Find index of an element matching a predicate in Clojure?

Question

Using keep-indexed you can get a sequence of indices for which a predicate is satisfied:

(defn indices [pred coll]
   (keep-indexed #(when (pred %2) %1) coll))

With this simple function you'll solve your problem with the expression

user=> (first (indices pos? [-1 0 99 100 101]))
2

Note that, due to the lazyness of keep-indexed (and indices), the entire sequence need not be realized so no extraneous calculations are performed.

(defn first-pos [x] 
  (loop [arr x n 0]
     (if (pos? (first arr))
     n
     (recur (next arr) (inc n)))))

This is a good example of using functional programming's powerful tail recursion.

(first (filter #(not (nil? %)) (map #(when (pos? %1) %2) [-1 1 0 99 100 101] (range))))

Map can take one or more collections and return one list,put condition on map,and filter nil.

(defn pred-idx [pred [idx hist] cur]
  (if (pred cur)
    [(inc idx) (conj hist idx)]
    [(inc idx) hist]))

(defn idx-filter [pred col]
  (second (reduce (partial pred-idx pred) [0 []] col)))

(first (idx-filter pos? [-1 0 99 100 101]))
2

Not sure if this is better, but it works. I think it forces evaluation of the entire sequence though, and if you need all indices that would be better. The correct thing to do is probably turn it into a lazy sequence somehow, but I'm done for the evening.

I'm a little late to the party, but I prefer:

(defn index-of-pred
  [pred coll]
  (ffirst (filter (comp pred second) (map-indexed list coll))))

;; example usage
(index-of-pred pos? [-1 -2 -5 0 3 4 1 -100])
;=> 4