search for: lm's

Hi R-helpers, I have a character object called dd that has 32 elements each of which is a model formula contained within quotation marks. Here's what it looks like: > dd [1] "lm(y ~ 1,data=Cement)" "lm(y ~ X,data=Cement)" "lm(y ~ X1,data=Cement)" [4] "lm(y ~ X2,data=Cement)" "lm(y ~ X3,data=Cement)" "lm(y ~ X4,data=Cement)" [7] "lm(y ~...

Greetings to r-help land. I've run into some program crashes and I've traced them back to methods() behavior after the package gdata is loaded. I provide now a minimal re-producible example. This seems bugish to me. How about you? dat <- data.frame(x = rnorm(100), y = rnorm(100)) lm1 <- lm(y ~ x, data = dat) methods(class = "lm") ## OK so far library(gdata) methods(class = "lm") ## epic fail ## OUTPUT. > dat <- data.frame(x = rnorm(100), y = rnorm(100)) > lm1 <- lm(y ~ x, data = dat) > > methods(class = "lm") [1] add1....

...# verify data with http://www.otago.ac.nz/sas/stat/chap30/sect52.htm (cbind(drug,disease,y)); ## make a big table drug <- as.factor(rep(drug,6)); disease <- as.factor(rep(disease,6)); y <- c(y); ## verify data through type I ANOVA to http://www.otago.ac.nz/sas/stat/chap30/sect52.htm anova(lm(y~drug*disease)); require(car); ## Type III ANOVA in package car is not Type III ANOVA in SAS Anova(lm(y~drug*disease),type='III'); ## Type III ANOVA in package car equates to wishful ## "anova(lm(y~INTERACTION +disease),lm(y~drug*disease))" ## However in R, df of lm(y~drug:disea...

Dear list, I want to retrieve the p-value of a two-polynomial regression. For a one-polynomial lm I can easily do this with: summary(lm(b~a, data=c)[[4]][[8]]. But how do I find the final p-value in the two-polynomial regression? Under $coefficients I don't find it Any suggestions? Patrick alt <-(2260,2183,2189,1930,2435, 2000,2100,2050,2020,2470, 1700,2310,2090,1560,2060, 1790,1940,...

...e done. My question is: how can I use fewer lines of code (for example using a for loop, a function or plyr) to achieve the same output as below? data(mtcars) # Apply the same model to the dataset but choosing different combinations of dependent (DV) and independent (IV) variables in each case: lm.mpg= lm(mpg~cyl+disp+hp, data=mtcars) lm.drat = lm(drat~wt+qsec, data=mtcars) lm.gear = lm(gear~carb+hp, data=mtcars) # Plot residuals against fitted values for each model plot(lm.mpg$fitted,lm.mpg$residuals, main = "lm.mpg") plot(lm.drat$fitted,lm.drat$residuals, main = "lm.drat&qu...

Hello all, Does anyone know how to constrain/force specific coefficients when running lm()? I need to run recresid() {strucchange package} on the residuals of forecast.lm, but forecast.lm's coefficients must be determined by parameter.estimation.lm I could estimate forecast.lm without lm() and use some other kind of optimisation, but recresid() requires an object with class...

...39; guide at http://guides.rails.info/getting_started.html, running ROR 3. I''m running Ruby 1.9.2 (as instructed by the guide), installed using RVM. I get an error when following "4.3 Setting the Application Home Page". When making a request to http://localhost:3000/ , I get: lm@office:~/blog$ rails server => Booting WEBrick => Rails 3.0.0.beta4 application starting in development on http://0.0.0.0:3000 => Call with -d to detach => Ctrl-C to shutdown server [2010-06-17 17:40:09] INFO WEBrick 1.3.1 [2010-06-17 17:40:09] INFO ruby 1.9.2 (2010-05-31) [i686-linux...

Hello dear r-help group I am turning for you for help with FAQ number 7.31: "Why doesn't R think these numbers are equal?" http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f *My story* is this: I wish to run many lm predictions and need to have them run fast. Using predict.lm is relatively slow, so I tried having it run faster by doing the prediction calculations manually. But doing that gave me problematic results (I won't go into the details of how I found that out). I then discovered that the problem w...

...00 0 0 0 0 0 0 0 0 0 0 1 15 20232 M 1 1X1 0.618689 0 0 0 0 0 0 0 0 0 1 0 ********************************************************************************************************************************************************* I need to store the coefficients using lm() for different combinations of the 4 factors, or different combinations of 3 factors or different combinations of 2 factors or differennt combinations of 1 factor. The formula remains fixed as: > Formula UncDmd ~ M1 + M2 + M3 + M4 + M5 + M6 + M7 + M8 + M9 + M10 + M11 So, different models I wan...

