[Qgis-user] Timestep for (Temporal) Mesh/netcdf layers?

[Richard Duivenvoorde]

Ok, found it. It is in the temporalCapabilities...

So given a loaded netcdf/mesh with time.
Easiest is the firstTimeStepDuration(i) of QgsMeshDataProviderTemporalCapabilities:

  >>> p = iface.activeLayer().dataProvider().temporalCapabilities()
  >>> p
  <qgis._core.QgsMeshDataProviderTemporalCapabilities object at 0x7fa958267e20>
  >>> p.timeExtent()
  <QgsDateTimeRange:[2022-11-07T07:00:00Z, 2022-11-08T06:00:00Z]>
  >>> p.temporalUnit()
  <TemporalUnit.Hours: 3>   <<< ?????
  # datasetTime() Returns the relative time in milliseconds of the dataset
  >>> p.datasetTime(QgsMeshDatasetIndex(0,0))
  3600000
  >>> p.datasetTime(QgsMeshDatasetIndex(0,1))
  7200000
  >>> p.datasetTime(QgsMeshDatasetIndex(0,2))
  10800000
  # firstTimeStepDuration() Returns the duration of the first time step of the dataset group with index \a group
  # in milliseconds
  p.firstTimeStepDuration(0)
  3600000  == 3600 s = 1 hours
  p.firstTimeStepDuration(0)
  21600000  == 21600 s == 360 minute == 6 hours

Only strange thing is that the temporalUnit() always returns 'hours', while the actual data time is relative in minutes...

Regards,

Richard Duivenvoorde

On 3/8/23 10:46, Richard Duivenvoorde via QGIS-User wrote:
> Hi,
> 
> (trying here too, sorry for cross posting)
> 
> When loading (temporal) Netcdf files (as MeshLayers), the 'timestep'
> that you see in the Temporal Controller does not change.
> 
> I had a look at the api, but could not find any indication of knowledge
> by the QgsMeshLayer of the size of the timesteps of the data.
> 
> Within netcdf's it is common to define the timesteps as something like:
> "minutes since 2022-11-27 16:00:00.0Z"
> then the first set has 0, and then the next one for example 60 for
> hourly data or 10 for 10-minute data.
> 
> Question: IS a temporal mesh layer aware of the timesteps used?
> 
> Or if not: I seem not to be able to inspect the data array of the 'time'
> dimension. Should/can I?
> 
> IF so, we could we could deduct from the first 2 or 3 timesteps, the
> size of the timestep.
> 
> Anybody a hint or idea?
> 
> Regards,
> 
> Richard Duivenvoorde
> 
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