very good point, walter. i think we could find some cool ways to leverage this intelligence for our users after serving up the flattened version based on the simple range that they're expecting to see. the clarity is helpful in getting some creative ideas moving, so thanks.
best, -- *John Blythe* Product Manager & Lead Developer 251.605.3071 | [email protected] www.curvolabs.com 58 Adams Ave Evansville, IN 47713 On Mon, Dec 19, 2016 at 4:17 PM, Walter Underwood <[email protected]> wrote: > Percentiles are far more useful than that linear approximation. That is > just slope and intercept, basically two numbers. > > With percentiles, I can answer the question “how fast is the search for > 95% of my visitors?” With that linear interpolation, I don’t know anything > about my customers. > > wunder > Walter Underwood > [email protected] > http://observer.wunderwood.org/ (my blog) > > > > On Dec 19, 2016, at 12:51 PM, John Blythe <[email protected]> wrote: > > > > gotcha. yup, that was the back up plan so i think i'll go that route for > > now. > > > > thanks for the info! > > > > best, > > > > -- > > *John Blythe* > > Product Manager & Lead Developer > > > > 251.605.3071 | [email protected] > > www.curvolabs.com > > > > 58 Adams Ave > > Evansville, IN 47713 > > > > On Mon, Dec 19, 2016 at 3:41 PM, Toke Eskildsen <[email protected]> > > wrote: > > > >> John Blythe <[email protected]> wrote: > >>> if the range is 0 to 100 then, for my current purposes, i don't care if > >> the > >>> vast majority of the values are 92, i would want 25%=>25, 50%=>50, and > >>> 75%=>75. so is there an out-of-the-box way to get the percentiles to > >>> correspond to the range itself rather than the concentration of > distinct > >>> values? > >> > >> Then it is not percentiles. And I don't know of any build-in function > that > >> returns them directly. > >> > >> But as you have the min and max, you can just do > >> 25%: (max-min)*0.25+min > >> 50%: (max-min)*0.5+min > >> 75%: (max-min)*0.75+min > >> > >> But of course, that won't guarantee that you match the distinct values. > If > >> you want that, you'll have to iterate the list of distinct values (hope > >> it's not too large) and pick out the nearest ones. > >> > >> - Toke Eskildsen > >> > >
