Why is computing ridge regression with a Cholesky decomposition much quicker than using SVD?Why is it much...
Are tamper resistant receptacles really safer?
How are showroom/display vehicles prepared?
Latex does not go to next line
Are babies of evil humanoid species inherently evil?
Why would one plane in this picture not have gear down yet?
Reverse string, can I make it faster?
Reversed Sudoku
Why doesn't this Google Translate ad use the word "Translation" instead of "Translate"?
Accepted offer letter, position changed
What wound would be of little consequence to a biped but terrible for a quadruped?
Are there historical instances of the capital of a colonising country being temporarily or permanently shifted to one of its colonies?
Is "conspicuously missing" or "conspicuously" the subject of this sentence?
When a wind turbine does not produce enough electricity how does the power company compensate for the loss?
Bash script should only kill those instances of another script's that it has launched
Is "history" a male-biased word ("his+story")?
They call me Inspector Morse
Why does liquid water form when we exhale on a mirror?
How is the wildcard * interpreted as a command?
Database Backup for data and log files
Does the nature of the Apocalypse in The Umbrella Academy change from the first to the last episode?
Vocabulary for giving just numbers, not a full answer
Does "Until when" sound natural for native speakers?
An alternative proof of an application of Hahn-Banach
Conservation of Mass and Energy
Why is computing ridge regression with a Cholesky decomposition much quicker than using SVD?
Why is it much quicker to compute ridge regression than regular linear regression?Computing ridge regression with prior different from 0SVD in linear regressionGaussian Process: Using partitions of a Cholesky decomposition solution for conditional deductionConfirming an understanding of SVDIll-conditioned covariance matrix in GP regression for Bayesian optimizationThe proof of shrinking coefficients using ridge regression through “spectral decomposition”Simulate Moderated Regression with Cholesky DecompositionCoefficients of Principal Components Regression in terms of original regressorsFor symmetric matrices, is the Cholesky decomposition better than the SVD?Why is it much quicker to compute ridge regression than regular linear regression?
$begingroup$
By my understanding, for a matrix with n samples and p features:
- Ridge regression using Cholesky decomposition takes O(p^3) time
- Ridge regression using SVD takes O(p^3) time
- Computing SVD when only the diagonal matrix is needed (and not u and v) takes O(np^2) time
I tested this out in scipy on both random and real-world data with p > n (p = 43624, n = 1750) and found ridge regression with a Cholesky decomposition to be much quicker than computing it using SVD and much quicker than just computing the diagonal matrix of the SVD. Why is this?
import numpy as np
from sklearn import linear_model
import scipy
import time
x = np.random.rand(1750, 43264)
y = np.random.rand(1750)
old_time = time.time()
clf = linear_model.Ridge(alpha=1.0, solver='cholesky')
clf.fit(x, y)
print("Time taken to solve Ridge with cholesky: ", time.time() - old_time)
old_time = time.time()
clf = linear_model.Ridge(alpha=1.0, solver='svd')
clf.fit(x, y)
print("Time taken to solve Ridge with SVD: ", time.time() - old_time)
old_time = time.time()
scipy.linalg.svd(x, full_matrices=False)
print("Time taken for SVD", time.time() - old_time)
old_time = time.time()
scipy.linalg.svd(x, full_matrices=False, compute_uv=False)
print("Time taken for SVD, just s", time.time() - old_time)
Output:
Time taken to solve Ridge with cholesky: 3.2336008548736572
Time taken to solve Ridge with SVD: 118.47378492355347
Time taken for SVD 92.01217150688171
Time taken for SVD, just s 44.7129647731781
regression ridge-regression svd scipy cholesky
New contributor
$endgroup$
add a comment |
$begingroup$
By my understanding, for a matrix with n samples and p features:
- Ridge regression using Cholesky decomposition takes O(p^3) time
- Ridge regression using SVD takes O(p^3) time
- Computing SVD when only the diagonal matrix is needed (and not u and v) takes O(np^2) time
I tested this out in scipy on both random and real-world data with p > n (p = 43624, n = 1750) and found ridge regression with a Cholesky decomposition to be much quicker than computing it using SVD and much quicker than just computing the diagonal matrix of the SVD. Why is this?