...icence()' for distribution details. R is a collaborative project with many contributors. Type `contributors()' for more information. Type `demo()' for some demos, `help()' for on-line help, or `help.start()' for a HTML browser interface to help. Type `q()' to quit R. > lm(c(1:10) ~ c(1:10)) Call: lm(formula = c(1:10) ~ c(1:10)) Coefficients: (Intercept) 5.5 > lm(c(1:10)+1 ~ c(1:10)) Call: lm(formula = c(1:10) + 1 ~ c(1:10)) Coefficients: (Intercept) c(1:10) 1 1 > lm(c(1:10)+1-1 ~ c(1:10)) Call: lm(formula = c(1:...

Hallo, I hope I am posting to the right place. I was advised to try this list by Ben Bolker (https://twitter.com/bolkerb/status/859909918446497795). I also posted this question to StackOverflow (http://stackoverflow.com/questions/43771269/lm-gives-different-results-from-lm-ridgelambda-0). I am a relative newcomer to R, but I wrote my first program in 1975 and have been paid to program in about 15 different languages, so I have some general background knowledge. I have a regression from which I extract the coefficients like this: lm...

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) > &...

How can I get the results of lm() into a list so I can loop through the results? e.g. myResults[1] <- lm(...) myResults[2] <- lm(...) myResults[3] <- lm(...) ... myResults[15] <- lm(...) myResults[16] <- lm(...) so far every attempt I've tried doesn't work throwing a "number of items to replace is n...

Hi Simon, Yes, if I uses coefficients() I get the same results for lm() and lm.ridge(). So that's consistent, at least. Interestingly, the "wrong" number I get from lm.ridge()$coef agrees with the value from SPSS to 5dp, which is an interesting coincidence if these numbers have no particular external meaning in lm.ridge(). Kind regards, Nick ----...

...ons on factors in regression: Q1. In a table, there a few categorical/factor variables, a few numerical variables and the response variable is numeric. Some factors are important but others not. How to determine which categorical variables are significant to the response variable? Q2. As we knew, lm can deal with categorical variables. I thought, when there is a categorical predictor, we may use lm directly without quantifying these factors and assigning different values to factors would not change the fittings as shown: x <- 1:20 ## numeric predictor yes.no <- c("yes","n...

Hi all, Suppose: y<-rnorm(100) x1<-rnorm(100) lm.yx<-lm(y~x1) To predict from a new data source, one can use: # works as expected dum<-data.frame(x1=rnorm(200)) predict(lm.yx, newdata=dum) Suppose lm.yx has been run and we have the lm object. And we have a dataframe that has columns that don't correspond by name to the original regre...

Am I asking for too much: for any object that a stat proc returns ( y <- lm( y~x) , etc ) ) , is there a super convenient function like give_all_extractors( y ) that lists all extractor functions , the datatype returned , and a text descriptor field ("pairwisepval" "lsmean" etc) That would just be so convenient. What are my options for querying an obj...

tmp <- data.frame(x=c(1,1), y=c(1,2)) tmp.lm <- lm(y ~ x, data=tmp) summary(tmp.lm) coef(summary(tmp.lm)) ## I consider this to be a bug. Since summary(tmp.lm) gives ## two rows for the coefficients, I believe the coef() function ## should also give two rows. > summary(tmp.lm) Call: lm(formula = y ~ x, data = tmp) Residuals:...

Hi, Here is (I hope) all the relevant output from R. > mean(s1$ZDEPRESSION, na.rm=T) [1] -1.041546e-16 > mean(s1$ZDIVERSITY_PA, na.rm=T) [1] -9.660583e-16 > mean(s1$ZMEAN_PA, na.rm=T) [1] -5.430282e-15 > lm.ridge(ZDEPRESSION ~ ZMEAN_PA * ZDIVERSITY_PA, data=s1)$coef ZMEAN_PA ZDIVERSITY_PA ZMEAN_PA:ZDIVERSITY_PA -0.3962254 -0.3636026 -0.1425772 ## This is what I thought was the problem originally. :-) > coefficients(lm(ZDEPRESSION ~ ZMEAN_PA * ZD...

Hi - I'm running R 1.3.0 on i686-pc-linux-gnu > rm(x, y, z) > df <- data.frame(x=1:20,y=1:20,z=factor(1:20 <= 10)) dummy.coef falls over: > dummy.coef.lm(lm(y ~ z * poly(x,1), data=df)) Error in poly(x, 1): Object "x" not found > dummy.coef.lm(lm(y ~ z * I(x), data=df)) Error in unique(c("AsIs", class(x))): Object "x" not found but works OK with: > dummy.coef.lm(lm(y ~ z * x, data=df)) > dummy.coef.lm(lm(...