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
example_lat_wn_file.nc (19.5 KB) example_lat_wn_file.nc |