similar to: Vectors with equal sd but different slope

Displaying 20 results from an estimated 1000 matches similar to: "Vectors with equal sd but different slope"

2012 Oct 13
4
Problems with coxph and survfit in a stratified model with interactions
I?m trying to set up proportional hazard model that is stratified with respect to covariate 1 and has an interaction between covariate 1 and another variable, covariate 2. Both variables are categorical. In the following, I try to illustrate the two problems that I?ve encountered, using the lung dataset. The first problem is the warning: To me, it seems that there are too many dummies
2012 Oct 14
1
Problems with coxph and survfit in a stratified model, with interactions
First, here is your message as it appears on R-help. On 10/14/2012 05:00 AM, r-help-request@r-project.org wrote: > I?m trying to set up proportional hazard model that is stratified with > respect to covariate 1 and has an interaction between covariate 1 and > another variable, covariate 2. Both variables are categorical. In the > following, I try to illustrate the two problems that
2023 Nov 15
2
Cannot calculate confidence intervals NULL
R-Experts, Here below my R code working without error message but I don't get the results I am expecting. Here is the result I get: [1] "All values of t are equal to 0.28611928397257 \n Cannot calculate confidence intervals" NULL If someone knows how to solve my problem, really appreciate. Best, S ######################################################### # Difference in Spearman
2003 Apr 28
2
stepAIC/lme problem (1.7.0 only)
I can use stepAIC on an lme object in 1.6.2, but I get the following error if I try to do the same in 1.7.0: Error in lme(fixed = resp ~ cov1 + cov2, data = a, random = structure(list( : unused argument(s) (formula ...) Does anybody know why? Here's an example: library(nlme) library(MASS) a <- data.frame( resp=rnorm(250), cov1=rnorm(250), cov2=rnorm(250),
2023 Nov 15
1
Cannot calculate confidence intervals NULL
I believe the problem is here: cor1 <- cor(x1, y1, method="spearman") cor2 <- cor(x2, y2, method="spearman") The x's and y's are not looked for in data (i.e. NSE) but in the environment where the function was defined, which is standard evaluation. Change the above to: cor1 <- with(d, cor(x1, y1, method="spearman")) cor2 <- with(d, cor(x2, y2,
2012 Jul 06
2
Anova Type II and Contrasts
the study design of the data I have to analyse is simple. There is 1 control group (CTRL) and 2 different treatment groups (TREAT_1 and TREAT_2). The data also includes 2 covariates COV1 and COV2. I have been asked to check if there is a linear or quadratic treatment effect in the data. I created a dummy data set to explain my situation: df1 <- data.frame( Observation =
2008 Aug 04
1
simulate data based on partial correlation matrix
Given four known and fixed vectors, x1,x2,x3,x4, I am trying to generate a fifth vector,z, with specified known and fixed partial correlations. How can I do this? In the past I have used the following (thanks to Greg Snow) to generate a fifth vector based on zero order correlations---however I'd like to modify it so that it can generate a fifth vector with specific partial
2007 Sep 26
2
generate fourth vector based on known correlations
I am trying to generate a fourth vector,z, given three known and fixed vectors, x1,x2,x3 with corresponding known and fixed correlations with themeselves and with z. That is, all correlations are known and prespecified. How can I do this? Thank you, ben
2024 Oct 04
3
apply
Homework questions are not answered on this list. Best, Uwe Ligges On 04.10.2024 10:32, Steven Yen wrote: > The following line calculates standard deviations of a column vector: > > se<-apply(dd,1,sd) > > How can I calculate the covariance matrix using apply? Thanks. > > ______________________________________________ > R-help at r-project.org mailing list -- To
2009 Nov 16
1
extracting values from correlation matrix
Hi! All, I have 2 correlation matrices of 4000x4000 both with same row names and column names say cor1 and cor2. I have extracted some information from 1st matrix cor1 which is something like this: rowname colname cor1_value a b 0.8 b a 0.8 c f 0.62 d k 0.59 - - -- -
2008 Aug 18
1
lmer syntax, matrix of (grouped) covariates?
