user> (for [n '(9) d '(2 3 5 7) :while (not= 0 (mod n d))] n)
It gives me,
user> (9)
And if I change the sequence in the list like this,
user> (for [n '(9) d '(2 5 3 7) :while (not= 0 (mod n d))] n)
Then it gives me,
user> (9 9)
Actually, I'm trying to get an empty list, if the integer n is divisible by any integer of sequence d; otherwise n. So, the question is, is there any seq function, or a combination of few, that can do the trick? Or do I have to use loop-recurr, or recursion, thing whatsoever?
I think that the function you are looking for is filter.
Returns a lazy sequence of the items in coll for which (pred item) returns true. pred must be free of side-effects.
The following function, divisible-by-fn, returns the number in the seq that is divisible by n
(defn divisible-by-fn [your-integer your-seq]
(filter #(zero? (mod your-integer %))your-seq))
(divisible-by-fn 9 '(2 5 3 7))
=>(3)
... but, in case, you need to get an empty list, if the integer n is divisible by any integer, d, in the seq; otherwise n, then
(defn your-fn [your-integer your-seq]
(if-not (not-any? #(zero? (mod your-integer %))your-seq)
()
your-integer)
)
(your-fn 9 '(2 5 3 7))
=> ()
(your-fn 11 '(2 5 3 7))
=> 11