On Mon, Apr 25, 2011 at 12:17 AM, <[email protected]> wrote:
> On Mon, Apr 25, 2011 at 2:50 AM, dileep kunjaai <[email protected]>
> wrote:
>> Dear sir,
>>
>> I am have 2 mxn numpy array say "obs" & "fcst". I have to
>> calculate sum of squre of (obs[i, j]-fcst[i, j]) using from cdms2
>> import MV2 as MV in CDAT without using "for" loop.
>>
>> For example:
>> obs=
>> [0.6 1.1 0.02 0.2 0.2
>> 0.8 0. 0. 0.4 0.8
>> 0.5 5.5 1.5 0.5 1.5
>> 3.5 0.5 1.5 5.0 2.6
>> 5.1 4.1 3.2 2.3 1.5
>> 4.4 0.9 1.5 2. 2.3
>> 1.1 1.1 1.5 12.6 1.3
>> 2.2 12 1.7 1.6 15
>> 1.9 1.5 0.9 2.5 5.5 ]
>>
>>
>>
>> fcst=
>>
>> [0.7 0.1 0.2 0.2 0.2
>> 0.3 0.8 0. 0. 0.
>> 0.5 0.5 0.5 0.5 0.5
>> 0.7 1. 1.5 2. 2.6
>> 5.1 4.1 3.2 2.3 1.5
>> 0.7 1. 1.5 2. 2.3
>> 1.1 1.1 1.1 12.7 1.3
>> 2.2 2. 1.7 1.6 1.5
>> 1.9 1.5 0.9 0.5 7.5]
>>
>> here "obs" and "fcst" are numpy array
>> I give
>>
>> obs=MV.array(obs)
>> fcst=MV.array(fcst)
>>
>> Then it become
>>
>>
>> sbst=obs-fcst
>>
>>>> subst=
>> [[ -0.1 1. -0.18 0. 0. ]
>> [ 0.5 -0.8 0. 0.4 0.8 ]
>> [ 0. 5. 1. 0. 1. ]
>> [ 2.8 -0.5 0. 3. 0. ]
>> [ 0. 0. 0. 0. 0. ]
>> [ 3.7 -0.1 0. 0. 0. ]
>> [ 0. 0. 0.4 -0.1 0. ]
>> [ 0. 10. 0. 0. 13.5 ]
>> [ 0. 0. 0. 2. -2. ]]
>>
>> But i dont know how to find sum of squre of each term....(Actually my aim is
>> to finding MEAN SQUARED ERROR)
>
> (sbst**2).sum()
>
> or with sum along columns
> (sbst**2).sum(0)
>
> explanation is in the documentation
If speed is an issue then the development version of bottleneck
(0.5.dev) has a fast sum of squares function:
>> a = np.random.rand(1000, 10)
>> timeit (a * a).sum(1)
10000 loops, best of 3: 143 us per loop
>> timeit bn.ss(a, 1)
100000 loops, best of 3: 14.3 us per loop
It is a faster replacement for scipy.stats.ss:
>> from scipy import stats
>> timeit stats.ss(a, 1)
10000 loops, best of 3: 159 us per loop
Speed up factor depends on the shape of the input array and the axis:
>> timeit stats.ss(a, 0)
10000 loops, best of 3: 42.8 us per loop
>> timeit bn.ss(a, 0)
100000 loops, best of 3: 17.3 us per loop
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