idiomatic way of counting loop iteration

[Joeyjoejoe]

Hi,

I'm just stating to learn clojure, i made a first read of "clojure programming" to get the big picture, and i'm starting to play with the repl, trying to solve some katas. A lot of theses katas involves returning the count of loop iterations. Most of the time, i end up with this kind of functions:

(defn my-function [a b & [n]]
 (if cond
   (my-function new-a new-b (inc (or n 0))
   (or n defaut-value)
  )
)

What are the pros/cons of doing this? Are there any idiomatic ways of doing this.

Thank you

[Stuart Sierra]

loop/recur is more typical for this kind of counting loop, as it avoids the risk of a stack-overflow when the number of iterations is high.

Also, I recommend against the [a b & [n]] argument pattern here:

https://stuartsierra.com/2015/06/01/clojure-donts-optional-arguments-with-varargs

–S

[Jason Felice]

Generally speaking, `loop`, `recur`, and writing recursive functions are not idiomatic in Clojure.  Using things like `iterate`, `map`, and `filter` are considered clearer.

If `n` is used just to count iterations, then `iterate` would be useful.  e.g.

(first (drop n (iterate (fn [[a b]] ... [new-a new-b]))))

If `n` is used in the computation to create new-a and new-b, then `reduce`  and `range` would be useful.

(reduce (fn [[a b] n]

                ...

                [new-a new-b])

             [a b]

             (range n))

It might be possible to use `map-indexed` with `repeat`, also.

-Jason

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[Joeyjoejoe]

Thanks, i like the "unambiguous intent" argument from your post, and indeed i met the stack-overflow issue. 

[Joeyjoejoe]

Thanks for helping me!

In your first example:

(first (drop n (iterate (fn [[a b]] ... [new-a new-b]))))

Given that iterate will return a sequence whose length is the number of iteration i'm looking for, the "(first (drop n" part will return one element of this sequence (depending on n value), and not the number of iteration. Am i right?

Maybe i could do this, to get the number of iteration:

Thank you again

(count (iterate (fn [[a b]] ... [new-a new-b]))))

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[Gary Johnson]

Almost right. The iterate function is an infinite sequence generator, so (count (iterate f x)) will never return.

If you want the iteration to terminate when cond is false (as in your original example), you're looking for this:

(count (take-while cond (iterate (fn [[a b]] ... [new-a new-b]) [init-a init-b])))

Happy hacking!

[Timothy Baldridge]

Also consider map-indexed if you just need to count how many things go through a lazy seq. It works like map, but takes a (fn [idx itm] ...) where idx is the index of the item in the overall seq. 

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