| Re: Histogram and bin sizes [message #58871 is a reply to message #58772] |
Thu, 21 February 2008 09:20  |
Conor
Messages: 138
Registered: February 2007
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Senior Member |
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On Feb 21, 12:19 pm, Conor <cmanc...@gmail.com> wrote:
> On Feb 21, 11:29 am, jeffnettles4...@gmail.com wrote:
>
>
>
>> On Feb 21, 9:05 am, Conor <cmanc...@gmail.com> wrote:
>
>>> On Feb 20, 2:43 pm, pgri...@gmail.com wrote:
>
>>>> jeffnettles4...@gmail.com wrote:
>>>> > I've always wondered why you have to use a constant bin size with
>>>> > HISTOGRAM().
>>>> > To quote J.D.'s famous tutorial: "a histogram
>>>> > represents nothing more than a fancy way to count." Doesn't an
>>>> > imposed constant bin size imply that this is the only way it's ok to
>>>> > count? I can think of several reasons i wouldn't want to do this - I
>>>> > used logarithmic bin sizes in my dissertation, for example (now i'm
>>>> > hoping someone isn't going to answer this post saying i screwed up in
>>>> > my dissertation :-) ).
>
>>>> I use logarithmic bins myself quite often, and the fact that a
>>>> logarithmic bin
>>>> size is the same as a constant bin size in log space, makes it is easy
>>>> to use histogram to get that. Less regulars binning don't work with
>>>> histogram, but nobody stops you from writing your own version to work
>>>> with them (it will not be as fast as histogram though).
>
>>>> Ciao,
>>>> Paolo
>
>>>> > And besides, Excel lets you use arbitrary bin
>>>> > sizes....and if Excel lets you do it, it has to be ok, right???? ;-)
>
>>>> > Jeff
>
>>> You can always do whatever binning you want, you just have to
>>> transform your data to the new space and then bin it constantly. Why
>>> doesn't histogram let you use aribtrary binsizes? Not being an IDL
>>> developer I don't know for sure, but I would guess it's a speed
>>> issue. The simpler a program is the faster it is. I use histogram
>>> all the time because it's one of the speedier programs in IDL. It
>>> would make me very sad if in order to make histogram more flexible, it
>>> also became much slower, especially since by transforming my data set
>>> I can use aribtrary bin sizes for histogram.
>
>> That actually sounds like what i've done in the past. For my
>> dissertation i needed two kinds of histograms: logarithmic bins
>> (which was fine, no trouble there) and bins that had arbitrary sizes.
>> For the latter, i would either do the histograms in Excel (yuck) or
>> compute two or three histograms in IDL using histogram() with
>> different bin sizes and sort of do some "mixing and matching" of the
>> resulting arrays to get what i wanted. Of course, David hadn't
>> written his awesome histoplot routine yet then either :( Anyway, i'm
>> up against the arbitrary bin sizes problem again for a project i'm
>> doing for someone, and it got me wondering whether this situation is
>> just so rare it wasn't worth supporting in histogram(). I wouldn't
>> want to lose histogram's speed either though.
>
>> Jeff
>
> Arbitrary bin sizes should be pretty easy to program. You just need
> to map your data points appropriately. For instance if you had the
> data set:
>
> x = randomu(seed,100)
>
> and you wanted bins from:
> [0-.1,.1-.3,.3-.35,.35-.8,.8-1]
>
> you might do something like this:
>
> x = randomu(seed,100)
> bins = [ [0,.1], [.1,.3], [.3,.35], [.35,.8], [.8,1] ]
> newx = fltarr(n_elements(x))
> for i=0,n_elements(bins[0,*])-1 do begin
> w = where( x ge bins[0,i] and x lt bins[1,i], c )
> if c gt 0 then newx[w] = i+.5
> endfor
>
> hist = histogram(newx,binsize=1.0,min=0)
> plothist,newx
Obviously you'll have to manaully set the x-axis labels... This might
need a little tweaking but it should give the general idea.
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