I'm trying to solve the problem
d = 0.5 * ||X - \Sigma||_{Frobenius Norm} + 0.01 * ||XX||_{1},
where X is a symmetric positive definite matrix, and all the diagnoal element should be 1. XX is same with X except the diagonal matrix is 0. \Sigma is known, I want minimum d with X.
My code is as following:
using Convex
m = 5;
A = randn(m, m);
x = Semidefinite(5);
xx=x;
xx[diagind(xx)].=0;
obj=vecnorm(A-x,2)+sumabs(xx)*0.01;
pro= minimize(obj, [x >= 0]);
pro.constraints+=[x[diagind(x)].=1];
solve!(pro)
MethodError: no method matching diagind(::Convex.Variable)
I just solve the optimal problem by constrain the diagonal elements in matrix, but it seems diagind function could not work here, How can I solve the problem.
I think the following does what you want:
m = 5
Σ = randn(m, m)
X = Semidefinite(m)
XX = X - diagm(diag(X))
obj = 0.5 * vecnorm(X - Σ, 2) + 0.01 * sum(abs(XX))
constraints = [X >= 0, diag(X) == 1]
pro = minimize(obj, constraints)
solve!(pro)
For the types of operations:
diag extracts the diagonal of a matrix, as a vector
diagm constructs a diagonal matrix out of a vector
So, to have XX be X with zero diagonal, we subtract the diagonal of X from it. And to constrain X having diagonal 1, we compare its diagonal with 1, using ==.
It is a good idea to keep immutable values as far as possible, instead of trying to modify things. I don't know whether Convex even supports that.
Related
I have an expression that contains several parts. However, for simplicity, consider only the following part as MWE:
Let's assume we have the inverse of a matrix Y that I want to differentiate w.r.t. x.
Y is given as I - (x * b * t(b)), where I is the identity matrix, x is a scalar, and b is a vector.
According to The Matrix Cookbook Equ. 59, the partial derivative of an inverse is:
Normally I would use the function D from the package stats to calculate the derivatives. But that is not possible in this case, because e.g. solve to specify Y as inverse and t() is not in the table of derivatives.
What is the best workaround to circumvent this problem? Are there any other recommended packages that can handle such input?
Example that doesn't work:
f0 <- expression(solve(I - (x * b %*% t(b))))
D(f0, "x")
Example that works:
f0 <- expression(x^3)
D(f0, "x")
3 * x^2
I assume that the question is how to get an explicit expression for the derivative of the inverse of Y with respect to x. In the first section we compute it and in the second section we double check it by computing it numerically and show that the two approaches give the same result.
b and the null space of b are both eigenspaces of Y which we can readily verify by noting that Yb = (1-(b'b)x)b and if z belongs to the nullspace of b then Yz = z. This also shows that the corresponding eigenvalues are 1 - x(b'b) with multiplicity 1 and 1 with multiplicity n-1 (since the nullspace of b has that dimension).
As a result of the fact that we can expand such a matrix into the sum of each eigenvalue times the projection onto its eigenspace we can express Y as the following where bb'/b'b is the projection onto the eigenspace spanned by b and the part pre-multiplying it is the eigenvalue. The remaining terms do not involve x because they involve an eigenvalue of 1 independently of x and the nullspace of b is independent of x as well.
Y = (1-x(b'b))(bb')/(b'b) + terms not involving x
The inverse of Y is formed by taking the reciprocals of the eigenvalues so:
Yinv = 1/(1-x(b'b)) * (bb')/(b'b) + terms not involving x
and the derivative of that wrt x is:
(b'b) / (1 - x(b'b))^2 * (bb')/(b'b)
Cancelling the b'b and writing the derivative in terms of R code:
1/(1 - x*sum(b*b))^2*outer(b, b)
Double check
Using specific values for b and x we can verify it against the numeric derivative as follows:
library(numDeriv)
x <- 1
b <- 1:3
# Y inverse as a function of x
Yinv <- function(x) solve(diag(3) - x * outer(b, b))
all.equal(matrix(jacobian(Yinv, x = 1), 3),
1/(1 - x*sum(b*b))^2*outer(b, b))
## [1] TRUE
I'm reading Deep Learning by Goodfellow et al. and am trying to implement gradient descent as shown in Section 4.5 Example: Linear Least Squares. This is page 92 in the hard copy of the book.
The algorithm can be viewed in detail at https://www.deeplearningbook.org/contents/numerical.html with R implementation of linear least squares on page 94.
