How to write multiple statements in one if block in clojure?

Question

I'm writing a function in clojure, that takes in 2 arguments (both are lists), and iterates over vehicles by recursion, until the vehicle list becomes empty. The function is like:

(defn v [vehicles locs]
    (if (= (count vehicles) 0)
        nil
        (if (> (count vehicles) 0)
            (split-at 1 locs)
            (v (rest vehicles) (rest locs))
        ))
    )

So, if I give the input as (v '(v1 v2 v3) '([1 2] [2 3] [4 2] [5 3])), then I want the output as [([1 2]) ([3 4]) ([5 6] [6 7])]. I know that the statement (v (rest vehicles) (rest locs)) is not executing because it's taking it in the else case i.e., when count(vehicles) not > 0. I want to know how can I make this statement be executed in the same if block, i.e., in (if (> (count vehicles) 0)

@Mark Thanks! Is there something like break? Because while iterating in the end it reaches (if (= (count vehicles) 0) and then prints nil only. — Erica Maine
– Erica Maine, Commented Oct 22, 2014 at 5:58

Diego Basch · Accepted Answer · 2014-10-22 06:54:31Z

3

It sounds like you're trying to do something like this:

(defn v [vehicles locs]
  (if (seq (rest vehicles))
    (concat  [[(first locs)]] (v (rest vehicles) (rest locs)))
    [locs]))

answered Oct 22, 2014 at 6:54

Diego Basch

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1 Comment

Erica Maine Over a year ago

Thanks. Is there a way I could get the output as ((v1 [1 2]) (v2 [0 2] [3 2])), if I give the input: (v '(v1 v2) '([1 2] [0 2] [3 2]))?

amalloy · Accepted Answer · 2014-10-22 17:37:30Z

2

When you are using recursion and need to return some result of execution there is good approach to accumulate all of your recursion step's results in one holder variable. Here result is such a holder:

(defn v [vehicles locs]
  (if (empty? vehicles)
    nil
    (loop [result [] vs vehicles ls locs]
      (if (= 1 (count vs))
        (conj result ls)
        (recur (conj result (first ls)) (rest vs) (rest ls))))))

It works like this:

(v '(v1 v2 v3) '([1 2] [2 3] [4 2] [5 3]))
=> [[1 2] [2 3] ([4 2] [5 3])]

If you really need to wrap location one-element vectors (but actually it seems like a strange requirement) then try the following:

(defn v [vehicles locs]
  (if (empty? vehicles)
    nil
    (loop [result [] vs vehicles ls locs]
      (if (= 1 (count vs))
        (conj result ls)
        (recur (conj result (list (first ls))) (rest vs) (rest ls))))))

edited Oct 22, 2014 at 17:37

amalloy

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answered Oct 22, 2014 at 6:40

hsestupin

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1 Comment

Diego Basch Over a year ago

No need for an accumulator in a very simple case like this one, see my answer. It could be made lazy trivially too.

Aleš Roubíček · Accepted Answer · 2014-10-23 07:25:11Z

1

You can write multiple statements in if branches with a do form:

(if (= 1 (count vs))
  (do
    (conj result ls)
    (recur (conj result (first ls)) (rest vs) (rest ls)))
  :else)

If you don't need else branch it's convenient to use when macro:

(when (= 1 (count vs))
  (conj result ls)
  (recur (conj result (first ls)) (rest vs) (rest ls)))

answered Oct 23, 2014 at 7:25

Aleš Roubíček

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