| Re: array manipulation (TOTAL-ing or MEDIAN-ing) in uneven bins [message #82490 is a reply to message #82403] |
Wed, 12 December 2012 14:19   |
Jeremy Bailin
Messages: 618
Registered: April 2008
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Senior Member |
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On 12/12/12 4:18 PM, Jeremy Bailin wrote:
> On 12/12/12 4:03 PM, Jeremy Bailin wrote:
>> On 12/12/12 10:16 AM, havok2063@gmail.com wrote:
>>>
>>> I have several unrelated problems that I'm solving in the same
>>> efficient way (with loops). I'm trying to perform some array
>>> operation on an array, according to a list of (let's call them) uneven
>>> bins.
>>>
>>> I have an array, say d, of 146 elements. I have a separate array that
>>> represents uneven bins that I want to perform the operation on, like
>>> MEDIAN, or TOTAL. For example,
>>>
>>> ntot = [15,45,56,90,116,146]
>>>
>>> I want as output an array, of 6 elements, that contains the MEDIAN (or
>>> TOTAL) of array d according to the indices listed in ntot.
>>>
>>> So the 1st element would contain median(d[0:14],/even), the 2nd
>>> median(d[15:44],/even), etc....
>>>
>>> Or the same thing with total....total(d[0:14]), total(d[15:44]) , etc...
>>>
>>> Right now I'm looping over the number of elements in ntot to do this
>>> and I don't much care for loops.
>>>
>>> I don't think this is quite the same thing as the example given in the
>>> "Horror and Disgust of Histogram" article nor does this sound like
>>> something I can do with value_locate, although I'm not too familiar
>>> with value_locate.
>>>
>>> Any ideas on this? Thanks a lot.
>>>
>>
>> As David says, this screams VALUE_LOCATE. And HISTOGRAM. They play very
>> nicely together for this sort of problem!
>>
>> First we need to label the bin for each element:
>>
>> nelements = 146
>> binlabel = value_locate(ntot, lindgen(nelements))
>>
>> Then use histogram to group the elements by bin label. Notice that the
>> way you've defined ntot, elements 0 through 14 will be labelled "-1" by
>> value_locate, so we start the histogram there:
>>
>> nbin = n_elements(ntot)
>> hist = histogram(binlabel, min=-1, max=nbin-1, reverse_indices=ri)
>>
>> And finally we do the usual loop through the reverse indices to
>> calculate the statistics:
>>
>> medianbin = fltarr(nbin)
>> totbin = fltarr(nbin)
>> for i=0L,nbin-1 do if hist[i] gt 0 then begin
>> these = ri[ri[i]:ri[i+1]-1]
>> medianbin[i] = median(d[these], /even)
>> totbin[i] = total(d[these])
>> endif
>>
>> -Jeremy.
>
> Actually, for the total you can do a lot better by using cumulative:
>
> runningtotal = total(d, /cumulative)
> totbin = runningtotal[ntot] - [0,runningtotal[ntot]]
>
> -Jeremy.
Ack, hit send to soon. That should be:
runningtotal = total(d, /cumulative)
totbin = runningtotal[ntot-1] - [0,runningtotal[ntot-1]]
-Jeremy.
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