I think you misunderstood me. I meant something (more efficiently written)
along the lines of below.
import numpy as np
X0 = [[-1, 0], [0,-1]]
X1 = [[ 1, 0], [0, 1]]
trData = { 0: X0, 1: X1 }
X = np.array( [v for v in trData.values()] ).reshape( -1, 2 )
Y = np.array( [np.ones( len(v) ) * k for k, v in trData.iteritems()]
).ravel()
On Fri, Apr 5, 2013 at 10:42 AM, Philipp Singer <[email protected]> wrote:
> Dictionaries do not have duplicate keys (labels). You could only make a
> list of datawithLabelX for each key label. But what is the benefit of this?
>
> Philipp
>
> Am 05.04.2013 11:37, schrieb Bill Power:
> > i know this is going to sound a little silly, but I was thinking there
> > that it might be nice to be able to do this with scikit learn
> >
> > clf = sklearn.anyClassifier()
> > clf.fit( { 0: dataWithLabel0,
> > 1: dataWithLabel1 } )
> >
> > instead of having to separate the data/labels manually. i guess fit
> > would do that internally, but it might be nice to have this
> >
> > bill
> >
> >
> >
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>
>
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> Reduce network management and security costs.Learn how to hire
> the most talented Cisco Certified professionals. Visit the
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Reduce network management and security costs.Learn how to hire
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