I have a fairly large model: > length(Y) [1] 3051 > dim(covariates) [1] 3051 211 All of these 211 covariates need to be nested hierarchically within a grouping "class", of which there are 8. I have an accessory vector, " cov2class" that specifies the mapping between covariates and the 8 classes. Now, I understand I can break all this information up into individual
2010 May 03
1
Comparing the correlations coefficient of two (very) dependent samples
Hello all, I believe this can be done using bootstrap, but I am wondering if there is some other way that might be used to tackle this. #Let's say I have two pairs of samples: set.seed(100) s1 <- rnorm(100) s2 <- s1 + rnorm(100) x1 <- s1[1:99] y1 <- s2[1:99] x2 <- x1 y2 <- s2[2:100] #And both yield the following two correlations: cor(x1,y1) # 0.7568969 (cor1) cor(x2,y2)
2010 Jan 30
2
Questions on Mahalanobis Distance
Hello, I am a new R user and trying to learn how to implement the mahalanobis function to measure the distance between to 2 population centroids. I have used STATISTICA to calculate these differences, but was hoping to learn to do the analysis in R. I have implemented the code as below, but my results are very different from that of STATISTICA, and I believe I may not have interpreted the help
2006 Jul 11
2
Multiple tests on 2 way-ANOVA
Dear r-helpers, I have a question about multiple testing. Here an example that puzzles me: All matrixes and contrast vectors are presented in treatment contrasts. 1. example: library(multcomp) n<-60; sigma<-20 # n = sample size per group # sigma standard deviation of the residuals cov1 <- matrix(c(3/4,-1/2,-1/2,-1/2,1,0,-1/2,0,1), nrow = 3, ncol=3, byrow=TRUE, dimnames =
2024 Oct 04
1
apply
On 10/4/2024 5:13 PM, Steven Yen wrote: > Pardon me!!! > > What makes you think this is a homework question? You are not > obligated to respond if the question is not intelligent enough for you. > > I did the following: two ways to calculate a covariance matrix but > wonder how I might replicate the results with "apply". I am not too > comfortable with the
2024 Oct 04
1
apply
Pardon me!!! What makes you think this is a homework question? You are not obligated to respond if the question is not intelligent enough for you. I did the following: two ways to calculate a covariance matrix but wonder how I might replicate the results with "apply". I am not too comfortable with the online do of apply. > set.seed(122345671) > n<-3 > x<-rnorm(n); x
2013 Oct 18
1
crr question‏ in library(cmprsk)
Hi all I do not understand why I am getting the following error message. Can anybody help me with this? Thanks in advance. install.packages("cmprsk") library(cmprsk) result1 <-crr(ftime, fstatus, cov1, failcode=1, cencode=0 ) one.pout1 = predict(result1,cov1,X=cbind(1,one.z1,one.z2)) predict.crr(result1,cov1,X=cbind(1,one.z1,one.z2)) Error: could not find function
2005 Jul 13
1
help: how to plot a circle on the scatter plot
Hello, I have a data set with 15 variables, and use "pairs" to plot the scatterplot of this data set. Then I want to plot some circles on the small pictures with high correlation(e.g. > 0.9). First, I use "cor" to obtain the corresponding correlation matrix (x) for this scatterplot. Second, use "seq(along = x)[x > 0.9]" to find the positions of the small
2024 Oct 04
1
apply
Even if this is not a homework question, it smells like one. If you read the Posting Guide it warns you that homework is off-topic, so when you impose an arbitrary constraint like "must use specific unrelated function" we feel like you are either cheating or wasting our time, and it is up to you to explain why we should follow you down this rabbit hole, keeping in mind that statistics
2024 Oct 04
3
apply
OK. Thanks to all. Suppose I have two vectors, x and y. Is there a way to do the covariance matrix with ?apply?. The matrix I need really contains the deviation products divided by the degrees of freedom (n-1). That is, the elements (1,1), (1,2),...,(1,n) (2,1), (2,2),...., (2,n) .... (n,1),(n,2),...,(n,n). > Hello, > > This doesn't make sense, if you have only one vector you