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
> > >> ----------------------------------------------------------------------
> > >> For information about J forums see
> > http://www.jsoftware.com/forums.htm
> > >>
> > >
> > >
> > ----------------------------------------------------------------------
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