>> [fexpr~ abs($x[0] - $x[-1]) > 0.5] >> [threshold~ 0.5 0 0.5 0]
sick On Sat, Mar 2, 2019 at 7:31 PM Alexandre Torres Porres <[email protected]> wrote: > connect output of [phasor~] into: > > [fexpr~ abs($x[0] - $x[-1]) > 0.5] > > then use: > > [threshold~ 0.5 0 0.5 0] > > Em sáb, 2 de mar de 2019 às 11:05, Claude Heiland-Allen < > [email protected]> escreveu: > >> On 2019-03-02 13:20, Orm Finnendahl wrote: >> > is there some easy way in vanilla pd to get a bang on each phasor~ >> > wraparound which doesn't involve polling with snapshot~? >> >> https://lists.puredata.info/pipermail/pd-list/2006-12/044634.html >> >> -- >> https://mathr.co.uk >> >> >> >> _______________________________________________ >> [email protected] mailing list >> UNSUBSCRIBE and account-management -> >> https://lists.puredata.info/listinfo/pd-list >> > _______________________________________________ > [email protected] mailing list > UNSUBSCRIBE and account-management -> > https://lists.puredata.info/listinfo/pd-list > -- ____________________ m.e.grimm, m.f.a, ed.m. cornell u., tc3 megrimm.net ____________________
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