I think I may be misreading the help pages, too, but misreading how? I agree that inverting the fitted model is simpler, but I worry that I'm misusing ordinary least squares regression by treating my response, with its error distribution, as a predictor with no such error. In practice, with my real data that includes about six independent peak area measurements per known concentration level, the diagnostic plots from plot.lm(inv.model) look very strange and worrisome.
Certainly predict.lm(..., type = "terms") must be able to do what I need. -John On Wed, Aug 29, 2012 at 6:50 AM, Rui Barradas <ruipbarra...@sapo.pt> wrote: > Hello, > > You seem to be misreading the help pages for lm and predict.lm, argument > 'terms'. > A much simpler way of solving your problem should be to invert the fitted > model using lm(): > > > model <- lm(area ~ concn, data) # Your original model > inv.model <- lm(concn ~ area, data = data) # Your problem's model. > > # predicts from original data > pred1 <- predict(inv.model) > # predict from new data > pred2 <- predict(inv.model, newdata = new) > > # Let's see it. > plot(concn ~ area, data = data) > abline(inv.model) > points(data$area, pred1, col="blue", pch="+") > points(new$area, pred2, col="red", pch=16) > > > Also, 'data' is a really bad variable name, it's already an R function. > > Hope this helps, > > Rui Barradas > > Em 28-08-2012 23:30, John Thaden escreveu: > >> Hello all, >> >> How do I actually use the output of predict.lm(..., type="terms") to >> predict new term values from new response values? >> >> I'm a chromatographer trying to use R (2.15.1) for one of the most >> common calculations in that business: >> >> - Given several chromatographic peak areas measured for control >> samples containing a molecule at known (increasing) concentrations, >> first derive a linear regression model relating the known >> concentration (predictor) to the observed peak area (response) >> - Then, given peak areas from new (real) samples containing >> unknown amounts of the molecule, use the model to predict >> concentrations of the >> molecule in the unknowns. >> >> In other words, given y = mx +b, I need to solve x' = (y'-b)/m for new >> data y' >> >> and in R, I'm trying something like this >> >> require(stats) >> data <- data.frame(area = c(4875, 8172, 18065, 34555), concn = c(25, >> 50, 125, 250)) >> new <- data.frame(area = c(8172, 10220, 11570, 24150)) >> model <- lm(area ~ concn, data) >> pred <- predict(model, type = "terms") >> #predicts from original data >> pred <- predict(model, type = "terms", newdata = new) >> #error >> pred <- predict(model, type = "terms", newdata = new, se.fit = TRUE) >> #error >> pred <- predict(model, type = "terms", newdata = new, interval = >> "prediction") #error >> new2 <- data.frame(area = c(8172, 10220, 11570, 24150), concn = 0) >> new2 >> pred <- predict(model, type = "terms", newdata = new2) >> #wrong results >> >> Can someone please show me what I'm doing wrong? >> >> ______________________________**________________ >> R-help@r-project.org mailing list >> https://stat.ethz.ch/mailman/**listinfo/r-help<https://stat.ethz.ch/mailman/listinfo/r-help> >> PLEASE do read the posting guide http://www.R-project.org/** >> posting-guide.html <http://www.R-project.org/posting-guide.html> >> and provide commented, minimal, self-contained, reproducible code. >> > > [[alternative HTML version deleted]] ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
