Open fsaad opened 2 years ago
The .name attribute of a frozen rv_discrete.dist object does not correspond to the underlying class, i.e.,
.name
rv_discrete.dist
>>> from scipy.stats import norm >>> d = norm(loc=0, scale=1) >>> d.dist.name 'norm' >>> from scipy.stats import rv_discrete >>> d = rv_discrete(values=((1, 2), (.5, .5))).freeze() >>> d.dist.name 'Distribution'
This behavior will cause issue in serializing, since we rely on the name attribute to correspond to a scipy class: https://github.com/probcomp/sppl/blob/efff34fb3d3703247dd7001c36970069c5ac3825/src/compilers/spe_to_dict.py#L48
name
Related #121
The constructor of rv_discrete does accept a name attribute, which will need to be handled correctly when loading the scipy dist.
rv_discrete
The
.name
attribute of a frozenrv_discrete.dist
object does not correspond to the underlying class, i.e.,This behavior will cause issue in serializing, since we rely on the
name
attribute to correspond to a scipy class: https://github.com/probcomp/sppl/blob/efff34fb3d3703247dd7001c36970069c5ac3825/src/compilers/spe_to_dict.py#L48Related #121
The constructor of
rv_discrete
does accept aname
attribute, which will need to be handled correctly when loading the scipy dist.