[SciPy-User] specifying range in scipy.stats.truncnorm
Fri Jul 27 12:58:40 CDT 2012
On Fri 27 Jul 2012 12:07:15 PM CDT, firstname.lastname@example.org wrote:
> On Fri, Jul 27, 2012 at 12:30 PM, Joon Ro <email@example.com> wrote:
>> I tried to use scipy.stats.truncnorm and found the way to specifying the
>> parameters of truncated normal very confusing.
>> I expected a, b parameter to be the specification of the interval where I
>> want to truncate the distribution at, but it is not the case when the normal
>> I want to use is not standard.
>> According to the documentation, I need to standardize my values - for
>> example, if I want to have a truncated normal with mean 0.5, variance 1, on
>> [0, 1] interval, I need to do:
>> myclip_a = 0
>> myclip_b = 1
>> my_std =1
>> a, b = (myclip_a - my_mean) / my_std, (myclip_b - my_mean) / my_std
>> rv = truncnorm(a, b, loc=my_mean, scale=my_std)
>> Which is unnecessarily complicated in my opinion. Since we have to provide
>> location and scale parameter anyway, why not make truncnorm to accept the
>> actual interval values (in this case, a, b = 0, 1) instead and do the
>> standardization internally? I think it would be more intuitive that way.
> I agree there are several cases of distributions where the
> parameterization is not very intuitive or common. The problem is loc
> and scale and the corresponding transformation of the support is done
> So, I don't think it's possible to change this without a change in the
> generic setup for the distributions or writing a specific dispatch
> function or class that does the conversion.
> I think, changing the generic setup would break the standard behavior
> of distributions that have a predefined finite support limit, like
> those that are defined for positive real numbers, a=0, or rdist with
> a=-1, b=1.
I just took a look at the code, and I agree.
I wonder if it would be possible to add a couple of more parameters (in
this case, representing the not-standardized interval) with default
None to the generic rv_continuous class and when they are passed
instead of a and b, let a distribution specific function do the
standardization and calculate a and b.
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