similar to: loop issues (r.squared)

I would like to compare every column in my matrix with every other column and get the r-squared. I tried using the following formula and looping through every column: > summary(lm(matrix[,x]~matrix[,y]))$r.squared If I have 10,000 columns, the loops (10,000 * 10,000) take forever even if there is no formula inside. Then, I attempted to vectorize my code: > cor(matrix)^2 With 10,000

I have been using the read.csv function for a while now without any problems. My files are usually 20-50 MBs and they take up to a minute to import. They have all been under 50,000 rows and under 100 columns. Recently, I tried importing a file of a similar size (which means about the same amount of data), but with ~500,000 columns and ~20 rows. The process is taking forever (~1 hour so far). In

Hello R-helpers, I need to generate standard variates normal to 'create' chi-squared variates. To make you more understand, (1)   a<-rnorm(3,0,1) *after do (1), I need to squared and summed the three values. My problem is, how am I going to continue the programming if I had to repeat the process for 15 times, which in the end I will get 15 values from the whole programme.Hope you can

I am pretty new to R. So this may be an easy question for most of you. ? I would like to calculate the squared distances of a large set (let's say 20000) of vectors (let's say dimension of 5) to a fixed vector. ? Say I have a data frame MY_VECTORS with 20000 rows and 5 columns, and one 5x1 vector y. I would like to efficiently calculate the squared distances?between each of the 20000

Greetings R Users, I have a system of equations for which I would like to output all the R-squares. Assume there are four equations in my system, the only way I found to output all the R-squares is by calling them out one by one as this: summary(fitSUR$eq[[1]])$r.squared summary(fitSUR$eq[[2]])$r.squared summary(fitSUR$eq[[3]])$r.squared summary(fitSUR$eq[[4]])$r.squared But isn't there a

Full_Name: Adriano Azevedo Filho Version: 1.9.1 OS: Windows, Linux Submission from: (NULL) (200.171.246.212) R-squared and Adjusted R-squared appear to be wrong when the formula in lm() is specified without intercept. Problem present in both Windows and Linux 1.9.1 version. Also in the 1.8.1 version for Windows (other versions not checked). Possible example which reproduces the problem:

Hello, I am using cbind in a lm-model. For standard lm-models the r.squared can be easily extracted with summary(model)$r.squared, but that is not working in in the case with cbind. Here an example to illustrate the problem: a <- c(1,3,5,2,5,3,1,6,7,2,3,2,6) b <- c(12,15,18,10,18,22,9,7,9,23,12,17,13) c <- c(22,26,32,33,32,28,29,37,34,29,30,32,29) data <- data.frame(a,b,c)

Hi I am trying to find the relationship between two variables. First I fitted a linear model between two variables and I found the following results: Residual standard error: 0.03253 on 2498 degrees of freedom Multiple R-squared: 0.5551, Adjusted R-squared: 0.5549 F-statistic: 3116 on 1 and 2498 DF, p-value: < 2.2e-16 Then I used the cor function to see the correlation between two variable

Hello all, I am using the function lm to do my weighted least square regression. model<-lm(Y~X1+X2, weight=w) What I am confused is the r.squared. It does not seem that the r.squared for the weighted case is an ordinary 1-RSS/TSS. What is that precisely? Is the r.squared measure comparable to that obtained by the ordinary least square? <I also notice that model$res is the unweighted

Greetings Listers! the R-squared value reported by summary of lm is calculated as 1 - RSS/RSS_m where RSS_m is the residual sum of squares of a minimal model. In most cases, the minimal model is simply y = mean(y), but when a constant is left out of the model, the minimal model is y = 0. However, if you manually add a constant, R still considers y = 0 the minimal model. This also causes

What is the exact formula used in R lm() for the Adjusted R-squared? How can I interpret it? There seem to exist several formula's to calculate Adjusted R-squared. Wherry’s formula [1-(1-R2)·(n-1)/(n-v)] McNemar’s formula [1-(1-R2)·(n-1)/(n-v-1)] Lord’s formula [1-(1-R2)(n+v-1)/(n-v-1)] Stein 1-(n-1/n-k-1)(n-2)/n-k-2) (n+1/n) Theil's formula (found here:

I may be missing something obvious here, but consider the following simple dataset simulating repeated measures on 5 individuals with pretty strong between-individual variance. set.seed(1003) n<-5 v<-rep(1:n,each=2) d<-data.frame(factor(v),v+rnorm(2*n)) names(d)<-c("id","y") Now consider the following two linear models that provide identical fitted values,

Hi, We got a question about interpretating R-suqared. The actual outputs for a test dataset is X=(x1,x2, ..., xn). model 1 predicted the outputs as Y1=(y11,y12,..., y1n) model n predicted the outputs as Y2=(y21,y22,..., y2n) ... model m predicted the outputs as Ym=(ym1,ym2,..., ymn) Now we have two ways to calculate R squared to evaluate the average performance of committee model. (a)

How i can perform a Pearson's Chi-squared Test in this data set: | Outcome -----------------+-----------+----------------------------------+ Treatment | Sex | None |Some | Marked | Total -----------------+------------+--------+--------+-------------+ Active | Female | 6 | 5 | 16 | 27

Hi, R Users I find a problem in extracting the R-squared and P-value from the lm results described below (in Italic), *Residual standard error: 2.25 on 17 degrees of freedom* *Multiple R-squared: 0.001069, Adjusted R-squared: -0.05769 * *F-statistic: 0.01819 on 1 and 17 DF, p-value: 0.8943 * * * Any suggestions will be appreciated. Thanks. Wenjun [[alternative HTML version deleted]]

Hi, I would consider the calculation of r-squared in the following to be a bug, but then, I've been wrong before. It seems that R looks to see if the model contains an intercept term, and if it does not, computes r-squared in a way I don't understand. To my mind, the following are two alternative parametrizations of the same model, and should yield the same r-squared. Any insight much

Hi, Sorry for the naive question, but what exactly does the 'Adjusted R-squared' coefficient in the summary of linear model adjust for? Sample code: > x <- rnorm(15) > y <- rnorm(15) > lmr <- lm(y~x) > summary(lmr) Call: lm(formula = y ~ x) Residuals: Min 1Q Median 3Q Max -1.7828 -0.7379 -0.4485 0.7563 2.1570 Coefficients:

Dear UseRs, I'm running a chi-squared test where the expected matrix is the same as the observed, after rounding. R reports a X-squared of zero with a p value of one. I can justify this because any other result will deviate at least as much from the expected because what we observe is the expected, after rounding. But the formula for X-squared, sum (O-E)^2/E gives a positive value. What

I am simulating an ANOVA model and get a strange behavior from the summary function. To be more specific: please run the following code and see for yourself: the summary()[["r.squared"]] values of two identical models are quite different!! ## 3 x 3 ANOVA of two factors x and z on outcome y s.size <- 300 # the sample size p.z <- c(0.25, 0.5, 0.25) # the probabilities of factor z ##

Hi, I'm trying to verify the calculation of coefficient of determination (r squared) for linear regression. I've done the calculation manually with a simple test case and using the definition of r squared outlined in summary(lm) help. There seems to be a discrepancy between the what R produced and the manual calculation. Does anyone know why this is so? What does the multiple r squared