import numpy as np
from sklearn import linear_model
import scipy
import time
x = np.random.rand(1750, 43264)
y = np.random.rand(1750)
old_time = time.time()
clf = linear_model.Ridge(alpha=1.0, solver='cholesky')
clf.fit(x, y)
print("Time taken to solve Ridge with cholesky: ", time.time() - old_time)
old_time = time.time()
clf = linear_model.Ridge(alpha=1.0, solver='svd')
clf.fit(x, y)
print("Time taken to solve Ridge with SVD: ", time.time() - old_time)
old_time = time.time()
scipy.linalg.svd(x, full_matrices=False)
print("Time taken for SVD", time.time() - old_time)
old_time = time.time()
scipy.linalg.svd(x, full_matrices=False, compute_uv=False)
print("Time taken for SVD, just s", time.time() - old_time)
Output:
Time taken to solve Ridge with cholesky: 3.2336008548736572
Time taken to solve Ridge with SVD: 118.47378492355347
Time taken for SVD 92.01217150688171
Time taken for SVD, just s 44.7129647731781
regression ridge-regression svd scipy cholesky
New contributor
$endgroup$
$begingroup$
Cholesky decomposes a square symmetric matrix (such as correlations) and is much faster itself than SVD which, in general, decomposes a rectangular matrix (such as dataset) into 3 matrices.
$endgroup$
– ttnphns
6 hours ago
add a comment |
$begingroup$
By my understanding, for a matrix with n samples and p features:
- Ridge regression using Cholesky decomposition takes O(p^3) time
- Ridge regression using SVD takes O(p^3) time
- Computing SVD when only the diagonal matrix is needed (and not u and v) takes O(np^2) time
I tested this out in scipy on both random and real-world data with p > n (p = 43624, n = 1750) and found ridge regression with a Cholesky decomposition to be much quicker than computing it using SVD and much quicker than just computing the diagonal matrix of the SVD. Why is this?
import numpy as np
from sklearn import linear_model
import scipy
import time
x = np.random.rand(1750, 43264)
y = np.random.rand(1750)
old_time = time.time()
clf = linear_model.Ridge(alpha=1.0, solver='cholesky')
clf.fit(x, y)
print("Time taken to solve Ridge with cholesky: ", time.time() - old_time)
old_time = time.time()
clf = linear_model.Ridge(alpha=1.0, solver='svd')
clf.fit(x, y)
print("Time taken to solve Ridge with SVD: ", time.time() - old_time)
old_time = time.time()
scipy.linalg.svd(x, full_matrices=False)
print("Time taken for SVD", time.time() - old_time)
old_time = time.time()
scipy.linalg.svd(x, full_matrices=False, compute_uv=False)
print("Time taken for SVD, just s", time.time() - old_time)
Output:
Time taken to solve Ridge with cholesky: 3.2336008548736572
Time taken to solve Ridge with SVD: 118.47378492355347
Time taken for SVD 92.01217150688171
Time taken for SVD, just s 44.7129647731781
regression ridge-regression svd scipy cholesky
New contributor
$endgroup$
By my understanding, for a matrix with n samples and p features:
- Ridge regression using Cholesky decomposition takes O(p^3) time
- Ridge regression using SVD takes O(p^3) time
- Computing SVD when only the diagonal matrix is needed (and not u and v) takes O(np^2) time
I tested this out in scipy on both random and real-world data with p > n (p = 43624, n = 1750) and found ridge regression with a Cholesky decomposition to be much quicker than computing it using SVD and much quicker than just computing the diagonal matrix of the SVD. Why is this?
