I have a been trying to figure that out but without much success. I am working with a table with binary data (0s and 1s). I managed to estimate a distance matrix from my data using the R function dist(x,method="binary"), but I am not quite sure how exactly this function estimates the distance matrix. Is it using the Jaccard coefficient J=(M11)/(M10+M01+M11)?
This is easily found in the help page ?dist:
This function computes and returns the distance matrix computed by using the specified distance measure to compute the distances between the rows of a data matrix.
[...]
binary: (aka asymmetric binary): The vectors are regarded as binary
bits, so non-zero elements are ‘on’ and zero elements are ‘off’. The
distance is the proportion of bits in which only one is on amongst
those in which at least one is on.
This is equivalent to the Jaccard distance as described in Wikipedia:
An alternate interpretation of the Jaccard distance is as the ratio of the size of the symmetric difference to the union.
In your notation, it is 1 - J = (M01 + M10)/(M01 + M10 + M11).
Related
I am trying to generate two vectors with a given cosine similarity. Input would be the degree of cosine similarity (or angle as it depends on it anyway) and the number of dimensions (D) in the vectors, and output would be two vectors of D dimensions with that given similarity between them Now, I know how to use the cosine similarity function to calculate the similarity but I'm lost when trying it the other way around.
Is there such a procedure or algorithm and how is it called?
For a given starting vector u and cosine similarity c:
Generate 2 points in n-D space; call them a and b
Project b onto the plane orthogonal to u and containing a
Subtract a from the result to obtain a vector orthogonal to u; call it h
Use [u,h] as a basis and basic trigonometry to generate the desired vector v
The above method is dimension-agnostic as it only uses dot products. The resultant vectors {v} are of unit length and uniformly distributed around u.
My problem is the following:
I'm trying to use R in order to compute numerically this problem.
So I've correctly setup the problem in my console, and then I tried to compute the eigenvectors.
But I expect that the eigenvector associated with lambda = 1 is (1,2,1) instead of what I've got here. So, the scaling is correct (0.4082483 is effectively half of 0.8164966), but I would like to obtain a consistent result.
My original problem is to find a stationary distribution for a Markov Chain using R instead of doing it on paper. So from a probabilistic point of view, my stationary distribution is a vector whose sum of the components is equal to 1. For that reason I was trying to change the scale in order to obtain what I've defined "a consistent result".
How can I do that ?
The eigen vectors returned by R are normalized (for the square-norm). If V is a eigen vector then s * V is a eigen vector as well for any non-zero scalar s. If you want the stationary distribution as in your link, divide by the sum:
V / sum(V)
and you will get (1/4, 1/2, 1/4).
So:
ev <- eigen(t(C))$vectors
ev / colSums(ev)
to get all the solutions in one shot.
C <- matrix(c(0.5,0.25,0,0.5,0.5,0.5,0,0.25,0.5),
nrow=3)
ee <- eigen(t(C))$vectors
As suggested by #Stéphane Laurent in the comments, the scaling of eigenvectors is arbitrary; only the relative value is specified. The default in R is that the sum of squares of the eigenvectors (their norms) are equal to 1; colSums(ee^2) is a vector of 1s.
Following the link, we can see that you want each eigenvector to sum to 1.
ee2 <- sweep(ee,MARGIN=2,STATS=colSums(ee),FUN=`/`)
(i.e., divide each eigenvector by its sum).
(This is a good general solution, but in this case the sum of the second and third eigenvectors are both approximately zero [theoretically, they are exactly zero], so this only really makes sense for the first eigenvector.)
Rao QE is a weighted Euclidian distance matrix. I have the vectors for the elements of the d_ijs in a data table dt, one column per element (say there are x of them). p is the final column. nrow = S. The double sums are for the lower left (or upper right since it is symmetric) elements of the distance matrix.
If I only needed an unweighted distance matrix I could simply do dist() over the x columns. How do I weight the d_ijs by the product of p_i and p_j?
And example data set is at https://github.com/GeraldCNelson/nutmod/blob/master/RaoD_example.csv with the ps in the column called foodQ.ratio.
You still start with dist for the raw Euclidean distance matrix. Let it be D. As you will read from R - How to get row & column subscripts of matched elements from a distance matrix, a "dist" object is not a real matrix, but a 1D array. So first do D <- as.matrix(D) or D <- dist2mat(D) to convert it to a complete matrix before the following.
Now, let p be the vector of weights, the Rao's QE is just a quadratic form q'Dq / 2:
c(crossprod(p, D %*% p)) / 2
Note, I am not doing everything in the most efficient way. I have performed a symmetric matrix-vector multiplication D %*% p using the full D rather than just its lower triangular part. However, R does not have a routine doing triangular matrix-vector multiplication. So I compute the full version than divide 2.
This doubles computation amount that is necessary; also, making D a full matrix doubles memory costs. But if your problem is small to medium size this is absolutely fine. For large problem, if you are R and C wizard, call BLAS routine dtrmv or even dtpmv for the triangular matrix-vector computation.
