Daily mean from 3-hourly data

[Dhirendra Kumar]

Hello all, 

I am a new user of Iris. I am trying to calculate daily mean from the 3-hourly data. This means I have to compute a mean across all the time steps in a particular date. 

The shape of the cube is:

time: 49672; latitude: 241; longitude: 480

and I am trying to reduce it to 

time: 6209; latitude: 241; longitude: 480  

is there a way to do this in the Iris. 

Looking forward to the answers. 

Regards

Dhirendra

[Simon Peatman]

Hi Dhirendra,

I would do the following:

import iris.coord_categorisation

iris.coord_categorisation.add_year(cube, 'time')

iris.coord_categorisation.add_day_of_year(cube, 'time')

daily_mean = cube.aggregated_by(['year', 'day_of_year'], iris.analysis.MEAN)

This adds two auxiliary coordinates, one showing the year and one showing the day number in the year.  Then it computes the means by aggregating together any times with the same year and day number.

Simon

[Lucia Deaconu]

Hello all,

I have two gridded cubes (AOD and AAOD) with shape [time: 2920; latitude: 96; longitude: 192], and I want to mask one in function of the other, eg: mask all AAOD values for values of AOD > 0.3.
Is there a way of doing this and keeping them gridded? I haven’t found an example or a clear way of doing it.

Thanks,
Lucia

[Simon Peatman]

Lucia,

You want to use the mask_cube function - see documentation here: https://scitools-iris.readthedocs.io/en/stable/generated/api/iris/util.html#iris.util.mask_cube

You can do:

aaod = iris.util.mask_cube(aaod, aod.data > 0.3)

Note that this function, confusingly, masks the existing cube in place and also returns it (i.e., even if you did new_cube = iris.util.mask_cube(aaod, aod.data > 0.3) you would find that aaod has been masked, and aaod and new_cube are actually the same thing).  See: https://github.com/SciTools/iris/issues/3717

Simon

[Lucia Deaconu]

Hi Simon,

Very helpful! Thank you!

Lucia

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