import numpy as np
from sklearn import linear_model
import scipy
import time
x = np.random.rand(1750, 43264)
y = np.random.rand(1750)
old_time = time.time()
clf = linear_model.Ridge(alpha=1.0, solver='cholesky')
clf.fit(x, y)
print("Time taken to solve Ridge with cholesky: ", time.time() - old_time)
old_time = time.time()
clf = linear_model.Ridge(alpha=1.0, solver='svd')
clf.fit(x, y)
print("Time taken to solve Ridge with SVD: ", time.time() - old_time)
old_time = time.time()
scipy.linalg.svd(x, full_matrices=False)
print("Time taken for SVD", time.time() - old_time)
old_time = time.time()
scipy.linalg.svd(x, full_matrices=False, compute_uv=False)
print("Time taken for SVD, just s", time.time() - old_time)
Output:
Time taken to solve Ridge with cholesky: 3.2336008548736572
Time taken to solve Ridge with SVD: 118.47378492355347
Time taken for SVD 92.01217150688171
Time taken for SVD, just s 44.7129647731781
regression ridge-regression svd scipy cholesky
regression ridge-regression svd scipy cholesky
New contributor
New contributor
New contributor
asked 8 hours ago
pighead10pighead10
1264
1264
New contributor
New contributor
$begingroup$
Cholesky decomposes a square symmetric matrix (such as correlations) and is much faster itself than SVD which, in general, decomposes a rectangular matrix (such as dataset) into 3 matrices.
$endgroup$
– ttnphns
6 hours ago
add a comment |
$begingroup$
Cholesky decomposes a square symmetric matrix (such as correlations) and is much faster itself than SVD which, in general, decomposes a rectangular matrix (such as dataset) into 3 matrices.
$endgroup$
– ttnphns
6 hours ago
$begingroup$
Cholesky decomposes a square symmetric matrix (such as correlations) and is much faster itself than SVD which, in general, decomposes a rectangular matrix (such as dataset) into 3 matrices.
$endgroup$
– ttnphns
6 hours ago
$begingroup$
Cholesky decomposes a square symmetric matrix (such as correlations) and is much faster itself than SVD which, in general, decomposes a rectangular matrix (such as dataset) into 3 matrices.
$endgroup$
– ttnphns
6 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First, we note that ridge regression can be transformed into an OLS problem via the data augmentation trick.. This adds $p$ rows to the design matrix $mathbf{X}$, so the new matrix $mathbf{X}_lambda$ has dimensions $ (n + p) times p $.
We can then look up the run time for various OLS algorithms, for example in the book Least Squares Data Fitting with Applications or in Golub's own book on Matrix Computations. (Golub was one of inventors of currently state-of-the-art Golub-Reinsch SVD algorithm.) Since linking to Google Books is not very reliable, these costs can also be found, for example, here.
We see that relevant costs for the non-ridge case, assuming $mathbf{X}$ is $n times p$, are:
$$ C_{LU} = 2 n p^2 - frac{2}{3} p^3 tag{1} $$
$$ C_{SVD} = 4 n p^2 + 8 p^3 tag{2} $$
To obtain equations for ridge regression, we need to substitute $n rightarrow n + p $ to correctly account for data augmentation:
$$ C_{LU} = 2 n p^2 + frac{4}{3} p^3 tag{3} $$
$$ C_{SVD} = 4 n p^2 + 12 p^3 tag{4} $$
Note that unlike your example, $n > p$ for most typical regression problems. If we say that $n approx p$, we can obtained simplified costs in only one variable:
$$ C_{LU} = frac{10}{3} p^3 tag{5} $$
$$ C_{SVD} = 16 p^3 tag{6} $$
This is probably where those rough estimates you reference came from, but as you yourself discovered, this simplification hides important detail and makes inappropriate assumptions that don't fit your case. Note, for example, that the constant for SVD is much worse than for Cholesky, an important fact hidden by the big $mathord{O}$ notation.
What you are doing is exactly the opposite - you have $n ll p$. That means the second term is much more important! By comparing the coefficients of the $p^3$ terms of equations (3) and (4) we can argue the SVD approach will require roughly 10 times as many floating point operations as the Cholesky approach when $n ll p$.
