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Meridional mean of latitude versus wavenumber data

Added by Oliver Watt-Meyer over 10 years ago

Hi,

I have some data that is in latitude versus zonal wavenumber space (and also has time and vertical dimensions). I wish to compute meridional means of this data, but when I use mermean, there is warning that the operator is using equal weights for all latitudes, presumably because my data does not have a well-defined horizontal grid. I can set the grid to a lonlat grid and then compute the meridional average, but then the coordinates for the wavenumber dimension are changed to longitudes.

So I guess is there any way to force mermean to compute meridional means with the typical cos(lat) weighting, without having a properly defined horizontal grid? Or is CDO not the right tool to use for data that is not in strictly real-space dimensions?

I have attached an example that has been restricted to one vertical level and one timestep.

Thanks!

Oliver