I've tried implementing in R, and the algorithm as implemented converges on a vector, but this vector does not seem to minimize the least squares function as required. Adding epsilon to the vector in question frequently produces a "minimum" less than the minimum outputted by my program.
options(digits = 15)
dim_square = 2 ### set dimension of square matrix
# Generate random vector, random matrix, and
set.seed(1234)
A = matrix(nrow = dim_square, ncol = dim_square, byrow = T, rlnorm(dim_square ^ 2)/10)
b = rep(rnorm(1), dim_square)
# having fixed A & B, select X randomly
x = rnorm(dim_square) # vector length of dim_square--supposed to be arbitrary
f = function(x, A, b){
total_vector = A %*% x + b # this is the function that we want to minimize
total = 0.5 * sum(abs(total_vector) ^ 2) # L2 norm squared
return(total)
}
f(x,A,b)
# how close do we want to get?
epsilon = 0.1
delta = 0.01
value = (t(A) %*% A) %*% x - t(A) %*% b
L2_norm = (sum(abs(value) ^ 2)) ^ 0.5
steps = vector()
while(L2_norm > delta){
x = x - epsilon * value
value = (t(A) %*% A) %*% x - t(A) %*% b
L2_norm = (sum(abs(value) ^ 2)) ^ 0.5
print(L2_norm)
}
minimum = f(x, A, b)
minimum
minimum_minus = f(x - 0.5*epsilon, A, b)
minimum_minus # less than the minimum found by gradient descent! Why?
On page 94 of the pdf appearing at https://www.deeplearningbook.org/contents/numerical.html
I am trying to find the values of the vector x such that f(x) is minimized. However, as demonstrated by the minimum in my code, and minimum_minus, minimum is not the actual minimum, as it exceeds minimum minus.
Any idea what the problem might be?
Original Problem
Finding the value of x such that the quantity Ax - b is minimized is equivalent to finding the value of x such that Ax - b = 0, or x = (A^-1)*b. This is because the L2 norm is the euclidean norm, more commonly known as the distance formula. By definition, distance cannot be negative, making its minimum identically zero.
This algorithm, as implemented, actually comes quite close to estimating x. However, because of recursive subtraction and rounding one quickly runs into the problem of underflow, resulting in massive oscillation, below:
Value of L2 Norm as a function of step size
Above algorithm vs. solve function in R
Above we have the results of A %% x followed by A %% min_x, with x estimated by the implemented algorithm and min_x estimated by the solve function in R.
The problem of underflow, well known to those familiar with numerical analysis, is probably best tackled by the programmers of lower-level libraries best equipped to tackle it.
To summarize, the algorithm appears to work as implemented. Important to note, however, is that not every function will have a minimum (think of a straight line), and also be aware that this algorithm should only be able to find a local, as opposed to a global minimum.
I have a (non-symmetric) probability matrix, and an observed vector of integer outcomes. I would like to find a vector that maximises the probability of the outcomes, given the transition matrix. Simply, I am trying to estimate a distribution of particles at sea given their ultimate distribution on land, and a matrix of probabilities of a particle released from a given point in the ocean ending up at a given point on the land.
The vector that I want to find is subject to the constraint that all components must be between 0-1, and the sum of the components must equal 1. I am trying to figure out the best optimisation approach for the problem.
My transition matrix and data set are quite large, but I have created a smaller one here:
I used a simulated known at- sea distribution of
msim<-c(.3,.2,.1,.3,.1,0) and a simulated probability matrix (t) to come up with an estimated coastal matrix (Datasim2), as follows:
t<-matrix (c(0,.1,.1,.1,.1,.2,0,.1,0,0,.3,0,0,0,0,.4,.1,.3,0,.1,0,.1,.4,0,0,0,.1,0,.1,.1),
nrow=5,ncol=6, byrow=T)
rownames(t)<-c("C1","C2","C3","C4","C5") ### locations on land
colnames(t)<-c("S1","S2","S3","S4","S5","S6") ### locations at sea
Datasim<-as.numeric (round((t %*% msim)*500))
Datasim2<-c(rep("C1",95), rep("C2",35), rep("C3",90),rep("C4",15),rep("C5",30))
M <-c(0.1,0.1,0.1,0.1,0.1,0.1) ## starting M
I started with a straightforward function as follows:
EstimateSource3<-function(M,Data,T){
EstEndProbsall<-M%*%T
TotalLkhd<-rep(NA, times=dim(Data)[1])
for (j in 1:dim(Data)[1]){
ObsEstEndLkhd<-0
ObsEstEndLkhd<-1-EstEndProbsall[1,] ## likelihood of particle NOT ending up at locations other than the location of interest
IndexC<-which(colnames(EstEndProbsall)==Data$LocationCode[j], arr.ind=T) ## likelihood of ending up at location of interest
ObsEstEndLkhd[IndexC]<-EstEndProbsall[IndexC]
#Total likelihood
TotalLkhd[j]<-sum(log(ObsEstEndLkhd))
}
SumTotalLkhd<-sum(TotalLkhd)
return(SumTotalLkhd)
}
DistributionEstimate <- optim(par = M, fn = EstimateSource3, Data = Datasim2, T=t,
control = list(fnscale = -1, trace=5, maxit=500), lower = 0, upper = 1)
To constrain the sum to 1, I tried using a few of the suggestions posted here:How to set parameters' sum to 1 in constrained optimization
e.g. adding M<-M/sum(M) or SumTotalLkhd<-SumTotalLkhd-(10*pwr) to the body of the function, but neither yielded anything like msim, and in fact, the 2nd solution came up with the error “L-BFGS-B needs finite values of 'fn'”
I thought perhaps the quadprog package might be of some help, but I don’t think I have a symmetric positive definite matrix…
Thanks in advance for your help!
