search for: squared

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 way of making this automatic, for example, something more along the way of summary(fitSUR$eq[[1:4]])$r.squared ? The above does not work, by the way. I have attached belo...

When the shape of all vertices is set to "square" and the size of the vertices is also set, one get following error (commands attached): Error in l[[which.min(sapply(l, function(p) (p[1] - x0)^2 + (p[2] - y0)^2))]] : attempt to select less than one element Is there a way to solve this problem? Robbie ## Load the igraph package library(igraph) ## Create and plot a small graph

Hello, I have a self-defined function to be computed on each column in a matrix. The basic idea is to ignore the elements that have value of 0 during computation. I should be able to write my own function but it could be computational expensive, so I'd love to ask if anyone may have suggestions on how to implement it more efficiently. Thanks in advance. For example, there are three

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 fo...

Hi everyone I want a square scatterplot with abline(0,1) going exactly through the SW and NE corners. By "square" I mean that the plotting region is exactly square, and that the axis limits are identical. x <- 1:20 y <- x+rep(c(-1,1),10) lims <- range(c(x,y)) None of the following do this: plot(x,y) ; abline(0,1) #not square plot(x,y,asp=1);abline(0,1) #diagonal

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 <-...

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$res is the unweighted residual while summary(model)$res is the weighted resi...

Hi, everybody, 3 questions about R-square: ---------(1)----------- Does R2 always increase as variables are added? ---------(2)----------- Does R2 always greater than 1? ---------(3)----------- How is R2 in summary(lm(y~x-1))$r.squared calculated? It is different from (r.square=sum((y.hat-mean (y))^2)/sum((y-mean(y))^2)) I will illustrate these problems by the following codes: ---------(1)----------- R2 doesn't always increase as variables are added > x=matrix(rnorm(20),ncol=2) > y=rnorm(10) > > lm=lm(y~1) &gt...

...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) Calculate R squared between (X, Y1), (X, Y2), ..., (X,Ym), and then averaging the R squared (b) Calculate average Y=(Y1+Y2, + ... Ym)/m, and then calculate the R squared between (X, Y). We found it seemed that R squared calculated in...

Dear all, I cannot find to explicitly get the R-squared or adjusted R-squared from summary(lm()) Thanks a lot! [[alternative HTML version deleted]]

Hello! It's a shoot in the dark, but I'll try. If one has a total of 100 (e.g., %), and three components of the total, e.g., mytotal=data.frame(x=50,y=30,z=20), - one could build a pie chart with 3 sectors representing x, y, and z according to their proportions in the total. I am wondering if it's possible to build something very similar, but not on a circle but in a square - such that

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 I get the following result -0.7450344 How can we interpret the result based on R-squared and correlation ? From the p-value we c...

...e while applying chisquare test to the following Data ( below the subject of this mail) ...when I wanted to test the significance using three different free statistical packages, here R, EpiInfo and OpenEpi. *Only OpenEpi accepts the test based on Cochran's Recommendations. * R says " chi squared approximation may be incorrect." Does it mean the same as what EpInfo saying " Chi square is not valid" Regards, Dr. B. Manoj Aravind. *Table for analysis* Number of STDs Identified Type of Health worker <2 2 or > ASH...

...9 Region_14 774 121 Region_15 561 24 Region_16 834 134 How well does the survey represent the employee population? Chi-square test says, not very well: > chisq.test(ByRegion) Pearson's Chi-squared test data: ByRegion X-squared = 163.6869, df = 15, p-value < 2.2e-16 By striking three under-represented regions (3,6, and 15), we get a more reasonable, although still not convincing, result: > chisq.test(ByRegion[setdiff(1:16,c(3,6,15)),]) Pearson's Chi-squared test data:...

This is a linear regression of Y onto factors... If I take log of Y, and regress onto the factors, I got: Multiple R-squared: 0.4023, Adjusted R-squared: 0.2731 If I don't take log of Y, and directly regress Y onto the factors, I got: Multiple R-squared: 0.1807, Adjusted R-squared: -0.001112 Is this negative adjusted R^2 a problem? What observation can I make here and what might be the problem? Thanks!

...3.0-3 C front end for LLVM C/C++ compiler ii llvm-runtime 2.9-7 Low-Level Virtual Machine (LLVM), bytecode interpreter I have a very simple example that appears to illustrate this problem /******** squared.h BEGIN *****/ int squared(int val); /********* END *****/ /******* squared.c BEGIN ****/ int squared(int val) { return val*val; } /********* END ****/ /**** main.c BEGIN *****/ #include <stdio.h> int main(int argc,char *argv[] ) { int i = 0; for( i = 0; i &...

...C"),c("A1","A2","A3"))) > ta A1 A2 A3 A 45 32 1 B 85 40 2 C 27 34 1 > fisher.test(ta) Fisher's Exact Test for Count Data data: ta p-value = 0.02337 alternative hypothesis: two.sided > chisq.test(ta, simulate=T, B=100000) Pearson's Chi-squared test with simulated p-value (based on 1e+05 replicates) data: ta X-squared = 9.6976, df = NA, p-value = 0.04037 > chisq.test(ta) Pearson's Chi-squared test data: ta X-squared = 9.6976, df = 4, p-value = 0.04584 Warning message: Chi-squared approximation may be incorrect in: chisq.te...

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: x<-1:10 y<-c(2,4,3,4...

Hello All- I found an inconsistency between the R-squared reported in Excel vs. that in R, and I am wondering which (if any) may be correct and if this is a known issue. While it certainly wouldn't surprise me if Excel is just flat out wrong, I just want to make sure since the R- squared reported in R seems surprisingly high. Please let me kn...

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 s...