Making current function of list recursive Clojure

Question

Hi i am looking for a bit of help with some Clojure code. I have written a function that will take in a list and calculate the qty*price for a list eg. '(pid3 6 9)

What i am looking for is to expand my current function so that it recursively does the qty*price calculation until it reaches the end of the list.

My current function is written like this:

(defn pid-calc [list] (* (nth list 1) (nth list 2)))

I have tried implementing it into a recursive function but have had no luck at all, i want to be able to call something like this:

(pid-calcc '( (pid1 8 5) (pid2 5 6))
 return==> 70

Thats as close as i have came to an answer and cannot seem to find one. If anyone can help me find a solution i that would be great. As so far i am yet to find anything that will compile.

(defn pid-calc [list]
   (if(empty? list)
    nil
    (* (nth list 1) (nth list 2)(+(pid-calc (rest list))))))

nha · Accepted Answer · 2015-11-01 17:24:55Z

2

You don't need a recursive function. Just use + and map:

(defn pid-calc [list]
  (letfn [(mul [[_ a b]] (* a b))]
    (apply + (map mul list))))

edited Nov 1, 2015 at 17:24

nha

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answered Oct 30, 2015 at 12:43

sloth

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2 Comments

Thumbnail Over a year ago

Define list-calc, say as (fn [[_ a b]] (* a b)).

sloth Over a year ago

@Thumbnail thanks. OP edited his question and removed the definition of list-calc, so I edited my answer to reflect that.

Community · Accepted Answer · 2017-05-23 11:51:57Z

1

@sloth's answer, suitably corrected, is a concise and fast enough way to solve your problem. It shows you a lot.

Your attempt at a recursive solution can be (a)mended to

(defn pid-calc [list]
   (if (empty? list)
    0
    (let [x (first list)]
      (+ (* (nth x 1) (nth x 2)) (pid-calc (next list))))))

This works on the example, but - being properly recursive - will run out of stack space on a long enough list. The limit is usually about 10K items.

We can get over this without being so concise as @sloth. You might find the following easier to understand:

(defn pid-calc [list]
  (let [line-total (fn [item] (* (nth item 1) (nth item 2)))]
    (apply + (map line-total list))))

edited May 23, 2017 at 11:51

CommunityBot

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answered Oct 30, 2015 at 15:51

Thumbnail

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1 Comment

Bean Shares Over a year ago

Thanks for the help, im pretty much just getting started with clojure. I have only ever really written in OO languages. There is so much more i need to learn, i knew i could simply most of my functions to one liners almost but i dont know what all of the built in methods do yet.

Dominykas Mostauskis · Accepted Answer · 2015-10-30 16:12:30Z

0

reduce fits your scenario quite well:

(def your-list [[:x 1 2] [:x 1 3]])

(reduce #(+ %1 (* (nth %2 1) (nth %2 2))) 0 your-list)

(reduce #(+ %1 (let [[_ a b] %2] (* a b)) 0 your-list)

answered Oct 30, 2015 at 16:12

Dominykas Mostauskis

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