Thanks, that seems quicker.
I have another question, which is how can i find the first value that appears
in the list that meets some condition - i.e. if the item > 0.5 - and return its
position?
On 25 Sep 2011, at 22:03, Henry Rich wrote:
> (% 4 * >:@i.@#) _4 {:\ +/\ 1&p: 3 5 7 9 13 17 21 25 31 37 43 49
> 0.75 0.625 0.666667
>
>
> Henry Rich
>
> On 9/25/2011 4:46 PM, Don Guinn wrote:
>> Apply the 1&p: to the list before doing the infixes. Cuts down the number of
>> times 1&p: is invoked. But why are you adding one to the number of items in
>> each infix? Seems like the 3 5 7 9 should give 0.75, not 0.6 .
>>
>> ((+/%#)&,)\_4,\1 p:3 5 7 9 13 17 21 25 31 37 43 49
>>
>> On Sun, Sep 25, 2011 at 1:47 PM, David Vaughan<[email protected]
>>> wrote:
>>
>>> I've done: verb&,\
>>>
>>> I arranged the list to be of shape 'x 4'. This seems to work. The problem
>>> I'm now having is that my verb is slow. Can this be refined (significantly)?
>>>
>>> +/@:(1&p:) % 1&+@:#
>>>
>>> It takes finds the percentage of items in the list which are prime.
>>>
>>> On 25 Sep 2011, at 20:04, Don Guinn wrote:
>>>
>>>> verb=.<
>>>> (verb&,)\_4,\3 5 7 9 13 17 21 25 31 37 43 49
>>>> +-------+-------------------+-------------------------------+
>>>> |3 5 7 9|3 5 7 9 13 17 21 25|3 5 7 9 13 17 21 25 31 37 43 49|
>>>> +-------+-------------------+-------------------------------+
>>>>
>>>>
>>>> On Sun, Sep 25, 2011 at 12:44 PM, David Vaughan<
>>>> [email protected]> wrote:
>>>>
>>>>> I'm trying to apply a verb to infixes of length 4 but I want it to be
>>>>> cumulative. i.e. in the list:
>>>>>
>>>>> 3 5 7 9 13 17 21 25 31 37 43 49
>>>>>
>>>>> I want:
>>>>>
>>>>> verb 3 5 7 9
>>>>> verb 3 5 7 9 13 17 21 25
>>>>> verb 3 5 7 9 13 17 21 25 31 37 43 49
>>>>>
>>>>> Any help? Thanks.
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