lm

It is possible to use a shortcut for formula in lm() m <- matrix(rnorm(100), ncol=5) lm(m[,1] ~ m[,2:5] here it would be the same as lm(m[,1] ~ m[,2] + m[,3] + m[,4] + m[,5] but in the case when variables are not of the same level (at least this is my assumption for now) this does not work and I get the error: Error in model.frame.default(formula = hm[, 1] ~ hm[, 2:4], drop.unused.levels = TRUE) : invalid type (list) for variable 'hm[, 2:4]' Data (hm): N cor.distance switches time 1 50 0.04707842 2 0.003 2 100 -0.10769441 2 0.004 3 200 -0.01278359 2 0.004 4 300 0.04229509 5 0.008 5 500 -0

I've read other postings regarding named variables and tried implementing the answers but still get too many values for my new data that I want to run my existing model on. Here is working example code: set.seed(123) mydata <- data.frame("y"=rnorm(100,mean=0, sd = 1),"x"=c(1:100)) mylm <- lm(y ~ x, data=mydata) # ok so mylm is a model on 100 points - lets look at it and the data par(mfrow=c(2,2)) plot(mylm) par(mfrow=c(1,1)) predvals <- predict(mylm, data=mydata) plot(mydata$x,mydata$y) lines(predvals) No surprises here - a straight line through generated points - both 100 observations in

I was trying out linear regression with R using categorical attributes and observe that I don't get a coefficient value for each of the different factor levels I have. Please see my code below, I have 5 factor levels for states, but see only 4 values of co-efficients. > states = c("WA","TE","GE","LA","SF") > population = c(0.5,0.2,0.6,0.7,0.9) > df = data.frame(states,population) > df states population 1 WA 0.5 2 TE 0.2 3 GE 0.6 4 LA 0.7 5 SF 0.9 > states=NULL > population=NULL > lm(formula=population~states,data=df) Call: lm(formula = population ~ states, data = df) Coefficients: (Intercept)

Say I have data.frame a I use m.fit <- lm(col2 ~ col3 * col4, na.action = na.exclude) col2 has some NA values, col3 and col4 have values less than 1. I keep getting Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) : NA/NaN/Inf in foreign function call (arg 1) I've checked the mailing list and it appears that it is because of the NA s in col2 but I tried using na.action=na.exclude/omit/pass but none of them seem to work. I've tested lm again on first 10 entries, definitely not because of the NA s. Problem with this warning is every google results seem to be pointing at NA

I'm trying to predict a value in R using the predict() function, by passing along variables into the model. I am getting the following error: Error in eval(predvars, data, env) : numeric 'envir' arg not of length one Here is my data frame , name df: df <- read.table(text = ' Quarter Coupon Total 1 "Dec 06" 25027.072 132450574 2 "Dec 07" 76386.820 194154767 3 "Dec 08" 79622.147 221571135 4 "Dec 09" 74114.416 205880072 5 "Dec 10" 70993.058 188666980 6 "Jun 06" 12048.162 139137919 7 "Jun 07" 46889.369 165276325 8 "Jun 08" 84732.537 207074374 9 "Jun 09" 83240.084 221945162 10 "Jun 10" 81970.143

In order to correct heteroskedasticity in error terms, I am running the following weighted least squares regression in R : #Call: #lm(formula = a ~ q + q2 + b + c, data = mydata, weights = weighting) #Weighted Residuals: # Min 1Q Median 3Q Max #-1.83779 -0.33226 0.02011 0.25135 1.48516 #Coefficients: # Estimate Std. Error t value Pr(>|t|) #(Intercept) -3.939440 0.609991 -6.458 1.62e-09 *** #q 0.175019 0.070101 2.497 0.013696 * #q2 0.048790 0.005613 8.693 8.49e-15 *** #b 0.473891 0.134918 3.512 0.000598 *** #c 0.119551 0.125430 0.953 0.342167 #--- #Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0