LSSTApplications
18.0.0+106,18.0.0+50,19.0.0,19.0.0+1,19.0.0+10,19.0.0+11,19.0.0+13,19.0.0+17,19.0.0+2,19.0.0-1-g20d9b18+6,19.0.0-1-g425ff20,19.0.0-1-g5549ca4,19.0.0-1-g580fafe+6,19.0.0-1-g6fe20d0+1,19.0.0-1-g7011481+9,19.0.0-1-g8c57eb9+6,19.0.0-1-gb5175dc+11,19.0.0-1-gdc0e4a7+9,19.0.0-1-ge272bc4+6,19.0.0-1-ge3aa853,19.0.0-10-g448f008b,19.0.0-12-g6990b2c,19.0.0-2-g0d9f9cd+11,19.0.0-2-g3d9e4fb2+11,19.0.0-2-g5037de4,19.0.0-2-gb96a1c4+3,19.0.0-2-gd955cfd+15,19.0.0-3-g2d13df8,19.0.0-3-g6f3c7dc,19.0.0-4-g725f80e+11,19.0.0-4-ga671dab3b+1,19.0.0-4-gad373c5+3,19.0.0-5-ga2acb9c+2,19.0.0-5-gfe96e6c+2,w.2020.01
LSSTDataManagementBasePackage
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The class lsstDebug can be used to turn on debugging output in a non-intrusive way.
For example, the variable lsstDebug.Info
("lsst.meas.astrom.astrom").debug is used to control debugging output from the lsst.meas.astrom.astrom module.
It is always safe to interrogate lsstDebug
; for example lsstDebug.Info
("Robert.Hugh.Lupton").isBadPerson will return False.
The convention is that the name ("lsst.meas.astrom.astrom") is the __name__
of the module, so the source code will typically look something like:
which will print False
unless lsstDebug.Info(__name__)
.display has somehow been set to True
.
Why is this interesting? Because you can replace lsstDebug.Info
with your own version, e.g. if you put
into a file debug.py and
into foo.py, then
but
The command line task interface supports a flag –debug
to import debug.py from your PYTHONPATH