Hi all, There seems to be a bug of some sort in evaluating the ppf method of the scipy.stats.triang.ppf function. Evaluating the distribution with a location parameter 1 or greater seems to problematic. I am looking for confirmation on this behavior and suggestions for work around. Thanks in advance -- Leon Adams
On Tue, May 25, 2010 at 20:33, Leon Adams <skorpio11@gmail.com> wrote:
Hi all, There seems to be a bug of some sort in evaluating the ppf method of the scipy.stats.triang.ppf function. Evaluating the distribution with a location parameter 1 or greater seems to problematic. I am looking for confirmation on this behavior and suggestions for work around.
Please show us exactly what you did, exactly what results you got, and what results you expected. Please copy-and-paste rather than summarizing. -- Robert Kern "I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth." -- Umberto Eco
On May 25, 2010, at 7:33 PM, Leon Adams wrote:
Hi all, There seems to be a bug of some sort in evaluating the ppf method of the scipy.stats.triang.ppf function. Evaluating the distribution with a location parameter 1 or greater seems to problematic. I am looking for confirmation on this behavior and suggestions for work around.
Make sure you understand the shape parameter for this distribution and how the location and scale parameter interact with it. The required shape parameter is the peak of the pdf as a percentage of the width. The location parameter is the start of the non-zero portion of the triangular-shaped pdf. The scale parameter is the width of the non-zero portion of the pdf.
from scipy.stats import triang
triang.ppf([0.1, 0.5, 0.8], 0.5, loc=20, scale=10) array([ 22.23606798, 25. , 26.83772234])
-Travis
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Leon Adams -
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Travis Oliphant