Since one cannot apply a macro to a list, e.g.
;; does not work
(apply -> [expr1 expr2 expr3])
How does one generate such expression:
(-> expr1
expr2
expr3)
Where
expr1 is generated by (generate-expr1 f g h)
expr2 is generated by (generate-expr2 f g h)
expr3 is generated by (generate-expr3 f g h)
Context
I'm trying to design an embedded DSL, e.g.
["increment" "increment" "increment"]
Which is then converted into code, e.g.
(fn [n] (-> n inc inc inc))
You can generate it with a macro e.g.
(defmacro opfun [op-names]
(let [m {"increment" 'inc}
ops (map m op-names)]
`(fn [n#] (-> n# ~#ops))))
then
(opfun ["increment" "increment" "increment"])
if you need to provide your argument list at runtime you can interpret your structure directly e.g.
(defn build-eval [op-names]
(let [m {"increment" inc}
ops (map m op-names)]
(fn [n] (reduce (fn [acc f] (f acc)) n ops))))
Related
Following is the Clojure code:
(reduce (fn [r x] (if (nil? x) r (conj r x)))
[]
[:mouse nil :duck nil :lory nil])
In REPL, it evaluates to [:mouse :duck :lory].
My question is, how is the code evaluating?
According to me, r is [] and x is [:mouse nil :duck nil :lory nil]. nil? x is false as so it evaluates to (conj r x). But x is a vector, not an element so how it will add an element to the empty vector r in conj? I don't know but I am wrong somewhere in my approach. The output is the animals' name vector without nil values.
Can anyone please explain me the execution of code. Thanks.
Your problem appears to be understanding how reduce works. I'd like to refer you to the source code, but it simply maps onto a Java implementation, so I have to fake it.
The kind of reduce you are doing - supplying the initial value - might have been coded as follows:
(defn reduce [f init coll]
(loop [acc init, tail coll]
(if (seq tail)
(recur (f acc (first tail)) (rest tail))
acc)))
As you can see, it works through the sequence coll, applying the function f to acc and the first value in the sequence, to generate a new acc. When there is no more sequence, it returns acc.
How does this apply to your example? The reducing function ...
(fn [r x] (if (nil? x) r (conj r x)))
... ignores nil xs but conjs anything else onto the end of the accumulating vector r. Your code is more or less equivalent to ...
(remove nil? [:mouse nil :duck nil :lory nil])
=> (:mouse :duck :lory)
... except you get a lazy sequence instead of a vector.
Courtesy of FredOverflow's Clojure Transducers from the ground up you can wrap the reducing fn in this logging function to print the different arguments (r and x) to every step:
(defn logging [f]
(fn [& args]
(prn args)
(apply f args)))
(reduce (logging (fn [r x] (if (nil? x) r (conj r x))))
[]
[:mouse nil :duck nil :lory nil])
user=> ([] :mouse) ([:mouse] nil) ([:mouse] :duck) ([:mouse :duck] nil) ([:mouse :duck] :lory) ([:mouse :duck :lory] nil)
So only the keywords are added (not nils), starting with []. At the end [:mouse :duck :lory] is returned.
I like my code to have a "top-down" structure, and that means I want to do exactly the opposite from what is natural in Clojure: functions being defined before they are used. This shouldn't be a problem, though, because I could theoretically declare all my functions first, and just go on and enjoy life. But it seems in practice declare cannot solve every single problem, and I would like to understand what is exactly the reason the following code does not work.
I have two functions, and I want to define a third by composing the two. The following three pieces of code accomplish this:
1
(defn f [x] (* x 3))
(defn g [x] (+ x 5))
(defn mycomp [x] (f (g x)))
(println (mycomp 10))
2
(defn f [x] (* x 3))
(defn g [x] (+ x 5))
(def mycomp (comp f g))
3
(declare f g)
(defn mycomp [x] (f (g x)))
(defn f [x] (* x 3))
(defn g [x] (+ x 5))
But what I would really like to write is
(declare f g)
(def mycomp (comp f g))
(defn f [x] (* x 3))
(defn g [x] (+ x 5))
And that gives me
Exception in thread "main" java.lang.IllegalStateException: Attempting to call unbound fn: #'user/g,
That would mean forward declaring works for many situations, but there are still some cases I can't just declare all my functions and write the code in any way and in whatever order I like. What is the reason for this error? What does forward declaring really allows me to do, and what are the situations I must have the function already defined, such as for using comp in this case? How can I tell when the definition is strictly necessary?
