You could also use a macro.
-- John
On Dec 26, 2013, at 9:33 AM, Kevin Squire <kevin.squ...@gmail.com> wrote:
> The only issue in merging these seems to be the use of floor and ceil vs.
> ifloor and iceil. One possiblilty would be to replace those two calls with
> oftype(T, floor(log10(bw)) and oftype(T, ceil(lo/step)-1), respectively.
>
> A perhaps better change would be to create versions of ceil and floor (and
> other such functions) which take a type as the first parameter and dispatch
> to ceil/floor or iceil/ifloor accordingly.
>
> Cheers,
> Kevin
>
>
>
> On Thu, Dec 26, 2013 at 4:47 AM, Michael Fox <415...@gmail.com> wrote:
> It's kind of painful to see codes repeating themselves like this. Here's one
> I just ran across in `statistics.jl`. Is there something about the type or
> dispatch system that encourages or even requires such repetition? Can you
> think of a better idea?
>
> ## nice-valued ranges for histograms
> function histrange{T<:FloatingPoint,N}(v::AbstractArray{T,N}, n::Integer)
> if length(v) == 0
> return Range(0.0,1.0,1)
> end
> lo, hi = minimum(v), maximum(v)
> if hi == lo
> step = 1.0
> else
> bw = (hi - lo) / n
> e = 10.0^floor(log10(bw))
> r = bw / e
> if r <= 2
> step = 2*e
> elseif r <= 5
> step = 5*e
> else
> step = 10*e
> end
> end
> start = step*(ceil(lo/step)-1)
> Range(start,step,1+iceil((hi - start)/step))
> end
>
> function histrange{T<:Integer,N}(v::AbstractArray{T,N}, n::Integer)
> if length(v) == 0
> return Range(0,1,1)
> end
> lo, hi = minimum(v), maximum(v)
> if hi == lo
> step = 1
> else
> bw = (hi - lo) / n
> e = 10^max(0,ifloor(log10(bw)))
> r = bw / e
> if r <= 1
> step = e
> elseif r <= 2
> step = 2*e
> elseif r <= 5
> step = 5*e
> else
> step = 10*e
> end
> end
> start = step*(iceil(lo/step)-1)
> Range(start,step,1+iceil((hi - start)/step))
> end
>
>
