RE: Empirical CDF and InterpolatingFunction
- To: mathgroup at smc.vnet.net
- Subject: [mg36602] RE: [mg36555] Empirical CDF and InterpolatingFunction
- From: "DrBob" <drbob at bigfoot.com>
- Date: Fri, 13 Sep 2002 01:15:00 -0400 (EDT)
- Reply-to: <drbob at bigfoot.com>
- Sender: owner-wri-mathgroup at wolfram.com
I was completely wrong about the pattern list_?(VectorQ[#, NumericQ] &). I had forgotten about that notation and didn't think to look it up. Sorry! Bobby -----Original Message----- From: Mark Fisher [mailto:mark at markfisher.net] To: mathgroup at smc.vnet.net Subject: [mg36602] [mg36555] Empirical CDF and InterpolatingFunction I'm trying to write a fast empirical cummulative distribution function (CDF). Empirical CDFs are step functions that can be expressed in terms of a Which statement. For example, given the list of observations {1, 2, 3}, f = Which[# < 1, 0, # < 2, 1/3, # < 3, 2/3, True, 1]& is the empirical CDF. Note that f /@ {1, 2, 3} returns {1/3, 2/3, 1} and f is continuous from the right. When the number of observations is large, the Which statement evaluates fairly slowly (even if it has been Compiled). Since InterpolationFunction evaluates so much faster in general, I've tried to use Interpolation with InterpolationOrder -> 0. The problem is that the resulting InterpolatingFunction doesn't behave the way (I think) it ought to. For example, let g = Interpolation[{{1, 1/3}, {2, 2/3}, {3, 1}}, InterpolationOrder -> 0] Then, g /@ {1, 2, 3} returns {2/3, 2/3, 1} instead of {1/3, 2/3, 1}. In addition, g is continuous from the left rather than from the right. Obviously I am not aware of the considerations that went into determining the behavior of InterpolationFunction when InterpolationOrder -> 0. So I have two questions: (1) Does anyone have any opinions about how InterpolatingFunction ought to behave with InterpolationOrder -> 0? (2) Does anyone have a faster way to evaluate an empirical CDF than a compiled Which function? By the way, here's my current version: CompileEmpiricalCDF[list_?(VectorQ[#, NumericQ] &)] := Block[{x}, Compile[{{x, _Real}}, Evaluate[ Which @@ Flatten[ Append[ Transpose[{ Thread[x < Sort[list]], Range[0, 1 - 1/#, 1/#] & @ Length[list] }], {True, 1}]] ]]] --Mark