Update
I just found this simple paper: Rao's quadratic entropy as a measure of functional diversity based on multiple traits for definition and use of Rao's EQ. It mentions that we can replace Euclidean distance with Mahalanobis distance. In case we want to do this, use my code in Mahalanobis distance of each pair of observations for fast computation of Mahalanobis distance matrix.
I'd like to calculate multivariate distance from a set of points to the centroid of those points. Mahalanobis distance seems to be suited for this. However, I get an error (see below).
Can anyone tell me why I am getting this error, and if there is a way to work around it?
If you download the coordinate data and the associated environmental data, you can run the following code.
require(maptools)
occ <- readShapeSpatial('occurrences.shp')
load('envDat.Rdata')
#standardize the data to scale the variables
dat <- as.matrix(scale(dat))
centroid <- dat[1547,] #let's assume this is the centroid in this case
#Calculate multivariate distance from all points to centroid
mahalanobis(dat,center=centroid,cov=cov(dat))
Error in solve.default(cov, ...) :
system is computationally singular: reciprocal condition number = 9.50116e-19
The Mahalanobis distance requires you to calculate the inverse of the covariance matrix. The function mahalanobis internally uses solve which is a numerical way to calculate the inverse. Unfortunately, if some of the numbers used in the inverse calculation are very small, it assumes that they are zero, leading to the assumption that it is a singular matrix. This is why it specifies that they are computationally singular, because the matrix might not be singular given a different tolerance.
The solution is to set the tolerance for when it assumes that they are zero. Fortunately, mahalanobis allows you to pass this parameter (tol) to solve:
mahalanobis(dat,center=centroid,cov=cov(dat),tol=1e-20)
# [1] 24.215494 28.394913 6.984101 28.004975 11.095357 14.401967 ...
mahalanobis uses the covariance matrix, cov, (more precisely the inverse of it) to transform the coordinate system, then compute Euclidian distance in the new coordinates. A standard reference is Duda & Hart "Pattern Classification and Scene Recognition"
Looks like your cov matrix is singular. Perhaps there are linearly-dependent columns in "dat" that are unnecessary? Setting the tolerance to zero won't help if
the covariance matrix is truly singular. The first thing to do, instead, is look for columns that might be a rescaling of some other column, or might be just a sum of 2 or more other columns and remove them. Such columns are redundant for the mahalanobis distance.
BTW, since mahalanobis distance is effectively a rescaling and rotation, calling the scaling function looks superfluous - any reason why you want that?
Now i have a 7000*7000 correlation matrix and I have to do PCA on this in R.
I used the
CorPCA <- princomp(covmat=xCor)
, xCor is the correlation matrix
but it comes out
"covariance matrix is not non-negative definite"
it is because i have some negative correlation in that matrix.
I am wondering which inbuilt function in R that i can use to get the result of PCA
One method to do the PCA is to perform an eigenvalue decomposition of the covariance matrix, see wikipedia.
The advantage of the eigenvalue decomposition is that you see which directions (eigenvectors) are significant, i.e. have a noticeable variation expressed by the associated eigenvalues. Moreover, you can detect if the covariance matrix is positive definite (all eigenvalues greater than zero), not negative-definite (which is okay) if there are eigenvalues equal zero or if it is indefinite (which is not okay) by negative eigenvalues. Sometimes it also happens that due to numerical inaccuracies a non-negative-definite matrix becomes negative-definite. In that case you would observe negative eigenvalues which are almost zero. In that case you can set these eigenvalues to zero to retain the non-negative definiteness of the covariance matrix. Furthermore, you can still interpret the result: eigenvectors contributing the significant information are associated with the biggest eigenvalues. If the list of sorted eigenvalues declines quickly there are a lot of directions which do not contribute significantly and therefore can be dropped.
The built-in R function is eigen
If your covariance matrix is A then
eigen_res <- eigen(A)
# sorted list of eigenvalues
eigen_res$values
# slightly negative eigenvalues, set them to small positive value
eigen_res$values[eigen_res$values<0] <- 1e-10
# and produce regularized covariance matrix
Areg <- eigen_res$vectors %*% diag(eigen_res$values) %*% t(eigen_res$vectors)
not non-negative definite does not mean the covariance matrix has negative correlations. It's a linear algebra equivalent of trying to take square root of negative number! You can't tell by looking at a few values of the matrix, whether it's positive definite.
Try adjusting some default values like tolerance in princomp call. Check this thread for example: How to use princomp () function in R when covariance matrix has zero's?
An alternative is to write some code of your own to perform what is called a n NIPLAS analysis. Take a look at this thread on the R-mailing list: https://stat.ethz.ch/pipermail/r-help/2006-July/110035.html
I'd even go as far as asking where did you obtain the correlation matrix? Did you construct it yourself? Does it have NAs? If you constructed xCor from your own data, do you think you can sample the data and construct a smaller xCor matrix? (say 1000X1000). All these alternatives try to drive your PCA algorithm through the 'happy path' (i.e. all matrix operations can be internally carried out without difficulties in diagonalization etc..i.e., no more 'non-negative definite error msgs)