On Aug 20, 3:40 pm, David Fanning <n...@dfanning.com> wrote:
> te...@atmsci.msrc.sunysb.edu writes:
>> Well, I am looking into the histogram procedure, but I am not getting
>> what I think the 25th and 75th quartiles should be. It seems
>> histogram is not so easy to master. What I am looking into is doing
>> the following:
>
>> data=randomu(sd,100)*100
>> box plot needs min, max, median which are straight forward:
>
>> min(data)
>> max(data)
>> median(data,/even)
>
>> For the quartiles I am trying:
>
>> lower_ind=where(data lt median(data,/even))
>> upper_ind=where(data gt median(data,/even))
>> qtr_25th=median(data[lower_ind(0):lower_ind(n_elements(lower _ind)-1)],/
>> even)
>> qtr_75th=median(data[upper_ind(0):upper_ind(n_elements(upper _ind)-1)],/
>> even)
>
>> However, I think this would work only for a monotonically increasing
>> array. I am not sure how to get 'data' like that. If anyone wants to
>> add to this, feel free.
>
> I calculate it like this:
>
> data=randomu(sd,100)*100
>
> minVal = min(data)
> maxVal = max(data)
> medianVal = median(data,/even)
>
> ; Find the quartiles.
> binsize = (maxVal - minVal) / 4.0
> h = Histogram(data, BINSIZE=binsize, REVERSE_INDICES=ri)
> qtr_25th = Median(data[ri[ri[0]:ri[2]-1]])
> qtr_75th = Median(data[ri[ri[2]:ri[4]-1]])
>
> Print, minVal, maxVal, medianVal, qtr_25th, qtr_75th
> END
>
> With 100 values I get this:
>
> 0.401314 98.0063 58.9402 20.0477 73.3419
>
> With 10000 values, I get this, which leads me to think the
> algorithm might be correct:
>
> 0.0249010 99.9960 49.9658 25.0268 74.8059
>
> Cheers,
>
> David
>
> --
> David Fanning, Ph.D.
> Fanning Software Consulting, Inc.
> Coyote's Guide to IDL Programming:http://www.dfanning.com/
> Sepore ma de ni thui. ("Perhaps thou speakest truth.")
Hi, Suppose data is something simple like:
data=[2,3,5,7,7,10,11,11,12,15,16,17,17]
I get a 75th quartile of 11.0. Shouldn't I get around 15? Maybe I
made a mistake. However, what about if I revise what I wrote before
to:
sorted_data=data(sort(data))
lower_ind=where(sorted_data lt median(sorted_data,/even))
upper_ind=where(sorted_data gt median(sorted_data,/even))
qtr_25th=median(sorted_data[lower_ind(0):lower_ind(n_element s(lower_ind)-1)],/
even)
qtr_75th=median(sorted_data[upper_ind(0):upper_ind(n_element s(upper_ind)-1)],/
even)
Howie
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