I am no expert, but how about something like this:
function histrange{T<:Number,N}(v::AbstractArray{T,N}, n::Integer)
# Shortcut function.
t(x) = oftype(T,x)
if length(v) == 0
return t(Range(0,1,1))
end
lo, hi = minimum(v), maximum(v)
if hi == lo
step = t(1)
else
bw = (hi - lo) / n
e = t(10)^t(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*t(ceil(lo/step)-1)
Range(start,step,1+iceil((hi - start)/step))
end
Cheers,
Daniel.
On Thursday, 26 December 2013 07:47:09 UTC-5, Michael Fox 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
>
>
