Piping data through arbitrary functions in Clojure

问题

I know that the -> form can be used to pass the results of one function result to another:

(f1 (f2 (f3 x))) 
(-> x f3 f2 f1) ; equivalent to the line above

(taken from the excellent Clojure tutorial at ociweb)

However this form requires that you know the functions you want to use at design time. I'd like to do the same thing, but at run time with a list of arbitrary functions.

I've written this looping function that does it, but I have a feeling there's a better way:

(defn pipe [initialData, functions]
  (loop [
      frontFunc (first functions)
      restFuncs (rest functions)
      data initialData ]
    (if frontFunc
      (recur (first restFuncs) (rest restFuncs) (frontFunc data) )
      data )
  ) )

What's the best way to go about this?

回答1:

I must admit I'm really new to clojure and I might be missing the point here completely, but can't this just be done using comp and apply?

user> (defn fn1 [x] (+ 2 x))
user> (defn fn2 [x] (/ x 3))
user> (defn fn3 [x] (* 1.2 x))
user> (defn pipe [initial-data my-functions] ((apply comp my-functions) initial-data))
user> (pipe 2 [fn1 fn2 fn3])
2.8

回答2:

You can do this with a plain old reduce:

(defn pipe [x fs] (reduce (fn [acc f] (f acc)) x fs))

That can be shortened to:

(defn pipe [x fs] (reduce #(%2 %1) x fs))

Used like this:

user> (pipe [1 2 3] [#(conj % 77) rest reverse (partial map inc) vec])
[78 4 3]

回答3:

If functions is a sequence of functions, you can reduce it using comp to get a composed function. At a REPL:

user> (def functions (list #(* % 5) #(+ % 1) #(/ % 3)))
#'user/my-list
user> ((reduce comp functions) 9)
20

apply also works in this case because comp takes a variable number of arguments:

user> (def functions (list #(* % 5) #(+ % 1) #(/ % 3)))
#'user/my-list
user> ((apply comp functions) 9)
20

来源：https://stackoverflow.com/questions/3683219/piping-data-through-arbitrary-functions-in-clojure

标签

clojure