Mixing
Distance in k-NN
$begingroup$
In the K-nearest-neighbors model, we use
begin{equation}
hat{f}(x)=Ave(y_i mid x_i in N_k(x))
end{equation}
where $N_k(x)$ is a neighborhood of a certain size of x.
What is the actual distance that we use to determine the neighborhood of x? At first i thought that it was $d((x_i,y_i),(x,y))=sqrt{(x_i-x)^2+(y_i -y)^2}$ but that's not the case. The distance used (i think) is $d(x_0,x)=left|x_0 - x right|$. It seems to me that using the first distance is more intuitive, but then it's not the one that is used. Why?
Thanks
statistics machine-learning
asked Jan 2 at 20:56
yjntyjnt
63
$endgroup$
|
show 8 more comments
$begingroup$
In the K-nearest-neighbors model, we use
begin{equation}
hat{f}(x)=Ave(y_i mid x_i in N_k(x))
end{equation}
where $N_k(x)$ is a neighborhood of a certain size of x.
What is the actual distance that we use to determine the neighborhood of x? At first i thought that it was $d((x_i,y_i),(x,y))=sqrt{(x_i-x)^2+(y_i -y)^2}$ but that's not the case. The distance used (i think) is $d(x_0,x)=left|x_0 - x right|$. It seems to me that using the first distance is more intuitive, but then it's not the one that is used. Why?
Thanks
statistics machine-learning
asked Jan 2 at 20:56
yjntyjnt
63
$endgroup$
$begingroup$
But if $ vec x_0 := (x_i, y_i)$ and $vec x := (x, y)$, then $| vec x_0 -vec x| = sqrt{(x_i - x)^2 + (y_i - y)^2}$.
$endgroup$
– Kenny Wong
Jan 2 at 21:05
$begingroup$
but here $x_0$ is the x-coordinate not the vector.
$endgroup$
– yjnt
Jan 2 at 21:06
$begingroup$
So your distance metric ignores the $y$-coordinates???
$endgroup$
– Kenny Wong
Jan 2 at 21:06
$begingroup$
Also, i've implemented knn on python using these two distances, and the latter one gives much better results
$endgroup$
– yjnt
Jan 2 at 21:07
1
$begingroup$
Because for a test point, you don't know the $y$! So if your distance metric uses $y$'s, then you won't be able to calculate the distance between your test point and the various training points.
$endgroup$
– Kenny Wong
Jan 2 at 21:30
|
show 8 more comments
$begingroup$
In the K-nearest-neighbors model, we use
begin{equation}
hat{f}(x)=Ave(y_i mid x_i in N_k(x))
end{equation}
where $N_k(x)$ is a neighborhood of a certain size of x.
What is the actual distance that we use to determine the neighborhood of x? At first i thought that it was $d((x_i,y_i),(x,y))=sqrt{(x_i-x)^2+(y_i -y)^2}$ but that's not the case. The distance used (i think) is $d(x_0,x)=left|x_0 - x right|$. It seems to me that using the first distance is more intuitive, but then it's not the one that is used. Why?
Thanks
statistics machine-learning
asked Jan 2 at 20:56
yjntyjnt
63
$endgroup$
In the K-nearest-neighbors model, we use
begin{equation}
hat{f}(x)=Ave(y_i mid x_i in N_k(x))
end{equation}
where $N_k(x)$ is a neighborhood of a certain size of x.
What is the actual distance that we use to determine the neighborhood of x? At first i thought that it was $d((x_i,y_i),(x,y))=sqrt{(x_i-x)^2+(y_i -y)^2}$ but that's not the case. The distance used (i think) is $d(x_0,x)=left|x_0 - x right|$. It seems to me that using the first distance is more intuitive, but then it's not the one that is used. Why?
Thanks
statistics machine-learning
statistics machine-learning
asked Jan 2 at 20:56
yjntyjnt
63
asked Jan 2 at 20:56
yjntyjnt
63
asked Jan 2 at 20:56
yjntyjnt
63
asked Jan 2 at 20:56
yjntyjnt
63
asked Jan 2 at 20:56
yjntyjnt
63
63
$begingroup$
But if $ vec x_0 := (x_i, y_i)$ and $vec x := (x, y)$, then $| vec x_0 -vec x| = sqrt{(x_i - x)^2 + (y_i - y)^2}$.
$endgroup$
– Kenny Wong
Jan 2 at 21:05
$begingroup$
but here $x_0$ is the x-coordinate not the vector.
