Measuring distance preservation in dimensionality reduction
$begingroup$
I am looking to compare the distance preserved during dimension reductions for several techniques. I have read some papers on similar topics here and here.
For example, I would like to use the Euclidean Distance to measure the distance preserved during PCA's dimension reduction. However, my point of confusion what are $X$ and $Y$ in
$$d(X, Y) = sqrt{sum^n_{i=1}left(x_i - y_iright)^2}$$
I understand how to calculate $dleft(X, Yright)$ given two vectors/matrices, but I don't understand with context to PCA. Let me try to explain.
Let $W_{k times n}$ be the matrix of $k$ eigenvectors, $X_{dtimes n}$ be the original data, and $Z_{ktimes n}$ be the projection of $X$ onto the reduced subspace.
$Z = W^{T}X$
Back to calculating $dleft(X, Yright)$. My guess is that the PCA's $X$ correspond to $X$ and $Y$ can correspond to $Z$. But how does this work since $X$ and $Y$ have different dimensions? I have to be oblivious to something here.
Also, I am not concerned if a Euclidean Distance measure is not a good choice for measuring PCA's distance preservation (unless they are incompatible). This is simply exploration.
pca dimensionality-reduction distance
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$begingroup$
I am looking to compare the distance preserved during dimension reductions for several techniques. I have read some papers on similar topics here and here.
For example, I would like to use the Euclidean Distance to measure the distance preserved during PCA's dimension reduction. However, my point of confusion what are $X$ and $Y$ in
$$d(X, Y) = sqrt{sum^n_{i=1}left(x_i - y_iright)^2}$$
I understand how to calculate $dleft(X, Yright)$ given two vectors/matrices, but I don't understand with context to PCA. Let me try to explain.
Let $W_{k times n}$ be the matrix of $k$ eigenvectors, $X_{dtimes n}$ be the original data, and $Z_{ktimes n}$ be the projection of $X$ onto the reduced subspace.
$Z = W^{T}X$
Back to calculating $dleft(X, Yright)$. My guess is that the PCA's $X$ correspond to $X$ and $Y$ can correspond to $Z$. But how does this work since $X$ and $Y$ have different dimensions? I have to be oblivious to something here.
Also, I am not concerned if a Euclidean Distance measure is not a good choice for measuring PCA's distance preservation (unless they are incompatible). This is simply exploration.
pca dimensionality-reduction distance
asked 11 mins ago
qq3254qq3254
1
New contributor
qq3254 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am looking to compare the distance preserved during dimension reductions for several techniques. I have read some papers on similar topics here and here.
For example, I would like to use the Euclidean Distance to measure the distance preserved during PCA's dimension reduction. However, my point of confusion what are $X$ and $Y$ in
$$d(X, Y) = sqrt{sum^n_{i=1}left(x_i - y_iright)^2}$$
I understand how to calculate $dleft(X, Yright)$ given two vectors/matrices, but I don't understand with context to PCA. Let me try to explain.
Let $W_{k times n}$ be the matrix of $k$ eigenvectors, $X_{dtimes n}$ be the original data, and $Z_{ktimes n}$ be the projection of $X$ onto the reduced subspace.
$Z = W^{T}X$
Back to calculating $dleft(X, Yright)$. My guess is that the PCA's $X$ correspond to $X$ and $Y$ can correspond to $Z$. But how does this work since $X$ and $Y$ have different dimensions? I have to be oblivious to something here.
Also, I am not concerned if a Euclidean Distance measure is not a good choice for measuring PCA's distance preservation (unless they are incompatible). This is simply exploration.
pca dimensionality-reduction distance
asked 11 mins ago
qq3254qq3254
1
New contributor
qq3254 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am looking to compare the distance preserved during dimension reductions for several techniques. I have read some papers on similar topics here and here.
For example, I would like to use the Euclidean Distance to measure the distance preserved during PCA's dimension reduction. However, my point of confusion what are $X$ and $Y$ in
$$d(X, Y) = sqrt{sum^n_{i=1}left(x_i - y_iright)^2}$$
I understand how to calculate $dleft(X, Yright)$ given two vectors/matrices, but I don't understand with context to PCA. Let me try to explain.
Let $W_{k times n}$ be the matrix of $k$ eigenvectors, $X_{dtimes n}$ be the original data, and $Z_{ktimes n}$ be the projection of $X$ onto the reduced subspace.
$Z = W^{T}X$
Back to calculating $dleft(X, Yright)$. My guess is that the PCA's $X$ correspond to $X$ and $Y$ can correspond to $Z$. But how does this work since $X$ and $Y$ have different dimensions? I have to be oblivious to something here.
Also, I am not concerned if a Euclidean Distance measure is not a good choice for measuring PCA's distance preservation (unless they are incompatible). This is simply exploration.
pca dimensionality-reduction distance
pca dimensionality-reduction distance
asked 11 mins ago
qq3254qq3254
1
New contributor
qq3254 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 11 mins ago
qq3254qq3254
1
New contributor
qq3254 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 11 mins ago
qq3254qq3254
1
New contributor
qq3254 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 11 mins ago
qq3254qq3254
1
asked 11 mins ago
qq3254qq3254
1
1
New contributor
qq3254 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
qq3254 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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