You can accomplish your goal if you take advantage of Clojure's (poorly documented) var behavior:
(declare f g)
(def mycomp (comp #'f #'g))
(defn f [x] (* x 3))
(defn g [x] (+ x 5))
(mycomp 10) => 45
Note that the syntax #'f is just shorthand (technically a "reader macro") that translates into (var f). So you could write this directly:
(def mycomp (comp (var f) (var g)))
and get the same result.
Please see this answer for a more detailed answer on the (mostly hidden) interaction between a Clojure symbol, such as f, and the (anonymous) Clojure var that the symbol points to, namely either #'f or (var f). The var, in turn, then points to a value (such as your function (fn [x] (* x 3)).
When you write an expression like (f 10), there is a 2-step indirection at work. First, the symbol f is "evaluated" to find the associated var, then the var is "evaluated" to find the associated function. Most Clojure users are not really aware that this 2-step process exists, and nearly all of the time we can pretend that there is a direct connection between the symbol f and the function value (fn [x] (* x 3)).
The specific reason your original code doesn't work is that
(declare f g)
creates 2 "empty" vars. Just as (def x) creates an association between the symbol x and an empty var, that is what your declare does. Thus, when the comp function tries to extract the values from f and g, there is nothing present: the vars exist but they are empty.
P.S.
There is an exception to the above. If you have a let form or similar, there is no var involved:
(let [x 5
y (* 2 x) ]
y)
;=> 10
In the let form, there is no var present. Instead, the compiler makes a direct connection between a symbol and its associated value; i.e. x => 5 and y => 10.
I think Alan's answer addresses your questions very well. Your third example works because you aren't passing the functions as arguments to mycomp. I'd reconsider trying to define things in "reverse" order because it works against the basic language design, requires more code, and might be harder for others to understand.
But... just for laughs and to demonstrate what's possible with Clojure macros, here's an alternative (worse) implementation of comp that works for your preferred syntax, without dealing directly in vars:
(defn- comp-fn-arity [variadic? args f & fs] ;; emits a ([x] (f (g x)) like form
(let [args-vec (if variadic?
(into (vec (butlast args)) ['& (last args)])
(apply vector args))
body (reduce #(list %2 %1)
(if variadic?
(apply list 'apply (last fs) args)
(apply list (last fs) args))
(reverse (cons f (butlast fs))))]
`(~args-vec ~body)))
(defmacro momp
([] identity)
([f] f)
([f & fs]
(let [num-arities 5
args-syms (repeatedly num-arities gensym)]
`(fn ~#(map #(apply comp-fn-arity (= % (dec num-arities)) (take % args-syms) f fs)
(range num-arities))))))
This will emit something kinda like comp's implementation:
(macroexpand '(momp f g))
=>
(fn*
([] (f (g)))
([G__1713] (f (g G__1713)))
([G__1713 G__1714] (f (g G__1713 G__1714)))
([G__1713 G__1714 G__1715] (f (g G__1713 G__1714 G__1715)))
([G__1713 G__1714 G__1715 & G__1716] (f (apply g G__1713 G__1714 G__1715 G__1716))))
This works because your (unbound) functions aren't being passed as values to another function; during compilation the macro expands "in place" as if you'd written the composing function by hand, as in your third example.
(declare f g)
(def mycomp (momp f g))
(defn f [x] (* x 3))
(defn g [x] (+ x 5))
(mycomp 10) ;; => 45
(apply (momp vec reverse list) (range 10)) ;; => [9 8 7 6 5 4 3 2 1 0]
This won't work in some other cases, e.g. ((momp - dec) 1) fails because dec gets inlined and doesn't have a 0-arg arity to match the macro's 0-arg arity. Again, this is just for the sake of example and I wouldn't recommend it.
Let's say you have a recursive function defined in a let block:
(let [fib (fn fib [n]
(if (< n 2)
n
(+ (fib (- n 1))
(fib (- n 2)))))]
(fib 42))
This can be mechanically transformed to utilize memoize:
Wrap the fn form in a call to memoize.
Move the function name in as the 1st argument.
