similar to: Extract r.squared using cbind in lm

# Hello, # I have a data set that looks something like the following: site<-c(rep('a',5),rep('b',2),rep('c',4),rep('d',11)) year<-c(1980, 1981, 1982, 1993, 1995, 1980, 1983, 1981, 1993, 1995, 1999, c(1980:1990)) count<-c(60,35,36,12,8,112,98,20,13,15,15,65,43,49,51,34,33,33,33,40,11,0) data<-data.frame(site, year, count) # > site year count # 1

Hello, I''m using find_with_ferret to search multiple models and it works great. The trouble is I need to filter the results using :include and :conditions. I get two errors depending on the syntax I use in the search. Reading the source, I see the retrieve_records method seems to filter the :include and :conditions so that they only apply to the relevant model when searching

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]]

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]]

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

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:

What is the difference between the two R-squareds returned for a linear regression by summary(lm...)? When might one report multiple vs. adjusted R-squared? Thank you, Ben Osborne -- Botany Department University of Vermont 109 Carrigan Drive Burlington, VT 05405 benjamin.osborne at uvm.edu phone: 802-656-0297 fax: 802-656-0440

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:

Hi all, I have used a hold out sample to predict a model but now I want to compute an R squared value for the prediction. Any help is appreciated. Best regards -- View this message in context: http://r.789695.n4.nabble.com/R-squared-for-lm-prediction-tp2955328p2955328.html Sent from the R help mailing list archive at Nabble.com.

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:

Hello, I've two data.frames (data1 and data4), dec="." and sep=";". http://r.789695.n4.nabble.com/file/n4199964/data1.txt data1.txt http://r.789695.n4.nabble.com/file/n4199964/data4.txt data4.txt When I do plot(data1$nx,data1$ny, col="red") points(data4$nx,data4$ny, col="blue") , results seem very similar (at least to me) but the R-squared of

Do you mean w <- z$residuals ? Type names(z) to see the list of item in your model. I ran your code on a lm and it work fine. You don't need the brackets around mss <- Michael Long On 04/07/2016 02:21 PM, Murray Efford wrote: > Following some old advice on this list, I have been reading the code for summary.lm to understand the computation of R-squared from a weighted

I have a situation with a large dataset (3000+ observations), where I'm doing lags as regressors, where I get: Call: lm(formula = rj ~ rM + rM.1 + rM.2 + rM.3 + rM.4) Residuals: 1990-06-04 1994-11-14 1998-08-21 2002-03-13 2005-09-15 -5.64672 -0.59596 -0.04143 0.55412 8.18229 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -0.003297 0.017603

On 07/04/2016 5:21 PM, Murray Efford wrote: > Following some old advice on this list, I have been reading the code for summary.lm to understand the computation of R-squared from a weighted regression. Usually weights in lm are applied to squared residuals, but I see that the weighted mean of the observations is calculated as if the weights are on the original scale: > > [...] > f

Dear all, I'm puzzled by the definition of r-squared used by 'lm' when the model is without intercept. The help for summary says: >r.squared: R^2, the ``fraction of variance explained by the model'', > > R^2 = 1 - Sum(R[i]^2) / Sum((y[i]- y*)^2), > > where y* is the mean of y[i] if there is an intercept and > zero

Following some old advice on this list, I have been reading the code for summary.lm to understand the computation of R-squared from a weighted regression. Usually weights in lm are applied to squared residuals, but I see that the weighted mean of the observations is calculated as if the weights are on the original scale: [...] f <- z$fitted.values w <- z$weights [...] m

On 08 Apr 2016, at 12:57 , Duncan Murdoch <murdoch.duncan at gmail.com> wrote: > On 07/04/2016 5:21 PM, Murray Efford wrote: >> Following some old advice on this list, I have been reading the code for summary.lm to understand the computation of R-squared from a weighted regression. Usually weights in lm are applied to squared residuals, but I see that the weighted mean of the

Hi, A have small technical question about the calculation of R-squared using lm(). In a study case with experimental values, it seems more logical to force the regression line to pass through origin with lm(y ~ x +0). However, R-squared values are higher in this case than when I compute the linear regression with lm(y ~ x). It seems to be surprising to me: is this result normal ? Is there

>>>>> Murray Efford <murray.efford at otago.ac.nz> >>>>> on Fri, 8 Apr 2016 18:45:33 +0000 writes: > Thanks for these perfectly consistent replies - I didn't > understand the purpose of m = sum(w * f/sum(w)) and saw it > merely as a weighted average of the fitted values. My > ultimate concern is how to compute an appropriate

Hello. Consider the response-variable of data.frame df is constant, so analytically perfect fit of a linear model is expected. Fitting a regression line using lm result in residuals, slope and std.errors not exactly zero, which is acceptable in some way, but errorneous. But if you use summary.lm it shows inacceptable error propagation in the calculation of the t value and the corresponding