Array as Variable using "from cdms2 import MV2 as MV"
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) Thanking you................... -- DILEEPKUMAR. R J R F, IIT DELHI
On Mon, Apr 25, 2011 at 2:50 AM, dileep kunjaai
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 Josef
Thanking you...................
-- DILEEPKUMAR. R J R F, IIT DELHI
_______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Thank you sir .......thanks a lot
Keep in touch
On Mon, Apr 25, 2011 at 12:47 PM,
On Mon, Apr 25, 2011 at 2:50 AM, dileep kunjaai
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
Josef
Thanking you...................
-- DILEEPKUMAR. R J R F, IIT DELHI
_______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
_______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
-- DILEEPKUMAR. R J R F, IIT DELHI
On Mon, Apr 25, 2011 at 12:17 AM,
On Mon, Apr 25, 2011 at 2:50 AM, dileep kunjaai
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
participants (3)
-
dileep kunjaai
-
josef.pktd@gmail.com
-
Keith Goodman