I am fitting a model with:
var4pca <- lm(lg[5:415,1] ~ pcalg1$x[, 1:8] + pcalg2$x[, 1:8] + pcalg3$x[, 1:8] + pcalg4$x[, 1:8])
I now want to predict values for a validation set(83 rows). How can I do this?
I am trying to use:
pred_pca<-predict(var4pca, va)
where va is my validation set. But this is returning me a vector with length 411, whereas I only want length 83
In my experience, lm is very fussy about prediction. It demands that the new data look exactly like the data used to create the model. By that I mean things like col names have to match. What typically will work is to create a data frame of all the data and then create df.train and df.test as the correct rows of the data frame. That should do the trick. As joran says be careful with formulas. One advantage of putting all the data into a df with named cols is that then one can use the formula depvar ~. - typically much easier to write.
Related
Working in R to develop regression models, I have something akin to this:
c_lm = lm(trainingset$dependent ~ trainingset$independent)
c_pred = predict(c_lm,testset$independent))
and every single time, I get a mysterious error from R:
Warning message:
'newdata' had 34 rows but variables found have 142 rows
which essentially translates into R not being able to find the independent column of the testset data.frame. This is simply because the exact name from the right-hand side of the formula in lm must be there in predict. To fix it, I can do this:
tempset = trainingset
c_lm = lm(trainingset$dependent ~ tempset$independent)
tempset = testset
c_pred = predict(c_lm,tempset$independent))
or some similar variation, but this is really sloppy, in my opinion.
Is there another way to clean up the translation between the two so that the independent variables' data frame does not have to have the exact same name in predict as it does in lm?
No, No, No, No, No, No! Do not use the formula interface in the way you are doing if you want all the other sugar that comes with model formulas. You wrote:
c_lm = lm(trainingset$dependent ~ trainingset$independent)
You repeat trainingset twice, which is a waste of fingers/time, redundant, and not least causing you the problem that you are hitting. When you now call predict, it will be looking for a variable in testset that has the name trainingset$independent, which of course doesn't exist. Instead, use the data argument in your call to lm(). For example, this fits the same model as your formula but is efficient and also works properly with predict()
c_lm = lm(dependent ~ independent, data = trainingset)
Now when you call predict(c_lm, newdata = testset), you only need to have a data frame with a variable whose name is independent (or whatever you have in the model formula).
An additional reason to write formulas as I show them, is legibility. Getting the object name out of the formula allows you to more easily see what the model is.
I am trying to convert Absorbance (Abs) values to Concentration (ng/mL), based on an established linear model & standard curve. I planned to do this by using the predict() function. I am having trouble getting predict() to return the desired results. Here is a sample of my code:
Standards<-data.frame(ng_mL=c(0,0.4,1,4),
Abs550nm=c(1.7535,1.5896,1.4285,0.9362))
LM.2<-lm(log(Standards[['Abs550nm']])~Standards[['ng_mL']])
Abs<-c(1.7812,1.7309,1.3537,1.6757,1.7409,1.7875,1.7533,1.8169,1.753,1.6721,1.7036,1.6707,
0.3903,0.3362,0.2886,0.281,0.3596,0.4122,0.218,0.2331,1.3292,1.2734)
predict(object=LM.2,
newdata=data.frame(Concentration=Abs[1]))#using Abs[1] as an example, but I eventually want predictions for all values in Abs
Running that last lines gives this output:
> predict(object=LM.2,
+ newdata=data.frame(Concentration=Abs[1]))
1 2 3 4
0.5338437 0.4731341 0.3820697 -0.0732525
Warning message:
'newdata' had 1 row but variables found have 4 rows
This does not seem to be the output I want. I am trying to get a single predicted Concentration value for each Absorbance (Abs) entry. It would be nice to be able to predict all of the entries at once and add them to an existing data frame, but I can't even get it to give me a single value correctly. I've read many threads on here, webpages found on Google, and all of the help files, and for the life of me I cannot understand what is going on with this function. Any help would be appreciated, thanks.
You must have a variable in newdata that has the same name as that used in the model formula used to fit the model initially.
You have two errors:
You don't use a variable in newdata with the same name as the covariate used to fit the model, and
You make the problem much more difficult to resolve because you abuse the formula interface.
Don't fit your model like this:
mod <- lm(log(Standards[['Abs550nm']])~Standards[['ng_mL']])
fit your model like this
mod <- lm(log(Abs550nm) ~ ng_mL, data = standards)
Isn't that some much more readable?
To predict you would need a data frame with a variable ng_mL:
predict(mod, newdata = data.frame(ng_mL = c(0.5, 1.2)))
Now you may have a third error. You appear to be trying to predict with new values of Absorbance, but the way you fitted the model, Absorbance is the response variable. You would need to supply new values for ng_mL.
The behaviour you are seeing is what happens when R can't find a correctly-named variable in newdata; it returns the fitted values from the model or the predictions at the observed data.
This makes me think you have the formula back to front. Did you mean:
mod2 <- lm(ng_mL ~ log(Abs550nm), data = standards)
?? In which case, you'd need
predict(mod2, newdata = data.frame(Abs550nm = c(1.7812,1.7309)))
say. Note you don't need to include the log() bit in the name. R recognises that as a function and applies to the variable Abs550nm for you.
If the model really is log(Abs550nm) ~ ng_mL and you want to find values of ng_mL for new values of Abs550nm you'll need to invert the fitted model in some way.
I am new to R and am trying to produce a vast number of diagnostic plots for linear models for a huge data set.
I discovered the lmList function from the nlme package.
This works a treat but what I now need is a means of passing in a fraction of this data into the plot function so that the resulting plots are not minute and unreadable.
