randomAffine3d() claim to take function handle for shear, but parser fails it
5 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
According to the doc of randomAffine3d:
Shear — Range of shear
[0 0] (default) | 2-element numeric vector | function handle
...
A function handle. The function must accept no input arguments and return the shear angle as a numeric scalar. Use a function handle to pick a shear angle from a disjoint interval or using a nonuniform probability distribution.
When using this function e.g.
function A = selectShear1
A = -(rand(1,1)*5+5);
end
the validator (below) returns with this error message:
'The value of 'Shear' is invalid. Expected Shear to be one of these types:
double, single, uint8, uint16, uint32, uint64, int8, int16, int32, int64
Instead its type was function_handle.'
which seems to be alright given:
function TF = validateShear(val)
iValidateNumericRange('Shear',val,'<',90,'>',-90);
TF = true;
end
That it after all doesn't take function handles...
1 Kommentar
Steven Lord
am 24 Mär. 2023
Can you show us the code you've written to call randomAffine3d using this function handle?
Antworten (1)
Siehe auch
Kategorien
Mehr zu Deep Learning Toolbox finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!