Closed jpsamaroo closed 6 years ago
Not sure exactly why this happens, but FWIW you can use an anonymous function expression to get the right thing:
@flow function (x, t)
return (x, t)
end
While that example does work, trying the same thing on my original example now throws a different error:
julia> @flow function (x, t)
hidden = σ( Wxh*x + Whh*hidden )
y = σ( Why*hidden + Wxy*x )
return (y, t)
end
ERROR: UndefVarError: σ not defined
The readme is just really old, and the package has changed quite a lot, so that stuff wasn't maintained. I've updated things to be more consistent (and fixed the readme as well).
Slightly-modified example from the README:
Returning either
y
ort
separately works just fine, of course.This is happening on the master branch of DataFlow with Julia 0.6.2