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Hi, I'm trying to build a xarray dataset with mine own data, and the code is as following:
`numpy_array = data
lons = np.linspace(-180, 180, 73)
lats = np.linspace(87.5, -87.5, 71)
# mydatetime=
# dims = ('batch', 'time', 'channel', 'longitude', 'latitude')
times = pd.to_timedelta([f'{i:2d}:00:00' for i in range(data.shape[1])])
datetimes = pd.date_range(start=f'{start_year}-01-01', end=f'{end_year}-12-31 23:00:00', freq='H')
coords = {
'batch': range(data.shape[0]),
'time': ('time',times),
'channel':[0],
'lon': ('lon', lons),
'lat': ('lat',lats),
'datetime': (('batch','time'), [datetimes])
}
data_vars = {'Temp': (('batch', 'time', 'channel', 'lat', 'lon'), numpy_array[:, :, 0:1, :, :])}
xr_dataset = xarray.Dataset(data_vars=data_vars, coords=coords)`
but in the result dataset ,
the coord "datetime" which type is xarray.DataArray has own coords and dims "batch" and "time" , such like:
<xarray.DataArray 'datetime' (batch: 1, time: 17544)> Size: 140kB
array([['2020-01-01T00:00:00.000000000', '2020-01-01T01:00:00.000000000',
'2020-01-01T02:00:00.000000000', ...,
'2021-12-31T21:00:00.000000000', '2021-12-31T22:00:00.000000000',
'2021-12-31T23:00:00.000000000']], dtype='datetime64[ns]')
Coordinates:
* batch (batch) int64 8B 0
* time (time) timedelta64[ns] 140kB 00:00:00 ... 730 days 23:00:00
datetime (batch, time) datetime64[ns] 140kB 2020-01-01 ... 2021-12-31T23...
However if I load the example dataset "source-era5_date-2022-01-01_res-0.25_levels-37_steps-01.nc", whose coord "datetime" has "batch" and "time" dims ,but only the "time" coord. such like :
<xarray.DataArray 'datetime' (batch: 1, time: 3)> Size: 24B
array([['2022-01-01T00:00:00.000000000', '2022-01-01T06:00:00.000000000',
'2022-01-01T12:00:00.000000000']], dtype='datetime64[ns]')
Coordinates:
* time (time) timedelta64[ns] 24B 00:00:00 06:00:00 12:00:00
datetime (batch, time) datetime64[ns] 24B 2022-01-01 ... 2022-01-01T12:0...
Dimensions without coordinates: batch
So my question is how to remove the coord "batch" of the datetime in my dataset?
Thanks a lot!
Sincerely, Qc
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