Can't automatically convert KeyedTuple (from sqlalchemy) to Julia's NamedTuple
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Beschreibung
I use PyCall to call sqlalchemy from Julia, which uses a class called `KeyedTuple` to represent rows resulting from a query and I want to convert those automatically to julia's `NamedTuple`. I expected something like this to work:
```julia
using PyCall
sa = pyimport("sqlalchemy")
pytype_mapping(sa."util"."KeyedTuple", NamedTuple)
Base.convert(::Type{NamedTuple}, o::PyObject) = NamedTuple{Symbol.(o._fields)}([getproperty(o, f) for f in Symbol.(o._fields)])
x = sa.util.KeyedTuple([1, 2, 3], labels=["one", "two", "three"]) # returns (1, 2, 3)
```
This is no surprise, since:
```julia
o = @pycall sa.util.KeyedTuple([1, 2, 3], labels=["one", "two", "three"])::PyObject
pytype_query(o) # Returns `Tuple{Integer,Integer,Integer}`
```
I guess a simple way to make my thing work would be to alter the order in which `pytype_query` does things> I mean first look for a match in the `pytype_queries` array and then proceed with the rest...?
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Rechercherichtung
Start with the pytype_query and pytype_queries entry points mentioned in the issue, then reproduce the Julia/PyCall example using SQLAlchemy's util.KeyedTuple. Determine whether the type-specific mapping should be checked before the existing tuple handling, and consider the conversion complete when the example produces a NamedTuple with the expected field names and values.
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Bewertung
- Tech-Stack
- julia, python, sqlalchemy
- Bereich
- databases
- Issue-Typ
- Feature
- Schwierigkeit
- 3/5
- Geschätzter Aufwand
- 1-2 Tage
- Aktivitätsstatus
- Veraltet
- Klarheit
- Größtenteils klar
- Anfängerfreundlichkeit
- 35/100