What about that: Let D = distribution at land, M = at sea, T the transition matrix. You know D, T, you want to calculate M. You have
D' = M' T
hence D' T' = M' (T T')
and accordingly D'T'(T T')^(-1) = M'
Basically you solve it as when doing linear regression (seems SO does not support math notation: ' is transpose, ^(-1) is ordinary matrix inverse.)
Alternatively, D may be counts of particles, and now you can ask questions like: what is the most likely distribution of particles at sea. That needs a different approach though.
Well, I have never done such models but think along the following lines. Let M be of length 3 and D of length 2, and T is hence 3x2. We know T and we observe D_1 particles at location 1 and D_2 particles at location 2.
What is the likelihood that you observe one particle at location D_1? It is Pr(D = 1) = M_1 T_11 + M_2 T_21 + M_3 T_32. Analogously, Pr(D = 2) = M_1 T_12 + M_2 T_22 + M_3 T_32. Now you can easily write the log-likelihood of observing D_1 and D_2 particles at locations 1 and 2. The code might look like this:
loglik <- function(M) {
if(M[1] < 0 | M[1] > 1)
return(NA)
if(M[2] < 0 | M[2] > 1)
return(NA)
M3 <- 1 - M[1] - M[2]
if(M3 < 0 | M3 > 1)
return(NA)
D[1]*log(T[1,1]*M[1] + T[2,1]*M[2] + T[3,1]*M3) +
D[2]*log(T[1,2]*M[1] + T[2,2]*M[2] + T[3,2]*M3)
}
T <- matrix(c(0.1,0.2,0.3,0.9,0.8,0.7), 3, 2)
D <- c(100,200)
library(maxLik)
m <- maxLik(loglik, start=c(0.4,0.4), method="BFGS")
summary(m)
I get the answer (0, 0.2, 0.8) when I estimate it but standard errors are very large.
As I told, I have never done it so I don't know it it makes sense.
I'm currently trying to recreate this Matlab function in R:
function X = uniform_sphere_points(n,d)
% X = uniform_sphere_points(n,d)
%
%function generates n points unformly within the unit sphere in d dimensions
z= randn(n,d);
r1 = sqrt(sum(z.^2,2));
X=z./repmat(r1,1,d);
r=rand(n,1).^(1/d);
X = X.*repmat(r,1,d);
Regarding the the right matrix division I installed the pracma package. My R code right now is:
uniform_sphere_points <- function(n,d){
# function generates n points uniformly within the unit sphere in d dimensions
z = rnorm(n, d)
r1 = sqrt(sum(z^2,2))
X = mrdivide(z, repmat(r1,1,d))
r = rnorm(1)^(1/d)
X = X * matrix(r,1,d)
return(X)
}
But it is not really working since I always end with a non-conformable arrays error in R.
This operation for sampling n random points from the d-dimensional unit sphere could be stated in words as:
Construct a n x d matrix with entries drawn from the standard normal distribution
Normalize each row so it has (2-norm) magnitude 1
For each row, compute a random value by taking a draw from the uniform distribution (between 0 and 1) and raise that value to the 1/d power. Multiply all elements in the row by that value.
The following R code does these operations:
unif.samp <- function(n, d) {
z <- matrix(rnorm(n*d), nrow=n, ncol=d)
z * (runif(n)^(1/d) / sqrt(rowSums(z^2)))
}
Note that in the second line of code I have taken advantage of the fact that multiplying a n x d matrix in R by a vector of length n will multiply each row by the corresponding value in that vector. This saves us the work of using repmat to construct matrices of exactly the same size as our original matrix for these sorts of row-specific operations.