Pass the function into itself wherever it is called.
Rebind the function symbol to do the same using partial.
Transforming the above code leads to:
(let [fib (memoize
(fn [fib n]
(if (< n 2)
n
(+ (fib fib (- n 1))
(fib fib (- n 2))))))
fib (partial fib fib)]
(fib 42))
This works, but feels overly complicated. The question is: Is there a simpler way?
I take risks in answering since I am not a scholar but I don't think so. You pretty much did the standard thing which in fine is a partial application of memoization through a fixed point combinator.
You could try to fiddle with macros though (for simple cases it could be easy, syntax-quote would do name resolution for you and you could operate on that). I'll try once I get home.
edit: went back home and tried out stuff, this seems to be ok-ish for simple cases
(defmacro memoize-rec [form]
(let [[fn* fname params & body] form
params-with-fname (vec (cons fname params))]
`(let [f# (memoize (fn ~params-with-fname
(let [~fname (partial ~fname ~fname)] ~#body)))]
(partial f# f#))))
;; (clojure.pprint/pprint (macroexpand '(memoize-rec (fn f [x] (str (f x))))))
((memoize-rec (fn fib [n]
(if (< n 2)
n
(+ (fib (- n 1))
(fib (- n 2)))))) 75) ;; instant
((fn fib [n]
(if (< n 2)
n
(+ (fib (- n 1))
(fib (- n 2))))) 75) ;; slooooooow
simpler than what i thought!
I'm not sure this is "simpler" per se, but I thought I'd share an approach I took to re-implement letfn for a CPS transformer I wrote.
The key is to introduce the variables, but delay assigning them values until they are all in scope. Basically, what you would like to write is:
(let [f nil]
(set! f (memoize (fn []
<body-of-f>)))
(f))
Of course this doesn't work as is, because let bindings are immutable in Clojure. We can get around that, though, by using a reference type — for example, a promise:
(let [f (promise)]
(deliver! f (memoize (fn []
<body-of-f>)))
(#f))
But this still falls short, because we must replace every instance of f in <body-of-f> with (deref f). But we can solve this by introducing another function that invokes the function stored in the promise. So the entire solution might look like this:
(let [f* (promise)]
(letfn [(f []
(#f*))]
(deliver f* (memoize (fn []
<body-of-f>)))
(f)))
If you have a set of mutually-recursive functions:
(let [f* (promise)
g* (promise)]
(letfn [(f []
(#f*))
(g []
(#g*))]
(deliver f* (memoize (fn []
(g))))
(deliver g* (memoize (fn []
(f))))
(f)))
Obviously that's a lot of boiler-plate. But it's pretty trivial to construct a macro that gives you letfn-style syntax.
Yes, there is a simpler way.
It is not a functional transformation, but builds on the impurity allowed in clojure.
(defn fib [n]
(if (< n 2)
n
(+ (#'fib (- n 1))
(#'fib (- n 2)))))
(def fib (memoize fib))
First step defines fib in almost the normal way, but recursive calls are made using whatever the var fib contains. Then fib is redefined, becoming the memoized version of its old self.
I would suppose that clojure idiomatic way will be using recur
(def factorial
(fn [n]
(loop [cnt n acc 1]
(if (zero? cnt)
acc
(recur (dec cnt) (* acc cnt))
;; Memoized recursive function, a mash-up of memoize and fn
(defmacro mrfn
"Returns an anonymous function like `fn` but recursive calls to the given `name` within
`body` use a memoized version of the function, potentially improving performance (see
`memoize`). Only simple argument symbols are supported, not varargs or destructing or
multiple arities. Memoized recursion requires explicit calls to `name` so the `body`
should not use recur to the top level."
[name args & body]
{:pre [(simple-symbol? name) (vector? args) (seq args) (every? simple-symbol? args)]}
(let [akey (if (= (count args) 1) (first args) args)]
;; name becomes extra arg to support recursive memoized calls
`(let [f# (fn [~name ~#args] ~#body)
mem# (atom {})]
(fn mr# [~#args]
(if-let [e# (find #mem# ~akey)]
(val e#)
(let [ret# (f# mr# ~#args)]
(swap! mem# assoc ~akey ret#)
ret#))))))
;; only change is fn to mrfn
(let [fib (mrfn fib [n]
(if (< n 2)
n
(+ (fib (- n 1))
(fib (- n 2)))))]
(fib 42))
Timings on my oldish Mac:
original, Elapsed time: 14089.417441 msecs
mrfn version, Elapsed time: 0.220748 msecs
4Clojure Problem 58 is stated as:
Write a function which allows you to create function compositions. The parameter list should take a variable number of functions, and create a function applies them from right-to-left.