In the example below 27 plots are nicely displayed. I want to produce diagnostics for much more data.
Is it necessary to subset the data first? (presumably with loops) or is it possible to subset within the plotting function (presumably with some kind of loop) rather than create 270 data frames and pass them all in separately?
I'm sorry to say that my R is so basic that I do not even know how to pass variables into names and values together in for loops (I tried using the paste function but it failed).
The data and function for the example are below – I would be picking values of Subject by their row numbers within the data frame. I grant that the 27 plots here show nicely but for sake of example it would be nice to split them into say into 3 sets of 9.
fm1 <- lmList(distance ~ age | Subject, Orthodont)
# observed versus fitted values by Subject
plot(fm1, distance ~ fitted(.) | Subject, abline = c(0,1))
Examples from:
https://stat.ethz.ch/R-manual/R-devel/library/nlme/html/plot.lmList.html
I would be most grateful for help and hope that my question isn't insulting to anyone's intelligence or otherwise annoying.
I can't see how to pass a subset to the plot.lmList function. But, here is a way to do it using standard split-apply-combine strategy. Here, the Subjects are just split into three arbitrary groups of 9, and lmList is applied to each group.
## Make 3 lmLists
fits <- lapply(split(unique(Orthodont$Subject), rep(1:3, each=3)), function(x) {
eval(substitute(
lmList(distance ~ age | Subject, # fit the data to subset
data=Orthodont[Orthodont$Subject %in% x,]), # use the subset
list(x=x))) # substitue the actual x-values so the proper call gets stored
})
## Make plots
for (i in seq_along(fits)) {
dev.new()
print(plot(fits[[i]], distance ~ fitted(.) | Subject, abline = c(0,1)))
}
Working in R to develop regression models, I have something akin to this:
c_lm = lm(trainingset$dependent ~ trainingset$independent)
c_pred = predict(c_lm,testset$independent))
and every single time, I get a mysterious error from R:
Warning message:
'newdata' had 34 rows but variables found have 142 rows
which essentially translates into R not being able to find the independent column of the testset data.frame. This is simply because the exact name from the right-hand side of the formula in lm must be there in predict. To fix it, I can do this:
tempset = trainingset
c_lm = lm(trainingset$dependent ~ tempset$independent)
tempset = testset
c_pred = predict(c_lm,tempset$independent))
or some similar variation, but this is really sloppy, in my opinion.
Is there another way to clean up the translation between the two so that the independent variables' data frame does not have to have the exact same name in predict as it does in lm?
No, No, No, No, No, No! Do not use the formula interface in the way you are doing if you want all the other sugar that comes with model formulas. You wrote:
c_lm = lm(trainingset$dependent ~ trainingset$independent)
You repeat trainingset twice, which is a waste of fingers/time, redundant, and not least causing you the problem that you are hitting. When you now call predict, it will be looking for a variable in testset that has the name trainingset$independent, which of course doesn't exist. Instead, use the data argument in your call to lm(). For example, this fits the same model as your formula but is efficient and also works properly with predict()
c_lm = lm(dependent ~ independent, data = trainingset)
Now when you call predict(c_lm, newdata = testset), you only need to have a data frame with a variable whose name is independent (or whatever you have in the model formula).
An additional reason to write formulas as I show them, is legibility. Getting the object name out of the formula allows you to more easily see what the model is.
I am trying to use tapply() function to run models by several categories with not much sucess. My data has 20 clinics and I want to run the models BY each clinic.
Heres my model:
attach(qregdata)
rq(logA~ dose+ chtcm + cage +raceth + sex,tau=.9)
My data as a variable clinic (with values 1-20). Does anybody know how to run this model BY clinic in R as in other statistical packages?
A very general way of accomplishing this is shown in the following. The ddply function runs a supplied function (in this case lm) for each clinic. You can also run it on more complex cross-sections of your data. E.g. .(clinic,level) would run a separate model on each combination of clinic and level. The term lm(y~x)$coef[1] gets the intercept of the linear model. I think there is no easy way to save all the output of each model fit at once.
n <- 10
clinic <- factor(rep(1:3,each=n))
x <- rep(0:(n-1),3)
y <- rnorm(3*n)*x
d <- data.frame(clinic,x,y)
# plot data and linear fits
library(ggplot2)
ggplot(d,aes(x,y)) + geom_point() + facet_wrap(~clinic) + stat_smooth(method='lm')
# run a separate model for each clinic
library(plyr)
ddply(d,.(clinic),summarize,intercept=lm(y~x)$coef[1],slope=lm(y~x)$coef[2])
You could use 'lappy' across the unique values of clinic, and the use subset to extract the section of your Dataset for that clinic. Then just fit the model to the subset.
This will return a list of models, which you can then further process.
I had a similar issue to this recently and wanted to share a response in case someone is still interested in this topic; sorry to dredge up an old post.
tapply is very convenient to work with when the input object (the object being "split") is a vector. If the input object being split is a rectangular data set, it can be much simpler to use the (aptly named, in this case) by function, which is a convenient wrapper for tapply intended for data.frame objects. The return object of the by function is of the class by which can be simplified to an array or a list using the argument simplify = TRUE.
Certainly there are more efficient ways to perform this operation, but if you are looking for a tapply-like solution - by is it.
Here's an example using lm to regress petal width on sepal width "by" species in the iris data set:
## Load iris data
data(iris)
## Fit a model to each species-specific subset of the data
fitBySpecies <- by(
data = iris,
INDICES = iris$Species,
FUN = function(speciesSubset)
lm(Petal.Width ~ Sepal.Width, data = speciesSubset)
)