EDIT 2: this post seems to have been moved from CrossValidated to StackOverflow due to it being mostly about programming, but that means by fancy MathJax doesn't work anymore. Hopefully this is still readable.
Say I want to to calculate the squared Mahalanobis distance between two vectors x and y with covariance matrix S. This is a fairly simple function defined by
M2(x, y; S) = (x - y)^T * S^-1 * (x - y)
With python's numpy package I can do this as
# x, y = numpy.ndarray of shape (n,)
# s_inv = numpy.ndarray of shape (n, n)
diff = x - y
d2 = diff.T.dot(s_inv).dot(diff)
or in R as
diff <- x - y
d2 <- t(diff) %*% s_inv %*% diff
In my case, though, I am given
m by n matrix X
n-dimensional vector mu
n by n covariance matrix S
and want to find the m-dimensional vector d such that
d_i = M2(x_i, mu; S) ( i = 1 .. m )
where x_i is the ith row of X.
This is not difficult to accomplish using a simple loop in python:
d = numpy.zeros((m,))
for i in range(m):
diff = x[i,:] - mu
d[i] = diff.T.dot(s_inv).dot(diff)
Of course, given that the outer loop is happening in python instead of in native code in the numpy library means it's not as fast as it could be. $n$ and $m$ are about 3-4 and several hundred thousand respectively and I'm doing this somewhat often in an interactive program so a speedup would be very useful.
Mathematically, the only way I've been able to formulate this using basic matrix operations is
d = diag( X' * S^-1 * X'^T )
where
x'_i = x_i - mu
which is simple to write a vectorized version of, but this is unfortunately outweighed by the inefficiency of calculating a 10-billion-plus element matrix and only taking the diagonal... I believe this operation should be easily expressible using Einstein notation, and thus could hopefully be evaluated quickly with numpy's einsum function, but I haven't even begun to figure out how that black magic works.
So, I would like to know: is there either a nicer way to formulate this operation mathematically (in terms of simple matrix operations), or could someone suggest some nice vectorized (python or R) code that does this efficiently?
BONUS QUESTION, for the brave
I don't actually want to do this once, I want to do it k ~ 100 times. Given:
m by n matrix X
k by n matrix U
Set of n by n covariance matrices each denoted S_j (j = 1..k)
Find the m by k matrix D such that
D_i,j = M(x_i, u_j; S_j)
Where i = 1..m, j = 1..k, x_i is the ith row of X and u_j is the jth row of U.
I.e., vectorize the following code:
# s_inv is (k x n x n) array containing "stacked" inverses
# of covariance matrices
d = numpy.zeros( (m, k) )
for j in range(k):
for i in range(m):
diff = x[i, :] - u[j, :]
d[i, j] = diff.T.dot(s_inv[j, :, :]).dot(diff)
First off, it seems like maybe you're getting S and then inverting it. You shouldn't do that; it's slow and numerically inaccurate. Instead, you should get the Cholesky factor L of S so that S = L L^T; then
M^2(x, y; L L^T)
= (x - y)^T (L L^T)^-1 (x - y)
= (x - y)^T L^-T L^-1 (x - y)
= || L^-1 (x - y) ||^2,
and since L is triangular L^-1 (x - y) can be computed efficiently.
As it turns out, scipy.linalg.solve_triangular will happily do a bunch of these at once if you reshape it properly:
L = np.linalg.cholesky(S)
y = scipy.linalg.solve_triangular(L, (X - mu[np.newaxis]).T, lower=True)
d = np.einsum('ij,ij->j', y, y)
Breaking that down a bit, y[i, j] is the ith component of L^-1 (X_j - \mu). The einsum call then does
d_j = \sum_i y_{ij} y_{ij}
= \sum_i y_{ij}^2
= || y_j ||^2,
like we need.
Unfortunately, solve_triangular won't vectorize across its first argument, so you should probably just loop there. If k is only about 100, that's not going to be a significant issue.
If you are actually given S^-1 rather than S, then you can indeed do this with einsum more directly. Since S is quite small in your case, it's also possible that actually inverting the matrix and then doing this would be faster. As soon as n is a nontrivial size, though, you're throwing away a lot of numerical accuracy by doing this.
