Plot residuals vs predicted response in R - r

Is Plot residuals vs predicted response equivalent to Plot residuals vs fitted ?
If so, then would be plotted by plot(lm) and plot(predict(lm)), where lm is the linear model ?
Am I correct?

Maybe little off-topic, but as an addition: package named ggfortify might come handy. Super easy to use, like this:
library(ggfortify)
autoplot(mod3)
Yields an output with the most important things you need to know, if your model violates the lm assumptions or not. An example output here:

Yes, the fitted values are the predicted responses on the training data, i.e. the data used to fit the model, so plotting residuals vs. predicted response is equivalent to plotting residuals vs. fitted.
As for your second question, the plot would be obtained by plot(lm), but before that you have to run par(mfrow = c(2, 2)). This is because plot(lm) outputs 4 plots, one of which is the one you want, i.e the residuals vs fitted plot. The command above divides the output screen into four facets, so each plot will be shown in one. The plot you are looking for will appear in the top left.

Related

Residual Plot for multivariate regression in Time Series, with time on X axis in R

I have a dataframe which is a time series. I am using the function lm to build a multivariate regression model.
linearmodel <- lm(Y~X1+X2+X3, data = data)
I want to plot the residuals of this linearmodel on the y-axis and time on the x-axis using a simple function, with the lm() object as the input.
Standard residual plotting functions like the one in car package (car::residualPlot) gives residuals on the Y-axis and fitted-values on the Y-axis.
Ideally, I need the residuals on the Y-axis and the timescale on the X-axis. But I understand that the function lm() is time agnositc. So, I can live with if the residuals are on Y-axis in the same order as the data input and nothing on the X-axis
Is there a plotting function which i can use by passing the linearmodel object into the function (not something where i can extract the residuals and use ggplot2). So for example: plot<- plotresidualsinorder(linearmodels) should give me the residuals on Y-axis in the same order of the data input?
I want to use this plot in R-shiny ultimately.
My research led me to car package, which is wonderful in its own right, but doesn't have the function to solve my problem.
Many thanks in advance for the help.
You can use the Residual Plot information. For the proposed solution, we need to apply the lm function to a formula that describes your Y variables by the variables X1+X2+X3, and save the linear regression model in a new linearmodel variable. Finally, we compute the residual with the resid function. In your case, the following solution can be representative for your problem.
Proposed solution:
linearmodel <- lm(Y~X1+X2+X3, data = data)
lm_resid <- resid(linearmodel)
plot(data$X1+X2+X3, lm_resid,
ylab="Residuals", xlab="Time",
main="Data")
abline(0, 0)
For any help concerning how does the resid function works, you can try:
help(resid)
Calisto's solution will work, but there is a more simple and straightforward solution. The lm function already give to you the regression residuals. So you may simply pass:
plot(XTime, linearmodel$residuals, main = "Residuals")
XTime is the Date variable of your dataset, maybe you may require to format that with POSIX functions: https://www.rdocumentation.org/packages/base/versions/3.6.2/topics/as.POSIX*
Add parameters as you need to share it on R-shiny.

Backtransform LMM output in R

When performing linear mixed models, I have had to square-root(log) transform the data to achieve a normal distribution. Having performed the LMMs, I now want to plot the results onto a graph, but on the original scale i.e. not square-root(log) transformed.
Apparently I can use my raw (untransformed data) on a graph, and to create the predicted regression line I can use the coefficients from my LMM output to get backtransformed predicted y-values for each of my x values. This is where I'm stuck - I have no idea how to do this. Can anyone help?

How to rectify heteroscedasticity for multiple linear regression model

I'm fitting a multiple linear regression model with 6 predictiors (3 continuous and 3 categorical). The residuals vs. fitted plot show that there is heteroscedasticity, also it's confirmed by bptest().
summary of sales_lm
rediduals vs. fitted plot
Also I calculated the sqrt for my train data and test data, as showed below:
sqrt(mean(sales_train_lm_pred-sales_train$SALES)^2)
2 3533.665
sqrt(mean(sales_test_lm_pred-sales_test$SALES)^2)
2 3556.036
I tried to fit glm() model, but still didn't rectify heteroscedasticity.
glm.test3<-glm(SALES~.,weights=1/sales_fitted$.resid^2,family=gaussian(link="identity"), data=sales_train)
resid vs. fitted plot for glm.test3
it looks weird.
glm.test3 plot
Could you please help me what should I do next?
Thanks in advance!
That you observe heteroscedasticity for your data means that the variance is not stationary. You can try the following:
1) Apply the one-parameter Box-Cox transformation (of the which the log transform is a special case) with a suitable lambda to one or more variables in the data set. The optimal lambda can be determined by looking at its log-likelihood function. Take a look at MASS::boxcox.
2) Play with your feature set (decrease, increase, add new variables).
2) Use the weighted linear regression method.

The 4 outputs of a linear model in R

In R, after creating a linear model using the function model <- lm() and plotting it using plot(model), you will get back 4 graphs each displaying your model differently. Can anyone explain what these graphs mean?
plot.lm can produce 6 different diagnostic plots, controlled by the which parameter. These are:
a plot of residuals against fitted values
a Normal Q-Q plot
a Scale-Location plot of sqrt(| residuals |) against fitted values
a plot of Cook's distances versus row labels
a plot of residuals against leverages
a plot of Cook's distances against leverage/(1-leverage)
By default it will produce numbers 1, 2, 3 and 5, pausing between plots in interactive mode.
You can see them all in one go if you set up the graphics device for multiple plots, eg:
mdl <- lm(hp~disp,mtcars)
par(mfrow=c(3,2))
plot(mdl,which=1:6)
Interpretation of these plots is a question for Cross Validated, though ?plot.lm gives some basic information.

Plotting backtransformed data with LS means plot

I have used the package lsmeans in R to get the average estimate for all observations for my treatment factor (across the levels of a block factor in the experimental design that has been included with systematic effect because it only had 3 levels). I have used a sqrt transformation for my response variable.
Thus I have used the following commands in R.
First defining model
model<-sqrt(response)~treatment+block
Then applying lsmeans
model_lsmeans<-lsmeans(model,~treatment)
Then plotting this
plot(model_lsmeans,ylab="treatment", xlab="response(with 95% CI)")
This gives a very nice graph with estimates and 95% confidense intervals for the different treatment.
The problems is just that this graph is for the transformed response.
How do I get this same plot with the backtransformed response (so the squared response)?
I have tried to create a new data frame and extract the lsmean, lower.CL, and upper.CL:
a<-summary(model_lsmeans)
New_dataframe<-as.data.frame(a[c("treatment","lsmean","lower.CL","upper.CL")])
And then make these squared
New_dataframe$lsmean<-New_dataframe$lsmean^2
New_dataframe$lower.CL<-New_dataframe$lower.CL^2
New_dataframe$upper.CL<-New_dataframe$upper.CL^2
New_dataframe
This gives me the estimates and CI boundaries squared that I need.
The problem is that I cannot make the same graph for thise estimates and CI as the one that I did in LS means above.
How can I do this? The reason that I ask is that I want to have graphs that are all of a similar style for my article. Since I very much like this LSmeans plot, and it is very convenient for me to use on the non-transformed response variables, I would like to have all my graphs in this style.
Thank you very much for your help! Hope everything is clear!
Kind regards
Ditlev

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