theta-v is wrong for two reasons. The first is pretty obvious: you accept f as a parameter and then ignore it. A more faithful translation would be to use def style, as you have for your other functions:
(def theta-v
"Turing fixed-point combinator by-value"
(let [A (fn [x] (fn [y] (y (fn [z] ((x x) y) z))))]
(A A)))
The second reason is a bit sneakier: you translated λz.xxyz to (fn [z] ((x x) y) z), remembering that lisps need more parentheses to denote function calls that are implicit in lambda-calculus notation. However, you missed one set: just as x x y z would have meant "evaluate x twice, then y once, then finally return z", what you wrote means "evaluate ((x x) y), then throw away that result and return z". Adding the extra set of parentheses yields a working theta-v:
(def theta-v
"Turing fixed-point combinator by-value"
(let [A (fn [x] (fn [y] (y (fn [z] (((x x) y) z)))))]
(A A)))
and we can demonstrate that it works by calculating some factorials:
user> (map (theta-v !') (range 10))
(1 1 2 6 24 120 720 5040 40320 362880)
As for U: to use the U combinator, functions being combined must change how they self-call, meaning you would need to rewrite !' as well:
(def U [f] (f f))
(def ! (U (fn [f]
(fn [n]
(if (zero? n)
1
(* n ((f f) (dec n))))))))
Note that I have changed (f (dec n)) to ((f f) (dec n)).
I want to write the function collect which can combine the sublists into a list, like:
user=> (collect '(a b c d e) 1)
((a)(b)(c)(d)(e))
user=> (collect '(a b c d e) 2)
((a b)(c d)(e))
user=> (collect '(a b c d e) 5)
(a b c d e))
this is my code:
(defn collect [lst num]
(loop [l lst res (atom ())]
(if (<= (count l) num) #res
(recur (drop num l) (swap! res conj (take num (drop num l)))))))
But when I run
user=> (collect '(a b c d e) 1)
I got the error:
ClassCastException clojure.lang.PersistentList cannot be cast to clojure.lang.IAtom clojure.core/swap!
why I cannot get the res when I use "swap!" ? Thank you.
It's blowing up in the second pass through the loop.
swap returns the value that was put into the atom, not the atom it's self. So the first pass is updating the atom, and then passing the value it just put into the atom to the second pass through the loop. in the second pass it's trying to use the value as the atom, and getting the exception above.
To "fix" this use a do to update the atom, then pass the atom to the next pass through the loop once it contains the correct value.
user> (defn collect [lst num]
(loop [l lst res (atom ())]
(if (<= (count l) num) #res
(recur (drop num l)
(do (swap! res conj (take num (drop num l)))
res)))))
#'user/collect
user> (collect '(a b c d e) 2)
((e) (c d))
You can also in this case, just remove the atom completely and get exactly the same result (I fixed on ordering problem from your example by using a [] instead of () in the initial value of res):
user> (defn collect [lst num]
(loop [l lst res []]
(if (<= (count l) num) res
(recur (drop num l)
(conj res (take num (drop num l)))))))
#'user/collect
user> (collect '(a b c d e) 2)
[(c d) (e)]
and of course you can also use partition-all as glts mentions above.
;; this would be a correct way to do it
(defn collect [coll n]
(partition-all n coll))
;; this would be a clumsy way to do it
(defn collect
"using a loop (there is not point to do that but at least you can see the logic working as in your example)"
[coll n]
(lazy-seq
(loop [res []
coll coll]
(if (empty? coll)
res
(recur (conj res (take n coll)) (drop n coll))))))
Regarding your error, on the second loop, res is a list-like value, not an atom anymore. That would lead us to :
(defn collect [coll n]
(lazy-seq (loop [res (atom [])
coll coll]
(if (empty? coll)
#res
(recur (do (swap! res conj (take n coll))
;; return the atom instead of the value'
res)
(drop n coll))))))
Note that in order to preserve the order in the solution, I use a vector (litteral []) instead of a list (litteral '()). This is because of the behaviour of conj described here.
I am having a problem running my program in Clojure. I just start learning Clojure a couple of weeks ago. So I don't know the quick and easy way to debug a Clojure program. My func2 raises an exception at (adj(a b)) as followed:
ClassCastException java.lang.Long cannot be cast to clojure.lang.IFn
user/func2.
I don't know what is wrong with it. Can someone point out the problem with my coding?
And in func3, I call func2 recursively, but it throws:
ArityException Wrong number of args (0) passed to: PersistentVector
clojure.lan g.AFn.throwArity (AFn.java:437)
What is wrong with func3? Thank you.
(defn adj [value1 value2]
(def result (+ (/ value1 2) (/ value2 2)))
(if (= (mod result 2) 1)
(+ result 1)
result
)
)
(defn func2 [list]
(let [[a b c d] list]
(inc d)
([(adj c a) (adj a b) (adj b c) d]))
)
(defn func3 [list]
(loop [v list r []]
(if(= (v 0) (v 1) (v 2))
(conj list r)
(func3(func2(list)))
))
)
What's the intended result of these functions? We probably need to see some sample inputs and expected results to really be able to help you.
