Ditto what everyone else said, plus let's get rid of the duplicated
call to Math/log by splitting the iterate into an iterate + a map:
(take-while
(fn [[_ second]] (< second 10))
(map (fn [x] [x (/ x (Math/log x))])
(iterate #(* % 2) 2)))
Stu
>
> On 02.06.2009, at 17:35, Wilson MacGyver wrote:
>
>> for example, the first 10 produces.
>>
>> ([2 2.8853900817779268] [4 2.8853900817779268] [8 2.8853900817779268]
>> [16 3.8471867757039027] [32 5.7707801635558535] [64
>> 9.233248261689365]
>> [128 15.38874710281561] [256 26.380709319112476] [512
>> 46.16624130844683] [1024 82.07331788168325])
>>
>> what if I want to filter so I only get pairs for which the 2nd value
>> is < 10. I couldn't figure out how to get
>> filter to work for pair values.
>
> Try this:
>
> (take 3
> (filter #(< (second %) 10)
> (iterate (fn [[a b]] [(* 2 a) (/ a (Math/log a))])
> [2 (/ 2 (Math/log 2))])))
>
> If you try to take 10 values, you will create an endless loop because
> with the given parameters your sequence actually has less than ten
> elements that satisfy the condition!
>
>> a 2nd question is more of general clojure idiom, in trying to covert
>> the following java code from michaelg's java 1 presentation
>>
>> private int calcSize(){
>> int max = 2;
>> while ((max/Math.log(max)) < size &&
>> max < Integer.MAX_VALUE && max > 0){
>> max *=2;
>> }
>> return max;
>> }
>>
>> My first reaction was to do it using a sequence. Is this the clojure
>> idiomatic way to convert a while loop from other languages?
>
> Clojure has loops as well:
>
> (let [size 10] ;made up
> (loop [max 2]
> (if (or (>= (/ max (Math/log max)) size)
> (>= max Integer/MAX_VALUE)
> (<= max 0))
> max
> (recur (* 2 max)))))
>
> Konrad.
>
> >
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