Big thx, it is good way, Intresting is also this dagr = diff(a) ./ diff(gradient(x))
and another offset avg ... Paul W dniu wtorek, 27 maja 2014 17:37:41 UTC+2 użytkownik Tomas Lycken napisał: > > If you have the y-values in a sorted vector, and the x-values in another > sorted vector (or if they're just the index you don't need them), you can > get the discrete derivative by using diff: > > da = diff(a) ./ diff(x) > > Now, the tangential tilt is less than 45 degrees whenever that derivative > is larger than -1, so > > a[da .< -1] > > will give you the portion of a for which the slope is steeper than 45 > degrees. (I might be off by one here, so you might want to experiment a > little and see if you need to include one more, or one less, data point to > get exactly what you want...) > > // T > > On Tuesday, May 27, 2014 4:53:52 PM UTC+2, paul analyst wrote: >> >> Thanks for the cool link. >> Generally this is the point where the tangential tilt is 45 degrees. >> Paul >> >> W dniu wtorek, 27 maja 2014 15:06:25 UTC+2 użytkownik Harlan Harris >> napisał: >>> >>> Paul, I don't believe that this is a well-posed question. Determining >>> what is the tail of a distribution and what isn't is very much >>> problem-dependent. "getting flat" depends entirely on the scale of your >>> data, and isn't meaningful. You could do what people constructing boxplots >>> do, and use some multiple of the inter-quartile range from the median, but >>> that still depends on the source, distribution, and meaning of your data. >>> In any case, you'd be better off asking this somewhere like CrossValidated ( >>> http://stats.stackexchange.com/) -- this isn't a Julia question. >>> >>> >>> On Tue, May 27, 2014 at 8:22 AM, paul analyst <[email protected]>wrote: >>> >>>> Does not work always, distributions are different. >>>> How to find the number of elements of the vector from which the chart is >>>> getting flat (where is the beginning of the tail?) >>>> Paul >>>> >>>> >>>> W dniu wtorek, 27 maja 2014 12:19:09 UTC+2 użytkownik Yuuki Soho >>>> napisał: >>>> >>>>> The simplest way to do it is probably to use a quantile: >>>>> >>>>> >>>>> a=[5 3 2 1.5 1.1 1 0.8 0.25 0.2 0.16] >>>>> >>>>> q = quantile(vec(a),0.1) >>>>> >>>>> a = a[1:cut] >>>>> >>>>> >>>>> >>>
