I used nls package to analyze non linear model (power curve, y= ax^b).
cal<- nls(formula= agw~a*area^b, data=calibration_6, start=list(a=1, b=1))
summary(cal)
What I want now is to force intercept (a) to zero to check something. In Excel, I can't set intercept for power curve. In R, is that possible to set intercept?
If possible, could you tell me how to do it?
y~ x^b model is what I considered first,
nls(formula= agw ~ area^b, data=calibration_6, start=list(b=1))
but, I also found other way, please check the below link.
How to calculate R-squared in nls package (non-linear model) in R?
Related
I analyzed non-linear regression using nls package.
power<- nls(formula= agw~a*area^b, data=calibration_6, start=list(a=1, b=1))
summary(power)
I heard in non-linear model, R-squared is not valid and rather than R-squared, we usually show residual standard error which R provides
However, I just want to know what R-squared is. Is that possible to check R-squared in nls package?
Many thanks!!!
OutPut
I found the solution. This method might not be correct in terms of statistics (As R^2 is not valid in non-linear model), but I just want see the overall goodness of fit for my non-linear model.
Step 1> to transform data as log (common logarithm)
When I use non-linear model, I can't check R^2
nls(formula= agw~a*area^b, data=calibration, start=list(a=1, b=1))
Therefore, I transform my data to log
x1<- log10(calibration$area)
y1<- log10(calibration$agw)
cal<- data.frame (x1,y1)
Step 2> to analyze linear regression
logdata<- lm (formula= y1~ x1, data=cal)
summary(logdata)
Call:
lm(formula = y1 ~ x1)
This model provides, y= -0.122 + 1.42x
But, I want to force intercept to zero, therefore,
Step 3> to force intercept to zero
logdata2<- lm (formula= y1~ 0 + x1)
summary(logdata2)
Now the equation is y= 1.322x, which means log (y) = 1.322 log (x),
so it's y= x^1.322.
In power curve model, I force intercept to zero. The R^2 is 0.9994
I am trying to fit a delayed entry parametric regression model for a Poisson process with Weibull baseline rate. It doesn't appear that R's survreg function supports left truncated data (I get the error: start-stop type Surv objects are not supported). Is there an alternate approach/R package that I could use to do this?
You may want to try something like:
flexsurv::flexsurvreg(formula = Surv(starttime, stoptime, status) ~ x1 + x2,
data=data, dist = "weibull")
Check the options the package offers which may fit your need.
The other way would be to define a truncated distribution and use interval values. See the survival packages example of using a truncated Gaussian distribution aka ”tobit”. https://www.rdocumentation.org/packages/survival/versions/2.44-1.1/topics/tobin
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.
I'm a R novice but I'm looking for a way to determine the three parameters A, B and C related by the following function in R:
y = A * (x1^B) * (x2^C)
Can someone give me some hints about R method(s) that would help me to achieve such a fitting?
One option is the nls function as #SvenHohenstein suggested. Another option is to convert your nonlinear regression into a linear regression. In the case of this equation just take the log of both sides of the equation and do a little algebra and you will have a linear equation. You can run the regression using something like:
fit <- lm( log(y) ~ log(x1) + log(x2), data=mydata)
The intercept will be log(A) so use exp to get the value, the B and C parameters will be the 2 slopes.
The big difference here is that nls will fit the model with normal errors added to the original equation and the lm fit with logs assumes that the errors in the original model are from a lognormal distribution and are multiplied instead of added to the model. Many datasets will give similar results for the 2 methods.
You can fit a nonlinear least-squares model with the function nls.
nls(y ~ A * (x1^B) * (x2^C))
Why don´t you use SVM (Suppor Vector Machines) Regression? there´s a package in CRAN named e1071 that can handle regression with SVM.
You can check this tutorial: http://www.svm-tutorial.com/2014/10/support-vector-regression-r/
I hope it can help you
I have a stationary time series to which I want to fit a linear model with an autoregressive term to correct for serial correlation, i.e. using the formula At = c1*Bt + c2*Ct + ut, where ut = r*ut-1 + et
(ut is an AR(1) term to correct for serial correlation in the error terms)
Does anyone know what to use in R to model this?
Thanks
Karl
The GLMMarp package will fit these models. If you just want a linear model with Gaussian errors, you can do it with the arima() function where the covariates are specified via the xreg argument.
There are several ways to do this in R. Here are two examples using the "Seatbelts" time series dataset in the datasets package that comes with R.
The arima() function comes in package:stats that is included with R. The function takes an argument of the form order=c(p, d, q) where you you can specify the order of the auto-regressive, integrated, and the moving average component. In your question, you suggest that you want to create a AR(1) model to correct for first-order autocorrelation in the errors and that's it. We can do that with the following command:
arima(Seatbelts[,"drivers"], order=c(1,0,0),
xreg=Seatbelts[,c("kms", "PetrolPrice", "law")])
The value for order specifies that we want an AR(1) model. The xreg compontent should be a series of other Xs we want to add as part of a regression. The output looks a little bit like the output of summary.lm() turned on its side.
Another alternative process might be more familiar to the way you've fit regression models is to use gls() in the nlme package. The following code turns the Seatbelt time series object into a dataframe and then extracts and adds a new column (t) that is just a counter in the sorted time series object:
Seatbelts.df <- data.frame(Seatbelts)
Seatbelts.df$t <- 1:(dim(Seatbelts.df)[1])
The two lines above are only getting the data in shape. Since the arima() function is designed for time series, it can read time series objects more easily. To fit the model with nlme you would then run:
library(nlme)
m <- gls(drivers ~ kms + PetrolPrice + law,
data=Seatbelts.df,
correlation=corARMA(p=1, q=0, form=~t))
summary(m)
The line that begins with "correlation" is the way you pass in the ARMA correlation structure to GLS. The results won't be exactly the same because arima() uses maximum likelihood to estimate models and gls() uses restricted maximum likelihood by default. If you add method="ML" to the call to gls() you will get identical estimates you got with the ARIMA function above.
What is your link function?
The way you describe it sounds like a basic linear regression with autocorrelated errors. In that case, one option is to use lm to get a consistent estimate of your coefficients and use Newey-West HAC standard errors.
I'm not sure the best answer for GLM more generally.