Hi Jacob
I went through the code. The 'fit' method in nearest neighbors does not do
any distance calculations. It only initializes the class variables. In that
case this is probably not a bug.
--
sp
On Wed, Feb 24, 2016 at 12:26 AM, Jacob Vanderplas <
jake...@cs.washington.edu> wrote:
> I have
>
> I have been experimenting with the above code. I have noticed the
> following things:
>
>
>1. If we set algorithm = 'brute' the algorithm does not enter the
>function tan, i.e., putting a breakpoint at the print statement does not
>stop execution on it during the fit method. It does
I have been experimenting with the above code. I have noticed the following
things:
1. If we set algorithm = 'brute' the algorithm does not enter the
function tan, i.e., putting a breakpoint at the print statement does not
stop execution on it during the fit method. It does however use t
I guess I got it now! This behavior (see below) is indeed a bit strange:
from sklearn.neighbors import NearestNeighbors
import numpy as np
X = np.array([[1.0, 0.0, 1.0, 1.0], [0.0, 0.0, 1.0, 0.0], [1.0, 1.0, 1.0, 1.0]])
def tan(x, y):
print(y)
return 1
nbrs = NearestNeighbors(n_neighbor
ps.
1. I printed the x,y array. And i thougtif these is the output:
[ 0.49178495 0.44239588 0.43451225 0.40576958 0.82022061 0.02921787
0.08832147 0.43397282 0.15083042 0.49916182] [ 0.49178495 0.44239588
0.43451225 0.40576958 0.82022061 0.02921787
0.08832147 0.43397282 0.15
Sorry that i coudln't explained it very well
I thought that
X = np.array([[1.0, 0.0, 1.0, 1.0], [0.0, 0.0, 1.0, 0.0], [1.0, 1.0, 1.0,
1.0]])
def tan(x, y):
print x,y
c=np.sum(x==y)
a1 = x[x == 1.0].shape[0]
b1 = y[y == 1.0].shape[0]
return float(c)/(a1 + b1 - c)
example
Hi, Herbert,
sorry, but I am still a bit confused about what you are trying to accomplish
when you say
> and the output is then what i mentioned
>
> x are only floats (0.573... ) and B are containing 1's and 0's like it should
When I run it on a small test dataset ...
from sklearn.neighbors
Here is an example code, where the failure occurs.
sorry for the big tests vector, couldn't show it otherwise.
import numpy as np
from sklearn.neighbors import NearestNeighbors
def tanimoto(x,y):
print "X OUTPUT\n ",x,"B OUTPUT\n",y
c=np.sum(x==y)
a1 = np.sum(x)
b1 = np.sum
The X array which I'm using to fit for the kNN algorithm is n_samples x n
features big.
but for calculating the distance or in this case the tanimoto.
i thought tanimot(x,y) if I call it with :
KNeighborsClassifier(metric=tanimoto).fit(X_train,y_train) is,
x= is in sample and y= is one sample
I see, as far as I know, your X input array should be a n_samples x n_features
array. Is this true in your case?
Btw. instead of converting the inputs to Python lists , you could also get the
counts via
a1 = x[x == 1.0].shape[0]
b1 = y[y == 1.0].shape[0]
And maybe a few extra lines
assert se
Yes, it is a number between 0 and 1.
but im calculating it, depended on 2 samples. So x and y. And if im
counting the 1's in x and the 1's in y, i should get the coef with return
float(c)/(a1 + b1 - c)
But i cannot count 1's in x if x is a skalar.
-
Isn't the tanimoto coeff a continuous number between 0 and 1?
> how are the skalars are occring in the X vektor?
You are returning a fraction -> "return float(c)/(a1 + b1 - c)"
instead of the count, maybe that's an misunderstanding?
> On Jan 12, 2016, at 1:24 PM, A neuman wrote:
>
> The
The custom metric, ist just calculating the tanimoto coef.
a=x.tolist()
b=y.tolist()
c=np.count_nonzero(x==y)
a1=a.count(1.0)
b1=b.count(1.0)
return float(c)/(a1 + b1 - c)
so im Just counting 1's in x and 1's in y
c= are the numer, where 1's are matching ( matching ==
Hi, I am not sure how your custom metric works, but would a np.where(x >= 0.5,
1., 0.) work in your case?
> On Jan 12, 2016, at 1:08 PM, A neuman wrote:
>
> Sorry, thats not right what I wrote:
> X:
> [ 0.6371319 0.54557285 0.30214217 0.14690307 0.49778446 0.89183238
> 0.52445514 0.633
Sorry, thats not right what I wrote:
X:
[ 0.6371319 0.54557285 0.30214217 0.14690307 0.49778446 0.89183238
0.52445514 0.63379164 0.71873681 0.55008567]
Y:
[ 0.6371319 0.54557285 0.30214217 0.14690307 0.49778446 0.89183238
0.52445514 0.63379164 0.71873681 0.55008567]
X:
[ 0.
Hey,
I Have an another problem,
if I'm using my own metric, there are not only the samples in x and y.
I'm using a 10 fold cv with k-NN Classifier.
My Attributes are only 1's and 0's, but if im printing them out, I'll get:
KNeighborsClassifier(metric=myFunc)
def myFunc(x,y):
print x,'\n'
Ah, that helped me a lot!!!
So i just write my own function that returns an skalar. This function is
used in the metric parameter of the kNN function.
Thank you!!!
On 9 January 2016 at 03:41, Sebastian Raschka wrote:
> You could just need “regular" Python function that outputs a scalar. For
>
You could just need “regular" Python function that outputs a scalar. For
example, consider the following example:
>>> from sklearn.neighbors import NearestNeighbors
>>> import numpy as np
>>> X = np.array([[-1, -1], [-2, -1], [-3, -2], [1, 1], [2, 1], [3, 2]])
>>> nbrs = NearestNeighbors(n_neighb
Hello everyone,
I actually want to use the KNeighboursClassifier, with my own distances.
in the Documentation stands the following:
[callable] : a user-defined function which accepts an array of distances,
and returns an array of the same shape containing the weights.
I just dont know, how shou
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