To figure out what to do with einsum, write everything in terms of components. I'll go straight to the bonus case, writing S_j^-1 = T_j for notational convenience:
D_{ij} = M^2(x_i, u_j; S_j)
= (x_i - u_j)^T T_j (x_i - u_j)
= \sum_k (x_i - u_j)_k ( T_j (x_i - u_j) )_k
= \sum_k (x_i - u_j)_k \sum_l (T_j)_{k l} (x_i - u_j)_l
= \sum_{k l} (X_{i k} - U_{j k}) (T_j)_{k l} (X_{i l} - U_{j l})
So, if we make arrays X of shape (m, n), U of shape (k, n), and T of shape (k, n, n), then we can write this as
diff = X[np.newaxis, :, :] - U[:, np.newaxis, :]
D = np.einsum('jik,jkl,jil->ij', diff, T, diff)
where diff[j, i, k] = X_[i, k] - U[j, k].
Dougal nailed this one with an excellent and detailed answer, but thought I'd share a small modification that I found increases efficiency in case anyone else is trying to implement this. Straight to the point:
Dougal's method was as follows:
def mahalanobis2(X, mu, sigma):
L = np.linalg.cholesky(sigma)
y = scipy.linalg.solve_triangular(L, (X - mu[np.newaxis,:]).T, lower=True)
return np.einsum('ij,ij->j', y, y)
A mathematically equivalent variant I tried is
def mahalanobis2_2(X, mu, sigma):
# Cholesky decomposition of inverse of covariance matrix
# (Doing this in either order should be equivalent)
linv = np.linalg.cholesky(np.linalg.inv(sigma))
# Just do regular matrix multiplication with this matrix
y = (X - mu[np.newaxis,:]).dot(linv)
# Same as above, but note different index at end because the matrix
# y is transposed here compared to above
return np.einsum('ij,ij->i', y, y)
Ran both versions head-to-head 20x using identical random inputs and recorded the times (in milliseconds). For X as a 1,000,000 x 3 matrix (mu and sigma 3 and 3x3) I get:
Method 1 (min/max/avg): 30/62/49
Method 2 (min/max/avg): 30/47/37
That's about a 30% speedup for the 2nd version. I'm mostly going to be running this in 3 or 4 dimensions but to see how it scaled I tried X as 1,000,000 x 100 and got:
Method 1 (min/max/avg): 970/1134/1043
Method 2 (min/max/avg): 776/907/837
which is about the same improvement.
I mentioned this in a comment on Dougal's answer but adding here for additional visibility:
The first pair of methods above take a single center point mu and covariance matrix sigma and calculate the squared Mahalanobis distance to each row of X. My bonus question was to do this multiple times with many sets of mu and sigma and output a two-dimensional matrix. The set of methods above can be used to accomplish this with a simple for loop, but Dougal also posted a more clever example using einsum.
I decided to compare these methods with each other by using them to solve the following problem: Given k d-dimensional normal distributions (with centers stored in rows of k by d matrix U and covariance matrices in the last two dimensions of the k by d by d array S), find the density at the n points stored in rows of the n by d matrix X.
The density of a multivariate normal distribution is a function of the squared Mahalanobis distance of the point to the mean. Scipy has an implementation of this as scipy.stats.multivariate_normal.pdf to use as a reference. I ran all three methods against each other 10x using identical random parameters each time, with d=3, k=96, n=5e5. Here are the results, in points/sec:
[Method]: (min/max/avg)
Scipy: 1.18e5/1.29e5/1.22e5
Fancy 1: 1.41e5/1.53e5/1.48e5
Fancy 2: 8.69e4/9.73e4/9.03e4
Fancy 2 (cheating version): 8.61e4/9.88e4/9.04e4
where Fancy 1 is the better of the two methods above and Fancy2 is Dougal's 2nd solution. Since the Fancy 2 needs to calculate the inverses of all the covariance matrices I also tried a "cheating version" where it was passed these as a parameter, but it looks like that didn't make a difference. I had planned on including the non-vectorized implementation but that was so slow it would have taken all day.
What we can take away from this is that using Dougal's first method is about 20% faster than however Scipy does it. Unfortunately despite its cleverness the 2nd method is only about 60% as fast as the first. There are probably some other optimizations that can be done but this is already fast enough for me.
I also tested how this scaled with higher dimensionality. With d=100, k=96, n=1e4:
Scipy: 7.81e3/7.91e3/7.86e3
Fancy 1: 1.03e4/1.15e4/1.08e4
Fancy 2: 3.75e3/4.10e3/3.95e3
Fancy 2 (cheating version): 3.58e3/4.09e3/3.85e3
Fancy 1 seems to have an even bigger advantage this time. Also worth noting that Scipy threw a LinAlgError 8/10 times, probably because some of my randomly-generated 100x100 covariance matrices were close to singular (which may mean that the other two methods are not as numerically stable, I did not actually check the results).