(= [3 2 1] ((__ rest reverse) [1 2 3 4]))
(= 5 ((__ (partial + 3) second) [1 2 3 4]))
(= true ((__ zero? #(mod % 8) +) 3 5 7 9))
(= "HELLO" ((__ #(.toUpperCase %) #(apply str %) take) 5 "hello world"))
Here __ should be replaced by the solution.
In this problem the function comp should not be employed.
A solution I found is:
(fn [& xs]
(fn [& ys]
(reduce #(%2 %1)
(apply (last xs) ys) (rest (reverse xs)))))
It works. But I don't really understand how the reduce works here. How does it represent (apply f_1 (apply f_2 ...(apply f_n-1 (apply f_n args))...)?
Let's try modifying that solution in 3 stages. Stay with each for a while and see if you get it. Stop if and when you do lest I confuse you more.
First, let's have more descriptive names
(defn my-comp [& fns]
(fn [& args]
(reduce (fn [result-so-far next-fn] (next-fn result-so-far))
(apply (last fns) args) (rest (reverse fns)))))
then factor up some
(defn my-comp [& fns]
(fn [& args]
(let [ordered-fns (reverse fns)
first-result (apply (first ordered-fns) args)
remaining-fns (rest ordered-fns)]
(reduce
(fn [result-so-far next-fn] (next-fn result-so-far))
first-result
remaining-fns))))
next replace reduce with a loop which does the same
(defn my-comp [& fns]
(fn [& args]
(let [ordered-fns (reverse fns)
first-result (apply (first ordered-fns) args)]
(loop [result-so-far first-result, remaining-fns (rest ordered-fns)]
(if (empty? remaining-fns)
result-so-far
(let [next-fn (first remaining-fns)]
(recur (next-fn result-so-far), (rest remaining-fns))))))))
My solution was:
(fn [& fs]
(reduce (fn [f g]
#(f (apply g %&))) fs))
Lets try that for:
((
(fn [& fs]
(reduce (fn [f g]
#(f (apply g %&))) fs))
#(.toUpperCase %)
#(apply str %)
take)
5 "hello world"))
fs is a list of the functions:
#(.toUpperCase %)
#(apply str %)
take
The first time through the reduce, we set
f <--- #(.toUpperCase %)
g <--- #(apply str %)
We create an anonymous function, and assign this to the reduce function's accumulator.
#(f (apply g %&)) <---- uppercase the result of apply str
Next time through the reduce, we set
f <--- uppercase the result of apply str
g <--- take
Again we create a new anonymous function, and assign this to the reduce function's accumulator.
#(f (apply g %&)) <---- uppercase composed with apply str composed with take
fs is now empty, so this anonymous function is returned from reduce.
This function is passed 5 and "hello world"
The anonymous function then:
Does take 5 "hello world" to become (\h \e \l \l \o)
Does apply str to become "hello"
Does toUppercase to become "HELLO"
Here's an elegent (in my opinion) definition of comp:
(defn comp [& fs]
(reduce (fn [result f]
(fn [& args]
(result (apply f args))))
identity
fs))
The nested anonymous functions might make it hard to read at first, so let's try to address that by pulling them out and giving them a name.
(defn chain [f g]
(fn [& args]
(f (apply g args))))
This function chain is just like comp except that it only accepts two arguments.
((chain inc inc) 1) ;=> 3
((chain rest reverse) [1 2 3 4]) ;=> (3 2 1)
((chain inc inc inc) 1) ;=> ArityException
The definition of comp atop chain is very simple and helps isolate what reduce is bringing to the show.
(defn comp [& fs]
(reduce chain identity fs))
It chains together the first two functions, the result of which is a function. It then chains that function with the next, and so on.
