I'm sweeping through all the Traits in a large program, and many of our traits are synchronized. For example, consider HasTrait objects of the structure:
a = Material1.ShellMaterial
b = Material2.CoreMaterial
c = Material3.MaterialX
In our application, it turns out that a and c are synchronized traits. In other words, Material3.MaterialX is the same as Material1.ShellMaterial, and they have set using sync_trait() (HasTraits API).
Is it possible to inspect a,b,c and dynamically determine that a and c are synchronized?
The goal is to plot all of these, but hide redundant plots from the user. Typical comparisons between these like a==c return False, despite these objects representing the same data.
As far as I know there is no official API that allows to inspect the synchronization state of traits.
Of course, you could simply call the sync_trait() method again to make sure that the traits are synchronized (or not synchronized, if you use remove=True). As a result, you will know the synchronization state of the traits.
If you do not want to change the synchronization state, you have to rely on non-official API functions, which are not documented and possibly subject to change -- so use them at your own risk.
from traits.api import HasTraits, Float
class AA(HasTraits):
a =Float()
class BB(HasTraits):
b = Float()
aa = AA()
bb = BB()
aa.sync_trait("a", bb, "b")
# aa.a and bb.b are synchronized
# Now we use non-official API functions
info = aa._get_sync_trait_info()
synced = info.has_key("a") # True if aa.a is synchronized to some other trait
if synced:
sync_info = info["a"] # fails if a is not a synchronized trait
# sync_info is a dictionary which maps (id(bb),"b") to a tuple (wr, "b")
# If you do not know the id() of the HasTraits-object and the name of
# the trait, you have to loop through all elements of sync_info and
# search for the entry you want...
wr, name = sync_info[(id(bb), "b")]
# wr is a weakref to the class of bb, and name is the name
# of the trait which aa.a is synced to
cls = wr() # <__main__.BB at 0x6923a98>
Again, use at your own risk, but it works for me.