Beyond that, it's hard to know exactly what scipy is doing under the hood - is it calculating both $mathbf{U}$ and $mathbf{V}$ when doing SVD? Is it using the randomized algorithm? Is it using data augmentation for ridge regression, or some other approach? And even once those details are known, it's hard to translate these theoretical operation counts to real-world run time because libraries like LAPACK (scipy is almost surely using LAPACK or ATLAS under the hood) are so highly vectorized (taking advantage of CPU SIMD instructions) and often multithreaded that's it's hard to estimate accurately from purely theoretical arguments. But it's safe to say that it's reasonable to expect Cholesky factorization to be much faster than SVD for a given problem, especially when $n ll p$.
$endgroup$
1
$begingroup$
Explicit data augmentation would be a very slow way to compute ridge solution when $nll p$... Everything that scales as $p^3$ can instead be rewritten to scale as $n^3$.
$endgroup$
– amoeba
6 hours ago
$begingroup$
Great answer. This problem arose because a paper I was following suggested, sincen ≪ p
, computing the SVD in order to grid-search for the best alpha, since the resulting computation for each alpha only takes O(pn^2). However, my results/your answer shows that this would only take less time if there are greater than ~10 different values of alpha to try, due to the initial overheard of the SVD being huge.
$endgroup$
– pighead10
6 hours ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "65"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
pighead10 is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f396914%2fwhy-is-computing-ridge-regression-with-a-cholesky-decomposition-much-quicker-tha%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First, we note that ridge regression can be transformed into an OLS problem via the data augmentation trick.. This adds $p$ rows to the design matrix $mathbf{X}$, so the new matrix $mathbf{X}_lambda$ has dimensions $ (n + p) times p $.
We can then look up the run time for various OLS algorithms, for example in the book Least Squares Data Fitting with Applications or in Golub's own book on Matrix Computations. (Golub was one of inventors of currently state-of-the-art Golub-Reinsch SVD algorithm.) Since linking to Google Books is not very reliable, these costs can also be found, for example, here.
We see that relevant costs for the non-ridge case, assuming $mathbf{X}$ is $n times p$, are:
$$ C_{LU} = 2 n p^2 - frac{2}{3} p^3 tag{1} $$
$$ C_{SVD} = 4 n p^2 + 8 p^3 tag{2} $$
To obtain equations for ridge regression, we need to substitute $n rightarrow n + p $ to correctly account for data augmentation:
$$ C_{LU} = 2 n p^2 + frac{4}{3} p^3 tag{3} $$
$$ C_{SVD} = 4 n p^2 + 12 p^3 tag{4} $$
Note that unlike your example, $n > p$ for most typical regression problems. If we say that $n approx p$, we can obtained simplified costs in only one variable:
$$ C_{LU} = frac{10}{3} p^3 tag{5} $$
$$ C_{SVD} = 16 p^3 tag{6} $$
This is probably where those rough estimates you reference came from, but as you yourself discovered, this simplification hides important detail and makes inappropriate assumptions that don't fit your case. Note, for example, that the constant for SVD is much worse than for Cholesky, an important fact hidden by the big $mathord{O}$ notation.
What you are doing is exactly the opposite - you have $n ll p$. That means the second term is much more important! By comparing the coefficients of the $p^3$ terms of equations (3) and (4) we can argue the SVD approach will require roughly 10 times as many floating point operations as the Cholesky approach when $n ll p$.
Beyond that, it's hard to know exactly what scipy is doing under the hood - is it calculating both $mathbf{U}$ and $mathbf{V}$ when doing SVD? Is it using the randomized algorithm? Is it using data augmentation for ridge regression, or some other approach? And even once those details are known, it's hard to translate these theoretical operation counts to real-world run time because libraries like LAPACK (scipy is almost surely using LAPACK or ATLAS under the hood) are so highly vectorized (taking advantage of CPU SIMD instructions) and often multithreaded that's it's hard to estimate accurately from purely theoretical arguments. But it's safe to say that it's reasonable to expect Cholesky factorization to be much faster than SVD for a given problem, especially when $n ll p$.
$endgroup$
1
$begingroup$
Explicit data augmentation would be a very slow way to compute ridge solution when $nll p$... Everything that scales as $p^3$ can instead be rewritten to scale as $n^3$.
