search for: corr

Dear R users, Does the results below make any sense? Can the the interval of the correlation coefficient be between *-1.0185* and -0.8265 at 95% confidence level? Liviu > library(boot) > data(mtcars) > with(mtcars, cor.test(mpg, wt, met="spearman")) Spearman's rank correlation rho data: mpg and wt S = 10292, p-value = 1.488e-11 alternative hypothesis: tr...

Dear list, I'd like to use path.analysis in the package agricolae in batch format on many files, retrieving the path coefficients for each run and appending them to a table. I don't see any posts in the help files about this package or the path.analysis package. I've tried creating an object out of the call to path.analysis, but no matter what I try, the function automatically prints

Hi, I have a list p with different size dataframes and length of over 8000. I'm trying to calculate correlations between the rows of dataframes of this list and columns of another dataset (type data.frame also) so that first column is correlated with all the rows in the list dataframe. Some information from another dataset is also included to the final output (all.corrs). This worked a couple of...

Hi all, I have a function rho.f which gives a list of estimators. I have the following problems. rho.f(0.3) gives me the right answer. However, if I use rho.f(corr[4]) give me a different answer, even though corr[4]==0.3. This prevents me from using a for loop. Can someone give me some help? Thank you very much in advance. Hannah > rho.f(0.3) $est.1 [1] 0 0 0 0 0 0 $est.2 [1] 0 0 0 0 0 0 $est.3 [1]...

Hi Group, Consider a data frame like this: mylabel1 <- rep(c("A","B","C"),each=3) mylabel2 <- rep(c("A","B","C"),3) corrs <- c(1,.8,.7,.8,1,.7,.7,.7,1) myData <- data.frame(mylabel1,mylabel2,corrs) myData mylabel1 mylabel2 corrs 1 A A 1.0 2 A B 0.8 3 A C 0.7 4 B A 0.8 5 B B 1.0 6 B C 0.7 7 C A 0....

Hi, I am a newbie of R, and playing with the "ifelse" statement. I have the following codes: ## first, for(i in 1:3) { for(j in 2:4) { cor.temp <- cor(iris.allnum[,i], iris.allnum[,j]) if(i==1 & j==2) corr.iris <- cor.temp else corr.iris <- c(corr.iris, cor.temp) } } this code is working fine. I also tried to perform the same thing in another way with "ifelse": for(i in 1:3) { for(j in 2:4) { cor.temp <- cor(iris.allnum[,i], iris.allnum[,j]) corr.iris <- ifelse(i==1 & j==2...

Hi all, I'd like to use the Hmisc::summarize function, but it uses a function (FUN) of a single vector argument to create the statistical summaries. Consider an easy case: I'd like to compute the correlation between two variables in my dataframe, grouped according to other variables in the same dataframe. For exemple, consider the following dataframe D: V1 V2 V3 A 1 -1 A 1 1 A -1 -1 B 1 1 B 1 1 I'd like to use Hmisc::summarize(X=D, by=llist(myvar=D$V...

hi everyone, Suppose I have three variables a, b, and c each with 10 values. Now I construct a corr matrix for them. Now I want to give the names of columns of corr matrix as a, b, c, i.e. the first column of corr matrix will have name as ‘a’ second column with ‘b’ and so on. Can anyone give me any code by which I can automatically assign the names of columns of corr matrix which are originally t...

I've a vector of pairwise correlations in the order low-index element precedes the high-index element, say: corr(1,2)=0.1, corr(1,3)=0.2, corr(2,3)=0.3, corr(3,4)=0.4 How can I construct the corresponding correlation matrix? I tried using the "combn"-function in "combinat" package: library(combinat) comb...

...(age) { indic=c(2,2) lage=log(age) xpars=c(2.1862908,7.5528077,8.5806697,2.3319461,8.8959507,9.1380187,0.3720293) beta<-c(0,0) alpha1<-c(0,0) alpha2<-c(0,0) beta[1]<-xpars[1] alpha1[1]<-xpars[2] alpha2[1]<-xpars[3] beta[2]<-xpars[4] alpha1[2]<-xpars[5] alpha2[2]<-xpars[6] corr<-matrix(rep(0,4),nc=2) corr[1,1]=1 corr[2,2]=1 corr[1,2]=xpars[7] corr[2,1]=corr[1,2] LLK<-0 L<-c(0,0) R<-c(0,0) for(j in 1:2){ if(indic[j]==1){ L[j]<--Inf R[j]<-alpha1[j]-beta[j]*lage} if(indic[j]==2){ L[j]<-alpha1[j]-beta[j]*lage R[j]<-alpha2[j]...

Two friend reported me a problem, which I can't solve: (I run R-1.0.0, Debian Linux) They hava a function "corr.matrix" (see end of mail), and when they create a 173x173 matrix with this function V <- corr.matrix(0.3, n=173) V1 <- qr.solve(V) reports: Error in qr(a, tol = tol) : NA/NaN/Inf in foreign function call (arg 1) For n < 173, qr.solve returns the correct result. Torsten ________...