$endgroup$
– yjnt
Jan 2 at 21:06
$begingroup$
So your distance metric ignores the $y$-coordinates???
$endgroup$
– Kenny Wong
Jan 2 at 21:06
$begingroup$
Also, i've implemented knn on python using these two distances, and the latter one gives much better results
$endgroup$
– yjnt
Jan 2 at 21:07
1
$begingroup$
Because for a test point, you don't know the $y$! So if your distance metric uses $y$'s, then you won't be able to calculate the distance between your test point and the various training points.
$endgroup$
– Kenny Wong
Jan 2 at 21:30
|
show 8 more comments
$begingroup$
But if $ vec x_0 := (x_i, y_i)$ and $vec x := (x, y)$, then $| vec x_0 -vec x| = sqrt{(x_i - x)^2 + (y_i - y)^2}$.
$endgroup$
– Kenny Wong
Jan 2 at 21:05
$begingroup$
but here $x_0$ is the x-coordinate not the vector.
$endgroup$
– yjnt
Jan 2 at 21:06
$begingroup$
So your distance metric ignores the $y$-coordinates???
$endgroup$
– Kenny Wong
Jan 2 at 21:06
$begingroup$
Also, i've implemented knn on python using these two distances, and the latter one gives much better results
$endgroup$
– yjnt
Jan 2 at 21:07
1
$begingroup$
Because for a test point, you don't know the $y$! So if your distance metric uses $y$'s, then you won't be able to calculate the distance between your test point and the various training points.
$endgroup$
– Kenny Wong
Jan 2 at 21:30
$begingroup$
But if $ vec x_0 := (x_i, y_i)$ and $vec x := (x, y)$, then $| vec x_0 -vec x| = sqrt{(x_i - x)^2 + (y_i - y)^2}$.
$endgroup$
– Kenny Wong
Jan 2 at 21:05
$begingroup$
But if $ vec x_0 := (x_i, y_i)$ and $vec x := (x, y)$, then $| vec x_0 -vec x| = sqrt{(x_i - x)^2 + (y_i - y)^2}$.
$endgroup$
– Kenny Wong
Jan 2 at 21:05
$begingroup$
but here $x_0$ is the x-coordinate not the vector.
$endgroup$
– yjnt
Jan 2 at 21:06
$begingroup$
but here $x_0$ is the x-coordinate not the vector.
$endgroup$
– yjnt
Jan 2 at 21:06
$begingroup$
So your distance metric ignores the $y$-coordinates???
$endgroup$
– Kenny Wong
Jan 2 at 21:06
$begingroup$
So your distance metric ignores the $y$-coordinates???
$endgroup$
– Kenny Wong
Jan 2 at 21:06
$begingroup$
Also, i've implemented knn on python using these two distances, and the latter one gives much better results
$endgroup$
– yjnt
Jan 2 at 21:07
$begingroup$
Also, i've implemented knn on python using these two distances, and the latter one gives much better results
$endgroup$
– yjnt
Jan 2 at 21:07
1
1
$begingroup$
Because for a test point, you don't know the $y$! So if your distance metric uses $y$'s, then you won't be able to calculate the distance between your test point and the various training points.
$endgroup$
– Kenny Wong
Jan 2 at 21:30
$begingroup$
Because for a test point, you don't know the $y$! So if your distance metric uses $y$'s, then you won't be able to calculate the distance between your test point and the various training points.
$endgroup$
– Kenny Wong
Jan 2 at 21:30
|
show 8 more comments
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$begingroup$
But if $ vec x_0 := (x_i, y_i)$ and $vec x := (x, y)$, then $| vec x_0 -vec x| = sqrt{(x_i - x)^2 + (y_i - y)^2}$.
$endgroup$
– Kenny Wong
Jan 2 at 21:05
$begingroup$
but here $x_0$ is the x-coordinate not the vector.
$endgroup$
– yjnt
Jan 2 at 21:06
$begingroup$
So your distance metric ignores the $y$-coordinates???
$endgroup$
– Kenny Wong
Jan 2 at 21:06
$begingroup$
Also, i've implemented knn on python using these two distances, and the latter one gives much better results
$endgroup$
– yjnt
Jan 2 at 21:07
1
$begingroup$
Because for a test point, you don't know the $y$! So if your distance metric uses $y$'s, then you won't be able to calculate the distance between your test point and the various training points.
$endgroup$
– Kenny Wong
Jan 2 at 21:30