$endgroup$
– amoeba
6 hours ago
$begingroup$
Great answer. This problem arose because a paper I was following suggested, sincen ≪ p
, computing the SVD in order to grid-search for the best alpha, since the resulting computation for each alpha only takes O(pn^2). However, my results/your answer shows that this would only take less time if there are greater than ~10 different values of alpha to try, due to the initial overheard of the SVD being huge.
$endgroup$
– pighead10
6 hours ago
add a comment |
$begingroup$
First, we note that ridge regression can be transformed into an OLS problem via the data augmentation trick.. This adds $p$ rows to the design matrix $mathbf{X}$, so the new matrix $mathbf{X}_lambda$ has dimensions $ (n + p) times p $.
We can then look up the run time for various OLS algorithms, for example in the book Least Squares Data Fitting with Applications or in Golub's own book on Matrix Computations. (Golub was one of inventors of currently state-of-the-art Golub-Reinsch SVD algorithm.) Since linking to Google Books is not very reliable, these costs can also be found, for example, here.
We see that relevant costs for the non-ridge case, assuming $mathbf{X}$ is $n times p$, are:
$$ C_{LU} = 2 n p^2 - frac{2}{3} p^3 tag{1} $$
$$ C_{SVD} = 4 n p^2 + 8 p^3 tag{2} $$
To obtain equations for ridge regression, we need to substitute $n rightarrow n + p $ to correctly account for data augmentation:
$$ C_{LU} = 2 n p^2 + frac{4}{3} p^3 tag{3} $$
$$ C_{SVD} = 4 n p^2 + 12 p^3 tag{4} $$
Note that unlike your example, $n > p$ for most typical regression problems. If we say that $n approx p$, we can obtained simplified costs in only one variable:
$$ C_{LU} = frac{10}{3} p^3 tag{5} $$
$$ C_{SVD} = 16 p^3 tag{6} $$
This is probably where those rough estimates you reference came from, but as you yourself discovered, this simplification hides important detail and makes inappropriate assumptions that don't fit your case. Note, for example, that the constant for SVD is much worse than for Cholesky, an important fact hidden by the big $mathord{O}$ notation.
What you are doing is exactly the opposite - you have $n ll p$. That means the second term is much more important! By comparing the coefficients of the $p^3$ terms of equations (3) and (4) we can argue the SVD approach will require roughly 10 times as many floating point operations as the Cholesky approach when $n ll p$.
Beyond that, it's hard to know exactly what scipy is doing under the hood - is it calculating both $mathbf{U}$ and $mathbf{V}$ when doing SVD? Is it using the randomized algorithm? Is it using data augmentation for ridge regression, or some other approach? And even once those details are known, it's hard to translate these theoretical operation counts to real-world run time because libraries like LAPACK (scipy is almost surely using LAPACK or ATLAS under the hood) are so highly vectorized (taking advantage of CPU SIMD instructions) and often multithreaded that's it's hard to estimate accurately from purely theoretical arguments. But it's safe to say that it's reasonable to expect Cholesky factorization to be much faster than SVD for a given problem, especially when $n ll p$.
$endgroup$
1
$begingroup$
Explicit data augmentation would be a very slow way to compute ridge solution when $nll p$... Everything that scales as $p^3$ can instead be rewritten to scale as $n^3$.
$endgroup$
– amoeba
6 hours ago
$begingroup$
Great answer. This problem arose because a paper I was following suggested, sincen ≪ p
, computing the SVD in order to grid-search for the best alpha, since the resulting computation for each alpha only takes O(pn^2). However, my results/your answer shows that this would only take less time if there are greater than ~10 different values of alpha to try, due to the initial overheard of the SVD being huge.
$endgroup$
– pighead10
6 hours ago
add a comment |
$begingroup$
First, we note that ridge regression can be transformed into an OLS problem via the data augmentation trick.. This adds $p$ rows to the design matrix $mathbf{X}$, so the new matrix $mathbf{X}_lambda$ has dimensions $ (n + p) times p $.
