search for: nondecreasing

Displaying 8 results from an estimated 8 matches for "nondecreasing".

2008 Dec 28
1
Logistic regression with rcs() and inequality constraints?
Dear guRus, I am doing a logistic regression using restricted cubic splines via rcs(). However, the fitted probabilities should be nondecreasing with increasing predictor. Example: predictor <- seq(1,20) y <- c(rep(0,9),rep(1,10),0) model <- glm(y~rcs(predictor,n.knots=3),family="binomial") print(1/(1+exp(-predict(model)))) The last expression should be a nondecreasing sequence, as fitted probabilities make no sense t...
2011 Oct 07
1
Strange behaviour with the Homals Package
...s. X is a matrix with ordinal 1-5 Likert scale entries on 24 questions from about 2500 respondents. So I run X.nlpc = homals(X, rank = 1, level = "ordinal", ndim = 10) to get the first 10 pc's (the number of dimensions is irrelevant to my problem). I happen to get eigenvalues in nondecreasing order and I am not sure how to interpret this. Can someone help me? By the way, I would also be interested in knowing if someone has ever compared the homals solutions with the ones from the categorical quantizations performed with the Aspect package. And, yes, I wrote to package maintainer but...
2006 Jan 29
1
mosaicplot() labels overlap (PR#8536)
...Neglecting edge effects, assume you have a vector of desired positions z, and a vector of minimum widths for each label w. Then, you can compute the space used up by the labels: s[i] = -0.5*w[1] + sum(j<i of w[i]) + 0.5*w[i] and compute y = M(z-s) + s where M() gives the best-fit monotonically nondecreasing fit to it's argument. Y should then be the correct place to put each label. If there's a likelyhood of getting a patch accepted, I could probably supply one. (Given the opportunity, I'd think about shifting the blocks up and down also, to do an overall alignment.)
2024 Jan 29
0
DescTools::Quantile
...bad <- length(x)!=length(w) bad <- bad || (length(p)!=1) bad <- bad || any(diff(x)<=0) if(bad) stop("Arguments 'x' and 'w' must be " %,% "vectors of the same legnth. Argument " %,% "'x' must be a vector of nondecreasing " %,% "values.") if(any(w<=0)||(sum(w)!=1)) stop("elements of 'w' must be positive " %,% "and sum to 1") ## the actual body of the function x[max(which(cumsum(w)<=p))] } Now we write a vectorization of the above tha...
2007 Sep 05
1
Monotone splines
Hello, i have a little problem with R and i hope you can help me. I want to use splines to estimate a function but i want to force the interpolation to be monotone. Is this possible with R ? Thank you, RĂ©mi. --------------------------------- [[alternative HTML version deleted]]
2011 Oct 08
0
Strange behaviour with Homals
...s. X is a matrix with ordinal 1-5 Likert scale entries on 24 questions from about 2500 respondents. So I run X.nlpc = homals(X, rank = 1, level = "ordinal", ndim = 10) to get the first 10 pc's (the number of dimensions is irrelevant to my problem). I happen to get eigenvalues in nondecreasing order and I am not sure how to interpret this. Can someone help me? By the way, I would also be interested in knowing if someone has ever compared the homals solutions with the ones from the categorical quantizations performed with the Aspect package. And, yes, I wrote to package maintainer but...
2011 Mar 21
2
rqss help in Quantreg
Dear All, I'm trying to construct confidence interval for an additive quantile regression model. In the quantreg package, vignettes section: Additive Models for Conditional Quantiles http://cran.r-project.org/web/packages/quantreg/index.html It describes how to construct the intervals, it gives the covariance matrix for the full set of parameters, \theta is given by the sandwich formula
2010 Sep 10
6
adding zeroes after old zeroes in a vector ??
Hello Imagine I have a vector with ones and zeroes I write it compactly: 1111111100001111111111110000000001111111111100101 I need to get a new vector replacing the "N" ones following the zeroes to new zeroes. For example for N = 3 1111111100001111111111110000000001111111111100101 becomes 1111111100000001111111110000000000001111111100000 I can do it with a for loop but I've read