[R] predict.lm(...,type="terms") question

[John Thaden]

Could it be that my newdata object needs to include a column for the
concn term even though I'm asking for concn to be predicted? If so,
what numbers would I fill it with? Or should my newdata object include
the original data, too? Is there another mailing list I can ask?
Thanks,
-John

On Wed, Aug 29, 2012 at 9:16 AM, John Thaden <johnthaden at gmail.com> wrote:
> 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 <ruipbarradas at 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 at 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.
>>
>>
>