We can then look up the run time for various OLS algorithms, for example in the book Least Squares Data Fitting with Applications or in Golub's own book on Matrix Computations. (Golub was one of inventors of currently state-of-the-art Golub-Reinsch SVD algorithm.) Since linking to Google Books is not very reliable, these costs can also be found, for example, here.
We see that relevant costs for the non-ridge case, assuming $mathbf{X}$ is $n times p$, are:
$$ C_{LU} = 2 n p^2 - frac{2}{3} p^3 tag{1} $$
$$ C_{SVD} = 4 n p^2 + 8 p^3 tag{2} $$
To obtain equations for ridge regression, we need to substitute $n rightarrow n + p $ to correctly account for data augmentation:
$$ C_{LU} = 2 n p^2 + frac{4}{3} p^3 tag{3} $$
$$ C_{SVD} = 4 n p^2 + 12 p^3 tag{4} $$
Note that unlike your example, $n > p$ for most typical regression problems. If we say that $n approx p$, we can obtained simplified costs in only one variable:
$$ C_{LU} = frac{10}{3} p^3 tag{5} $$
$$ C_{SVD} = 16 p^3 tag{6} $$
This is probably where those rough estimates you reference came from, but as you yourself discovered, this simplification hides important detail and makes inappropriate assumptions that don't fit your case. Note, for example, that the constant for SVD is much worse than for Cholesky, an important fact hidden by the big $mathord{O}$ notation.
What you are doing is exactly the opposite - you have $n ll p$. That means the second term is much more important! By comparing the coefficients of the $p^3$ terms of equations (3) and (4) we can argue the SVD approach will require roughly 10 times as many floating point operations as the Cholesky approach when $n ll p$.
Beyond that, it's hard to know exactly what scipy is doing under the hood - is it calculating both $mathbf{U}$ and $mathbf{V}$ when doing SVD? Is it using the randomized algorithm? Is it using data augmentation for ridge regression, or some other approach? And even once those details are known, it's hard to translate these theoretical operation counts to real-world run time because libraries like LAPACK (scipy is almost surely using LAPACK or ATLAS under the hood) are so highly vectorized (taking advantage of CPU SIMD instructions) and often multithreaded that's it's hard to estimate accurately from purely theoretical arguments. But it's safe to say that it's reasonable to expect Cholesky factorization to be much faster than SVD for a given problem, especially when $n ll p$.
$endgroup$
First, we note that ridge regression can be transformed into an OLS problem via the data augmentation trick.. This adds $p$ rows to the design matrix $mathbf{X}$, so the new matrix $mathbf{X}_lambda$ has dimensions $ (n + p) times p $.
We can then look up the run time for various OLS algorithms, for example in the book Least Squares Data Fitting with Applications or in Golub's own book on Matrix Computations. (Golub was one of inventors of currently state-of-the-art Golub-Reinsch SVD algorithm.) Since linking to Google Books is not very reliable, these costs can also be found, for example, here.
We see that relevant costs for the non-ridge case, assuming $mathbf{X}$ is $n times p$, are:
$$ C_{LU} = 2 n p^2 - frac{2}{3} p^3 tag{1} $$
$$ C_{SVD} = 4 n p^2 + 8 p^3 tag{2} $$
To obtain equations for ridge regression, we need to substitute $n rightarrow n + p $ to correctly account for data augmentation:
$$ C_{LU} = 2 n p^2 + frac{4}{3} p^3 tag{3} $$
$$ C_{SVD} = 4 n p^2 + 12 p^3 tag{4} $$
Note that unlike your example, $n > p$ for most typical regression problems. If we say that $n approx p$, we can obtained simplified costs in only one variable:
$$ C_{LU} = frac{10}{3} p^3 tag{5} $$
$$ C_{SVD} = 16 p^3 tag{6} $$
This is probably where those rough estimates you reference came from, but as you yourself discovered, this simplification hides important detail and makes inappropriate assumptions that don't fit your case. Note, for example, that the constant for SVD is much worse than for Cholesky, an important fact hidden by the big $mathord{O}$ notation.
