iris.analysis.maths.mean?

[Paul Halloran]

Hi All,

I've recently been making a lot of use of 'cube_out = iris.analysis.maths.divide/add/multiply/subtract(cube_a,cube_b)' etc. to look at differences/similarities between fields from different models.

It would be great to be able to do the same but mean the data - e.g.:

cube_out iris.analysis.maths.mean(cube_a,cube_b,cube_c)

but this is not an option. I know this can be got round easily, but the iris.analysis.maths... framework is nice and easy, does anyone know of a similar way to simply mean two separate cubes - there may well be something obvious that I'm missing...?

Thanks in advance,
Paul

[Andrew Dawson]

Currently what you are describing does not exist. However, you may not know that cubes override the basic maths operators to use iris.analysis.add etc., so you can use the normal operators with cubes and things just work. So to take the mean of three cubes you just need to do:

mean = (cube_a + cube_b + cube_c) / 3.

If this is too repetitive you could always write a simple wrapper:

def mean_cube(*cubes):
    n = float(len(cubes))
    return reduce(iris.analysis.maths.add, cubes) / n

which can be called with any number of cubes as the argument, for example:

mean = mean_cube(cube_a, cube_b, cube_c)

Have I addressed your question? I wasn't sure if that was what you meant...

[Paul Halloran]

Hi Andrew, that answers my question exactly, thank you very much.

The reduce function is also new to me, so that is really useful.

Thanks!
Paul

[esc24]

It's getting late on a friday afternoon, so I may be talking nonsense, but does scipy.mean([cube_a, cube_b, cube_b]) not work? Or even scipy.mean(cubelist)?

[Stephan Hoyer]

You can't use scipy.mean (or np.mean, which is actually what scipy.mean calls under the covers), because it expects the arguments to be numpy arrays.

However, you can use Python's built-in "sum", which works on any data type, since sum([a,b,c]) is effectively expanded to a+b+c.

So my preferred solution is actually:

def mean(*args):
    return sum(args) / float(len(args))

You can call this mean function on any arguments, whether they are numpy arrays or cube or whatever.

Cheers,

Stephan

[Andrew Dawson]

It's getting late on a friday afternoon, so I may be talking nonsense, but does scipy.mean([cube_a, cube_b, cube_b]) not work? Or even scipy.mean(cubelist)?

Actually I think you are right, something like numpy.mean(cubelist) seems to work absolutely fine.

You can't use scipy.mean (or np.mean, which is actually what scipy.mean calls under the covers), because it expects the arguments to be numpy arrays.

It does work because the input list/cubelist will be turned into a numpy array where each element is a cube. 

[Stephan Hoyer]

Here's what I get when I tried this:

>>> print np.mean(cubes)

---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
<ipython-input-255-b689c50cf263> in <module>()
----> 1 print np.mean(cubes)


/Users/shoyer/dev/climatology-research/library/python/target/climatology/lib/python2.7/site-packages/numpy/core/fromnumeric.pyc in mean(a, axis, dtype, out, keepdims)
   2714
   2715     return _methods._mean(a, axis=axis, dtype=dtype,
-> 2716                             out=out, keepdims=keepdims)
   2717
   2718 def std(a, axis=None, dtype=None, out=None, ddof=0, keepdims=False):


/Users/shoyer/dev/climatology-research/library/python/target/climatology/lib/python2.7/site-packages/numpy/core/_methods.pyc in _mean(a, axis, dtype, out, keepdims)
     65                 ret, rcount, out=ret, casting='unsafe', subok=False)
     66     else:
---> 67         ret = ret.dtype.type(ret / rcount)
     68
     69     return ret


AttributeError: 'Cube' object has no attribute 'dtype'


> /Users/shoyer/dev/climatology-research/library/python/target/climatology/lib/python2.7/site-packages/numpy/core/_methods.py(67)_mean()
     66     else:
---> 67         ret = ret.dtype.type(ret / rcount)
     68


>>> print np.sum(cubes)
unknown / (Celsius)                 (time: 23351; station: 3)
     Dimension coordinates:
          time                           x               -
          station                        -               x
     Auxiliary coordinates:
          latitude                       -               x
          longitude                      -               x
          station_id                     -               x

So bizarrely, it appears that np.sum works but np.mean doesn't.

[Andrew Dawson]

That is odd because this example works for me:

import numpy as np
import numpy.ma as ma
import iris


a1 = ma.array([[1., 2., 3.], [4., 5., 6.]], mask=[[0, 1, 0], [0, 0, 1]])
a2 = ma.array([[8., 2., 1.], [3., 1., 4.]], mask=[[0, 1, 0], [0, 0, 1]])
print 'desired solution:'
print '-----------------'
print (a1 + a2) / 2.

print
print 'now with cubes'
print '--------------'
c1 = iris.cube.Cube(a1)
c2 = iris.cube.Cube(a2)
cs = np.mean([c1, c2])
print c1
print c2
print cs
print cs.data

It also works with a CubeList. I'm using numpy 1.7.1.

[Stephan Hoyer]

I'm running NumPy 1.8.0 (and the dev version of Iris), so that could very well explain why it doesn't work for me. I agree that this is very bizarre!

Cheers,

Stephan

[Andrew Dawson]

Well I guess that means it isn't a good idea then! It seems like writing a wrapper around the built-in sum function might be the best choice here.

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