Here's my attempt at cleaning them up. I've noted the changes I made as comments. func3 has the most serious problem in that it's an infinite recursion - there's no end condition. What should cause it to stop working and return a result?
(defn adj [value1 value2]
;; don't use def within functions, use let
(let [result (+ (/ value1 2) (/ value2 2))]
(if (= (mod result 2) 1)
(+ result 1)
result)))
(defn func2 [list]
(let [[a b c d] list]
;; The extra parens around this vector were causing it
;; to be called as a function, which I don't think is
;; what you intended:
[(adj c a) (adj a b) (adj b c) d]))
;; This needs an end condition - it's an infinite recursion
(defn func3 [list]
(loop [v list r []]
(if (= (v 0) (v 1) (v 2))
(conj list r)
;; Removed extra parens around list
(func3 (func2 list)))))
The reason I say not to use def within functions is that it always creates a global function. For local bindings you want let.
Regarding the extra parens, the difference between [1 2 3] and ([1 2 3]) is that the former returns a vector containing the numbers 1, 2, and 3, whereas the latter tries to call that vector as a function. You had excess parens around the literal vector in func2 and around list in func3, which was causing exceptions.
As a style note, the name list isn't a good choice. For one thing, it's shadowing clojure.core/list, and for another you're probably using vectors rather than lists anyway. It would be more idiomatic to use coll (for collection) or s (for sequence) as the name.
This would suggest at least one other change. In func3 you use a vector-only feature (using the vector as a function to perform lookup by index), so to be more general (accept other data structures) you can convert to a vector with vec:
(defn func3 [coll]
(loop [v (vec coll) r []]
(if (= (v 0) (v 1) (v 2))
(conj v r)
(func3 (func2 v)))))
Oh, there is no need to debug that. I suggest you have a look at LightTable.
The first two functions are easily fixed:
(defn adj [value1 value2]
;(def result (+ (/ value1 2) (/ value2 2))) def creates a global binding in current namespace !!!
(let [result (+ (/ value1 2) (/ value2 2))]
(if
(= (mod result 2) 1)
(inc result)
result)))
(defn func2 [xx]
(let [[a b c d] xx]
[ (adj c a) (adj a b) (adj b c) (inc d)]
))
The third function is not clear to me. I don't read your intent. What I understand is: "Keep applying func2 to itself until the first three elements of its result are equal." But I'm afraid this condition is never met, so I replaced it with a true in order to see just one result without blowing the stack.
(defn func3 [xx]
(loop [ v (func2 xx) ]
(if
;(= (v 0) (v 1) (v 2))
true
v
(recur (func2 v))
)))
Useful link: http://clojure.org/cheatsheet
Cheers -
I am working on the clojure koans and one of the questions in the functions needs further explanation for me to “get it” and have an aha moment. I am able to write the function that satisfies the question. But I don't fully understand why all the bits work.
Clojure> (= 25 ((fn [a b] (b a)) 5 (fn [n] (* n n))))
true
Question 1.
I do not understand why this throws an error:
Clojure> (= 25 ((fn [b a] (b a)) 5 (fn [n] (* n n))))
java.lang.ClassCastException: java.lang.Long cannot be cast to clojure.lang.IFn
So the only change in the above is switching the order of b and a.
In my brain I read “a function that takes an a and b” or a “b and an a” but how they are used is up to the subsequent statement in the parens. Why does the order matter at this point?
Questions 2.
Clojure> (= 25 ((fn [a] (5 a)) (fn [n] (* n n))))
java.lang.ClassCastException: java.lang.Long cannot be cast to clojure.lang.IFn
Why when I substitute out the value of b for the int it represents do I get an error?
Quentions 3.
((fn [a b] (b a)) 5 (fn [n] (* n n))))
Why does this not throw an error (b a) b in this instance is 5 which is a symbol.
The first item in brackets is expected to be a function or a special form unless it is a list?
look at your expression in the first function:
(b a)
since b is first, b has to be a function. In your second example, you're trying to pass 5 to b, but 5 is not a function. With descriptive names you can see that you're trying to use 5 as a function:
((fn [argument function] (argument function)) ;; problem!!
5
(fn [n] (* n n)))
Remember the evaluation rules for lisps: given this s-expression:
(f x y z)
evaluate f, x, y, and z, and apply f as a function to x, y, and z.
see answer to 1 -- '5' is not a function. Use descriptive names, and easily spot the problem:
((fn [function] (5 function)) ;; problem!!
(fn [n] (* n n)))
Change to:
((fn [a] (a 5)) ;; 'a' and '5' flipped
(fn [n] (* n n)))
to get this to run.
this isn't a problem: a is 5, b is a function in (b a). With descriptive names it makes more sense:
((fn [argument function] (function argument))
5
(fn [n] (* n n)))