I fitted an ARIMA model like follows:
arima_pri <- Arima(prits, order=c(7,1,0), xreg = t2, seasonal=list(order=c(1,1,1), period=12))
And want to look at the residuals vs fitted values plot:
plot(fitted(arima_pri), arima_pri$residuals)
then got this
Also tried ts.plot:
ts.plot(fitted(arima_pri), arima_pri$residuals)
then got this
It's still weird! I want to get a plot like
See http://rstudio-pubs-static.s3.amazonaws.com/21465_653278de4ce44fefa846002156e9b10a.html
Or,
(See http://rstudio-pubs-static.s3.amazonaws.com/21465_653278de4ce44fefa846002156e9b10a.html)
I'm sure all the libraries have been loaded correctly like "forecast". What should I do now? Thank you!
Related
I am applying a GAM model to my data: cell abundance over time.
The model works just fine (although I am aware of a pattern in my resiudals, but this is a different issue not relevant here).
It just fails to display the partial residuals in the final plot, although i set residuals = TRUE. Here is my output:
https://i.stack.imgur.com/C1MlY.png
also I used mgcv package.
Previously this code worked as I wanted, but on different data. Any ideas on why it is not working are welcome!
GAM_EA <- mgcv::gam(EUB_FISH ~ s(Day, by = Heatwave), data = HnH, method = "REML")
gam.check(GAM_EA) #Checking the model
mgcv::anova.gam(GAM_EA) #Retrieving the statistical results. See ?anova.gam
summary.gam(GAM_EA)
plot(GAM_EA, shift = coef(GAM_EA)[1], residuals = TRUE)
See argument by.resid in ?plot.gam. They way these are used in plot.gam would been meaningless for factor by terms unless you were to subset the partial residuals and plot only the residuals for observations in the specific level of the by factor.
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.
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.
My goal: I want to understand a time series, a strongly auto-regressive one (ACF and PACF output told me that) and make a forecast.
So what I did was I first transformed my data into a ts, then decomposed the time series, checked its stationarity (the series wasn't stationary). Then I conducted a log transformation and found an Arima model that fits the data best - I checked the accuracy with accuracy(x) - I selected the model with the accuracy output closest to 0.
Was this the correct procedure? I'm new to statistics and R and would appreciate some criticism if that wasn't correct.
When building the Arima model I used the following code:
ARIMA <- Arima(log(mydata2), order=c(2,1,2), list(order=c(0,1,1), period=12))
The result I received was a log function and the data from the past (the data I used to build the model) wasn't displayed in the diagram. So then to transform the log into the original scale I used the following code:
ARIMA_FORECAST <- forecast(ARIMA, h=24, lambda=0)
Is that correct? I found it somewhere on the web and don't really understand it.
Now my main question: How can I plot the original data and the ARIMA_FORECAST in one diagram? I mean displaying it the way the forecasts are displayed if no log transformation is undertaken - the forecast should be displayed as the extension of the data from the past, confidence intervals should be there too.
The simplest approach is to set the Box-Cox transformation parameter $\lambda=0$ within the modelling function, rather than take explicit logarithms (see https://otexts.org/fpp2/transformations.html). Then the transformation will be automatically reversed when the forecasts are produced. This is simpler than the approach described by #markus. For example:
library(forecast)
# estimate an ARIMA model to log data
ARIMA <- auto.arima(AirPassengers, lambda=0)
# make a forecast
ARIMA_forecast <- forecast(ARIMA)
# Plot forecasts and data
plot(ARIMA_forecast)
Or if you prefer ggplot graphics:
library(ggplot2)
autoplot(ARIMA_forecast)
The package forecast provides the functions autolayer and geom_forecast that might help you to draw the desired plot. Here is an example using the AirPassengers data. I use the function auto.arima to estimate the model.
library(ggplot2)
library(forecast)
# log-transform data
dat <- log(AirPassengers)
# estimate an ARIMA model
ARIMA <- auto.arima(dat)
# make a forecast
ARIMA_forecast <- forecast(ARIMA, h = 24, lambda = 0)
Since your data is of class ts you can use the autoplot function from ggplot2 to plot your original data and add the forecast with the autolayer function from forecast.
autoplot(AirPassengers) + forecast::autolayer(ARIMA_forecast)
The result is shown below.
I would like to do a regression in R
The formula is y_t = alpha +beta* x_t-1 & x_t = theta + rho * x_t-1.
Since I would like to estimate the covariance matrix of the error. I do not know how to run regression for both equation together. Thank you.
I tried
lm(c(y[2:756],x[2:756])~c(x[1:755],x[1:755]),data=data1)
756 is the length of vector, it does not work.
Your example looks like you are trying to fit an autoregressive model with lm. Try autoregressive models instead. For multivariate autoregressive models I suggest using the MTS package. Something like the following should work:
require("MTS")
VAR(data.frame(x=x, y=y))
For more detail, check out ?VAR. You may also want to have a look at the time series task view on CRAN.