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] >>>> >>>> >>>> >>
