Isn't
   ([*[+])/\. 1 0 1 1 1 0 1
the same as 

   ([*+)/\. 1 0 1 1 1 0 1
?



Den 21:10 torsdag den 23. oktober 2014 skrev Joe Bogner <[email protected]>:
 

>
>
>This can also answer the question in a backwards, slower sense:
>
>([*[+])/\. 1 0 1 1 1 0 1
>
>
>1 0 3 2 1 0 1
>
>
>
>On Wed, Oct 22, 2014 at 2:12 PM, R.E. Boss <[email protected]> wrote:
>
>>
>>    }.;([:<+/\);.1[ 0,x
>> 0 1 2 3 0 1 2 3 0 1 2 0 0 1 0 0 0 0 1 2
>>
>>
>> R.E. Boss
>>
>> (Add your info to http://www.jsoftware.com/jwiki/Community/Demographics )
>>
>>
>> > -----Original Message-----
>> > From: [email protected] [mailto:programming-
>> > [email protected]] On Behalf Of Roger Hui
>> > Sent: woensdag 22 oktober 2014 19:25
>> > To: Programming forum
>> > Subject: Re: [Jprogramming] count of consecutive 1s
>> >
>> >    x
>> > 0 1 1 1 0 1 1 1 0 1 1 0 0 1 0 0 0 0 1 1
>> >    s->./\(-.x)*s=.+/\x
>> > 0 1 2 3 0 1 2 3 0 1 2 0 0 1 0 0 0 0 1 2
>> >
>> >
>> >
>> > On Wed, Oct 22, 2014 at 10:21 AM, Roger Hui <[email protected]>
>> > wrote:
>> >
>> > > You can probably do better than the following, but it'd be useful as a
>> > > result checker:
>> > >
>> > >    ] x=: 0<20 ?@$ 3
>> > > 0 1 1 1 0 1 1 1 0 1 1 0 0 1 0 0 0 0 1 1
>> > >    }. ; +/\&.> <;.1 ] 0,x
>> > > 0 1 2 3 0 1 2 3 0 1 2 0 0 1 0 0 0 0 1 2
>> > >
>> > >
>> > >
>> > >
>> > > On Wed, Oct 22, 2014 at 10:10 AM, Joe Bogner <[email protected]>
>> > wrote:
>> > >
>> > >> This is probably easy but I can't figure it out. How can I count the
>> > >> number
>> > >> of consecutive 1s?
>> > >>
>> > >> Another way to think about it is a running sum that resets upon
>> hitting
>> a
>> > >> zero
>> > >>
>> > >> input=:1 0 1 1 1 1 0 1
>> > >>
>> > >> expected=: 1 0 1 2 3 4 0 1
>> > >> ----------------------------------------------------------------------
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>> > >>
>> > >
>> > >
>> > ----------------------------------------------------------------------
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