Re: Use repeat with a function argument
`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 -- You received this message because you are subscribed to the Google Groups "Clojure" group. To post to this group, send email to clojure@googlegroups.com Note that posts from new members are moderated - please be patient with your first post. To unsubscribe from this group, send email to clojure+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/clojure?hl=en
Re: Use repeat with a function argument
I deleted my answer minutes after posting. The phrase "function argument"
is ambiguous ("argument to function", "argument which is a function") and I
jumped to the wrong conclusion.
On Monday, December 24, 2012 7:00:48 PM UTC+1, Sean Corfield wrote:
>
> He's repeating a function argument, not a function.
>
> Meikel is correct that the second expression causes (some #{0} ...) to
> return nil when number is not a multiple of 3 or 5, and then zero? fails.
> As he suggests...
>
> (reduce + (filter (fn [number] (some zero? (map mod (take 2 (repeat
> number)) [3 5]))) (range 1 1000)))
>
> ...works (and returns 233168)
>
> Sean
>
> On Sun, Dec 23, 2012 at 11:56 PM, Marko Topolnik
>
> > 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 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 use (repeat 9) and not (repeat number)?
>>>
>>
>>
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Re: Use repeat with a function argument
He's repeating a function argument, not a function.
Meikel is correct that the second expression causes (some #{0} ...) to
return nil when number is not a multiple of 3 or 5, and then zero? fails.
As he suggests...
(reduce + (filter (fn [number] (some zero? (map mod (take 2 (repeat
number)) [3 5]))) (range 1 1000)))
...works (and returns 233168)
Sean
On Sun, Dec 23, 2012 at 11:56 PM, Marko Topolnik
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 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 use (repeat 9) and not (repeat number)?
>>
>
>
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Re: Use repeat with a function argument
of course repeat works with functions:
user=> (take 4 (map #(% 4) (repeat inc)))
(5 5 5 5)
On Sun, Dec 23, 2012 at 11:56 PM, Marko Topolnik
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 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 use (repeat 9) and not (repeat number)?
>
> --
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--
Ben Wolfson
"Human kind has used its intelligence to vary the flavour of drinks,
which may be sweet, aromatic, fermented or spirit-based. ... Family
and social life also offer numerous other occasions to consume drinks
for pleasure." [Larousse, "Drink" entry]
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Re: Use repeat with a function argument
*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] (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 use (repeat 9) and not (repeat number)?
>
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Re: Use repeat with a function argument
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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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Use repeat with a function argument
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)))
Why can I use (repeat 9) and not (repeat number)?
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