What you are doing is exactly the opposite - you have $n ll p$. That means the second term is much more important! By comparing the coefficients of the $p^3$ terms of equations (3) and (4) we can argue the SVD approach will require roughly 10 times as many floating point operations as the Cholesky approach when $n ll p$.
Beyond that, it's hard to know exactly what scipy is doing under the hood - is it calculating both $mathbf{U}$ and $mathbf{V}$ when doing SVD? Is it using the randomized algorithm? Is it using data augmentation for ridge regression, or some other approach? And even once those details are known, it's hard to translate these theoretical operation counts to real-world run time because libraries like LAPACK (scipy is almost surely using LAPACK or ATLAS under the hood) are so highly vectorized (taking advantage of CPU SIMD instructions) and often multithreaded that's it's hard to estimate accurately from purely theoretical arguments. But it's safe to say that it's reasonable to expect Cholesky factorization to be much faster than SVD for a given problem, especially when $n ll p$.
answered 6 hours ago
olooneyolooney
1,336614
1,336614
1
$begingroup$
Explicit data augmentation would be a very slow way to compute ridge solution when $nll p$... Everything that scales as $p^3$ can instead be rewritten to scale as $n^3$.
$endgroup$
– amoeba
6 hours ago
$begingroup$
Great answer. This problem arose because a paper I was following suggested, sincen ≪ p
, computing the SVD in order to grid-search for the best alpha, since the resulting computation for each alpha only takes O(pn^2). However, my results/your answer shows that this would only take less time if there are greater than ~10 different values of alpha to try, due to the initial overheard of the SVD being huge.
$endgroup$
– pighead10
6 hours ago
add a comment |
1
$begingroup$
Explicit data augmentation would be a very slow way to compute ridge solution when $nll p$... Everything that scales as $p^3$ can instead be rewritten to scale as $n^3$.
$endgroup$
– amoeba
6 hours ago
$begingroup$
Great answer. This problem arose because a paper I was following suggested, sincen ≪ p
, computing the SVD in order to grid-search for the best alpha, since the resulting computation for each alpha only takes O(pn^2). However, my results/your answer shows that this would only take less time if there are greater than ~10 different values of alpha to try, due to the initial overheard of the SVD being huge.
$endgroup$
– pighead10
6 hours ago
1
1
$begingroup$
Explicit data augmentation would be a very slow way to compute ridge solution when $nll p$... Everything that scales as $p^3$ can instead be rewritten to scale as $n^3$.
$endgroup$
– amoeba
6 hours ago
$begingroup$
Explicit data augmentation would be a very slow way to compute ridge solution when $nll p$... Everything that scales as $p^3$ can instead be rewritten to scale as $n^3$.
$endgroup$
– amoeba
6 hours ago
$begingroup$
Great answer. This problem arose because a paper I was following suggested, since
n ≪ p
, computing the SVD in order to grid-search for the best alpha, since the resulting computation for each alpha only takes O(pn^2). However, my results/your answer shows that this would only take less time if there are greater than ~10 different values of alpha to try, due to the initial overheard of the SVD being huge.$endgroup$
– pighead10
6 hours ago
$begingroup$
Great answer. This problem arose because a paper I was following suggested, since
n ≪ p
, computing the SVD in order to grid-search for the best alpha, since the resulting computation for each alpha only takes O(pn^2). However, my results/your answer shows that this would only take less time if there are greater than ~10 different values of alpha to try, due to the initial overheard of the SVD being huge.$endgroup$
– pighead10
6 hours ago
add a comment |
pighead10 is a new contributor. Be nice, and check out our Code of Conduct.
pighead10 is a new contributor. Be nice, and check out our Code of Conduct.
pighead10 is a new contributor. Be nice, and check out our Code of Conduct.
pighead10 is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Cross Validated!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f396914%2fwhy-is-computing-ridge-regression-with-a-cholesky-decomposition-much-quicker-tha%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Cholesky decomposes a square symmetric matrix (such as correlations) and is much faster itself than SVD which, in general, decomposes a rectangular matrix (such as dataset) into 3 matrices.
$endgroup$
– ttnphns
6 hours ago