...ntially larger than published. The authors of the published example used a different method to calculate the CIs around lambda, but I would have expected the results of the vitalsim to be closer. (The authors used the 'delta' method from Caswell (2001)). My main questions are: 1. Have I correctly calculated stochastic lambda and its 95% CIs? 2. Why am I unable to use any distributions other than beta? What can I do about that? Thank you, Shawn Morrison ##################################################### #My example: ##################################################### rm(list =...

...nrow=2, ncol=2, byrow=T) beta <- seq(0, 1, length.out=n+1) alpha <- sort (sapply(1-beta, qnorm)) x <- array(0, dim=c(n, n)) for (s in 1:n) { for (t in 1:n){ if (s>=t) x[s,t] <- pmvnorm(lower=c(alpha[s], alpha[t]),upper=c(alpha[s+1], alpha[t+1]), mean=c(0,0), corr=var)/(s*(s+1)) else x[s,t] <- pmvnorm(lower=c(alpha[s], alpha[t]),upper=c(alpha[s+1], alpha[t+1]), mean-c(0,0), corr=var)/(t*(t+1)) } } n*beta[2]-(n-1)*pmvnorm(lower=c(alpha[n-1],alpha[n-1]), upper=rep(Inf,2), corr=var, mean=c(0,0))+n*(n-1)*sum(x) } I got the error info...

...ment (med.hyper==1) table(med.hyper) med.hyper 0 1 472 97 I want to subtract 5 mmHg (rr.dia.2m-5) from the measured diastolic blood pressure (rr.dia.2m) if not treated - the value of the measured diastolic blood pressure should remain the same ->my code (data frame is attached !) rr.dia2.corr<-if(med.hyper==1) { rr.dia.2m-5 } else { rr.dia2.corr==rr.dia.2m } R warning Warnmeldung: In if (med.hyper == 1) { : Bedingung hat L?nge > 1 und nur das erste Element wird benutzt "Only first condition is use" which you can see from the result (simply 5 mm was subtracted and the...

Hi, I defined corr to be a function, not realising that this was also the name for correlation in package=boot. How do I explicitly call the corr function within package boot (so, scope up over the current frame, I guess is another way of saying it) without removing my new corr function ? Thanks, Tolga Please fol...

I have two binary files(rasters) with the same dimensions. The first file is called `over` and the second is `corr`. I want to replace values in `over` by `NA` whenever `corr` is greater than 0.5. to read the two files we can use: conne <- file("C:corr.bin","rb") over <- readBin(conne, numeric(), size=4, n=1440*720, signed=TRUE) frf <- file("C:cor206...

...panel.border = element_rect( color = "red", size = 3, linetype = "solid", fill = "transparent" ) ) } p } Basically, this function draws red squares over pairwise corr plots with p<0.05. Now, since I need to use the spearman rank corr, I tried to modify the lm function by adding ?method=spearman? but this did not work at al. Could anybody suggest the way to add the spearman rank corr function in this particular function ? Thanks. -------------- next pa...

...n. I wrote a tiny function to give me back from the original matrix, a matrix showing only the values smaller -0.8 and bigger 0.8. y<-c(0.1,0.2,0.3,-0.8,-0.4,0.9) x<-c(0.5,0.3,0.9,-0.9,-0.7,0.3) XY<-rbind(x,y) extract.values<-function (x) { if(x>=0.8|x<=-0.8)x else("low corr.") } works: Test<-sapply(XY,extract.values,simplify=FALSE) but now I try to solve the problem of having NA in the matrix. I tried like that: extract.values<-function (x) { if(x>=0.8|x<=-0.8|x==NA)x else("low corr.") } woks not: x<-c(0.5,0.3,0.9,-0.9,-0.7,0....

-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 First of all, I'm a physicist and therefore I'm not much used to use statistics. So, please forgive me if this is a FAQ or stupid, but I failed to find the answer by myself. I want to use corr.test to test for the correlation of two data sets (actually I have a lot of data set and perform pairwise testing). But I wanted to find sets where the correlation is less than a certain limit, i.e. |r_exp| < r_0. As far as I can see corr.test is designed to work only for the null hypothesis r_e...

...the minimisation I'm using nlminb and to satisfy the first 2 conditions the logistic transformation is used with box constraints for condition 3. However, I don't seem to be able to get the values x that i'm expecting from fitting a simpler model. For example, set.seed(0) # creating a correlation matrix corr <- diag(5) corr[lower.tri(corr)] <- 0.5 corr[upper.tri(corr)] <- 0.5 # Data for the minimisation mat <- rmvnorm(10000, mean=c(3, 2.75, 2.8, 3, 2.9), sd=c(0.1, 0.1,0.12, 0.15, 0.10), cov=corr) # here is the simple optimisation function that allows the 5th element to...