So using your last example:
((comp #(.toUpperCase %) #(apply str %) take) 5 "hello world") ;=> "HELLO"
The equivalent only using chain (no reduce) is:
((chain identity
(chain (chain #(.toUpperCase %)
#(apply str %))
take))
5 "hello world")
;=> "HELLO"
At a fundamental level, reduce is about iteration. Here's what an implementation in an imperative style might look like (ignoring the possibility of multiple arities, as Clojure's version supports):
def reduce(f, init, seq):
result = init
for item in seq:
result = f(result, item)
return result
It's just capturing the pattern of iterating over a sequence and accumulating a result. I think reduce has a sort of mystique around it which can actually make it much harder to understand than it needs to be, but if you just break it down you'll definitely get it (and probably be surprised how often you find it useful).
Here is my solution:
(defn my-comp
([] identity)
([f] f)
([f & r]
(fn [& args]
(f (apply (apply my-comp r) args)))))
I like A. Webb's solution better, though it does not behave exactly like comp because it does not return identity when called without any arguments. Simply adding a zero-arity body would fix that issue though.
Consider this example:
(def c (comp f1 ... fn-1 fn))
(c p1 p2 ... pm)
When c is called:
first comp's rightmost parameter fn is applied to the p* parameters ;
then fn-1 is applied to the result of the previous step ;
(...)
then f1 is applied to the result of the previous step, and its result is returned
Your sample solution does exactly the same.
first the rightmost parameter (last xs) is applied to the ys parameters:
(apply (last xs) ys)
the remaining parameters are reversed to be fed to reduce:
(rest (reverse xs))
reduce takes the provided initial result and list of functions and iteratively applies the functions to the result:
(reduce #(%2 %1) ..init.. ..functions..)
I'm trying to implement the factorial lambda expression as described in the book Lambda-calculus, Combinators and Functional Programming
The way it's described there is :
fact = (Y)λf.λn.(((is-zero)n)one)((multiply)n)(f)(predecessor)n
Y = λy.(λx.(y)(x)x)λx.(y)(x)x
where
(x)y is equivalent to (x y) and
(x)(y)z is equivalent to (x (y x)) and
λx.x is equivalent to (fn [x] x)
and is-zero, one, multiply and predecessor are defined for the standard church numerals. Actual definitions here.
I translated that to the following
(defn Y-mine [y] ; (defn Y-rosetta [y]
((fn [x] (y (x x))) ; ((fn [f] (f f))
(fn [x] ; (fn [f]
(y ; (y (fn [& args]
(x x))))) ; (apply (f f) args))))))
and
(def fac-mine ; (def fac-rosetta
(fn [f] ; (fn [f]
(fn [n] ; (fn [n]
(is-zero n ; (if (zero? n)
one ; 1
(multiply n (f (predecessor n))))))) ; (* n (f (dec n)))))))
The commented out versions are the equivalent fac and Y functions from Rosetta code.
Question 1:
I understand from reading up elsewhere that the Y-rosetta β-reduces to Y-mine. In which case why is it preferable to use that one over the other?
Question 2:
Even if I use Y-rosetta. I get a StackOverflowError when I try
((Y-rosetta fac-mine) two)
while
((Y-rosetta fac-rosetta) 2)
works fine.
Where is the unguarded recursion happening?
I suspect it's something to do with how the if form works in clojure that's not completely equivalent to my is-zero implementation. But I haven't been able to find the error myself.
Thanks.
Update:
Taking into consideration #amalloy's answer, I changed fac-mine slightly to take lazy arguments. I'm not very familiar with clojure so, this is probably not the right way to do it. But, basically, I made is-zero take anonymous zero argument functions and evaluate whatever it returns.
(def lazy-one (fn [] one))
(defn lazy-next-term [n f]
(fn []
(multiply n (f (predecessor n)))))
(def fac-mine
(fn [f]
(fn [n]
((is-zero n
lazy-one
(lazy-next-term n f))))))
I now get an error saying:
=> ((Y-rosetta fac-mine) two)
ArityException Wrong number of args (1) passed to: core$lazy-next-term$fn clojure.lang.AFn.throwArity (AFn.java:437)
Which seems really strange considering that lazy-next-term is always called with n and f
The body of fac-mine looks wrong: it's calling (is-zero n one) for side effects, and then unconditionally calling (multiply n (f (predecessor n))). Presumably you wanted a conditional choice here (though I don't see why this doesn't throw an arity exception, given your definition of is-zero).