`repeat` takes a value and returns a sequence of that value.
There's another function, `repeatedly`, which takes a function and returns
a sequence of values returned by calling that function.
-S
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To
, Andrew Care wrote:
I'm trying to use repeat with a function argument. This works:
(reduce + (filter (fn [number] (zero? (some #{0} (map mod (take 2 (repeat
9)) [3 5] (range 1 1000)))
This doesn't:
(reduce + (filter (fn [number] (zero? (some #{0} (map mod (take 2 (repeat
number)) [3 5
to use repeat with a function argument. This works:
(reduce + (filter (fn [number] (zero? (some #{0} (map mod (take 2 (repeat
9)) [3 5] (range 1 1000)))
This doesn't:
(reduce + (filter (fn [number] (zero? (some #{0} (map mod (take 2 (repeat
number)) [3 5] (range 1 1000)))
Why can I
(and returns 233168)
Sean
On Sun, Dec 23, 2012 at 11:56 PM, Marko Topolnik
marko.t...@gmail.comjavascript:
wrote:
*repeat* is not supposed to work with functions, but there's *repeatedly.
*
On Monday, December 24, 2012 4:20:23 AM UTC+1, Andrew Care wrote:
I'm trying to use repeat
I'm trying to use repeat with a function argument. This works:
(reduce + (filter (fn [number] (zero? (some #{0} (map mod (take 2 (repeat
9)) [3 5] (range 1 1000)))
This doesn't:
(reduce + (filter (fn [number] (zero? (some #{0} (map mod (take 2 (repeat
number)) [3 5] (range 1 1000
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Hi,
Am 24.12.12 04:20, schrieb Andrew Care:
Why can I use (repeat 9) and not (repeat number)?
Because you call zero? on nil, which doesn't work. Instead of (zero?
(some #{0} ...)) use (some zero? ...).
Kind regards
Meikel
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*repeat* is not supposed to work with functions, but there's *repeatedly.*
On Monday, December 24, 2012 4:20:23 AM UTC+1, Andrew Care wrote:
I'm trying to use repeat with a function argument. This works:
(reduce + (filter (fn [number] (zero? (some #{0} (map mod (take 2 (repeat
9)) [3 5
