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LSSTApplications
16.0-10-g0ee56ad,16.0-11-ga33d1f2+2,16.0-12-g3ef5c14+1,16.0-12-g71e5ef5+14,16.0-12-gbdf3636+2,16.0-13-g118c103+2,16.0-13-g8f68b0a,16.0-15-gbf5c1cb,16.0-16-gfd17674+2,16.0-17-g7c01f5c+2,16.0-18-g0a50484,16.0-20-ga20f992+3,16.0-21-g0e05fd4+2,16.0-21-g15e2d33,16.0-22-g62d8060,16.0-22-g847a80f,16.0-24-gfa57b64+1,16.0-28-g3990c221,16.0-3-gf928089+2,16.0-32-g88a4f23,16.0-34-gd7987ad,16.0-36-g6d07a18,16.0-4-g10fc685,16.0-4-g18f3627+24,16.0-4-g5f3a788+25,16.0-5-gaf5c3d7+1,16.0-5-gcc1f4bb,16.0-6-g3b92700,16.0-6-g4412fcd+2,16.0-6-g7235603+1,16.0-68-gb45e8e9d,16.0-7-g0913a87,16.0-7-g56728dd,16.0-8-g14ebd58,16.0-8-g2df868b,16.0-8-g4cec79c+1,16.0-81-g9604f5049,16.0-9-g5400cdc,16.0-9-ge6233d7+1,master-g2880f2d8cf+1,w.2019.05
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
1.8.13