i want to find the correlation of the adjuste of the curve plotted and the data, anyone knows how to do that?
library(drc)
S.alba.m1 <- drm(DryMatter~Dose, Herbicide, data = S.alba, fct = LL.4())
plot(S.alba.m1)
Use fitted:
cor(S.alba$DryMatter, fitted(S.alba.m1))
Also check out the modelFit function: ?modelFit for various tests.
Related
I was wondering what would be the best way to calculate and present standardized coefficients using fixest. Here is what I tried using easystats
library(parameters)
library(effectsize)
library(fixest)
m <- lm(rating ~ complaints, data = attitude)
standardize_parameters(m, method="basic")# works
m <- feols(rating ~ complaints, data = attitude)
standardize_parameters(m, method="basic")# Error in stats::model.frame(model)[[1]] : subscript out of bounds
I also tried the modelsummary approach, but it shows unstandardized coefficients with no error.
library(parameters)
library(effectsize)
m <- lm(rating ~ complaints, data = attitude)
modelsummary(m, standardize="refit") # works, coeffs are different
m <- feols(rating ~ complaints, data = attitude)
modelsummary(m, standardize="refit")# doesn't work, coeffs are the same
Any insight or advice on how to elegantly and easily pull standardized coefficients out of fixest estimation results would be greatly appreciated. My goal is to replicate the venerable to use listcoef package in Stata. Many thanks to the authors of the packages mentioned in this post!
Edit: ``` > packageVersion("modelsummary")
[1] ‘1.1.0.9000’
One potential solution is to just manually calculate the standardized coefficients yourself, as [detailed here][1]. As an example, below I scale your predictor and outcome, then calculate the standardized beta coefficient of the only predictor in your model.
#### Scale Predictor and Outcome ####
scale.x <- sd(attitude$complaints)
scale.y <- sd(attitude$rating)
#### Obtain Standardized Coefficients ####
sb <- coef(m)[2] * scale.x / scale.y
sb
Which gives you this (you can ignore the column name, as it is just borrowing it from the original coef vector):
complaints
0.8254176
[1]: https://www.sciencedirect.com/topics/mathematics/standardized-regression-coefficient#:~:text=The%20standardized%20regression%20coefficient%2C%20found,one%20of%20its%20standardized%20units%20(
I'm trying to model raw data by an asymptotic function with the equation $$f(x) = a + (b-a)(1-\exp(-c x))$$ using R. To do so I used the following code:
rawData <- import("path/StackTestData.tsv")
# executing regression
X <- rawData$x
Y <- rawData$y
model <- drm(Y ~ X, fct = DRC.asymReg())
# creating the regression function
f_0_ <- model$coefficients[1] #value for y if x=0
steepness <- model$coefficients[2]
plateau <- model$coefficients[3]
eq <- function(x){f_0_+(plateau-f_0_)*(1-exp(-steepness*x))}
# plotting the regression function together with the raw data
ggplot(rawData,aes(x=x,y=y)) +
geom_line(col="red") +
stat_function(fun=eq,col="blue") +
ylim(10,12.5)
In some cases, I got a proper regression function. However, with the attached data I don't get one. The regression function is not showing any correlation with the raw data whatsoever, as shown in the figure below. Can you perhaps offer a better solution for performing the asymptotic regression or do you know where the error lies?
Best Max
R4.1.2 was used using R Studio 1.4.1106. For ggplot the package ggpubr, for DRC.asymReg() the packages aomisc and drc were load.
First, i run a regression model. Then, i extract robust standard errors. However, i am not sure how to extract the confidence interval afterwards, coeftest() seems to include only the standard errors. Is there a way to do it automatically?
Here is the reproducible data and code:
library(plm)
library(lmtest)
library(broom)
data(Cigar)
model<- plm(price ~ sales + cpi, index=c("state", "year"), model = 'within',
data = Cigar)
#Extract the robust standard errors
plot_coeftest = tidy(coeftest(model))
as #deschen proposed, this is the solution:
plot_coeftest= broom::tidy(lmtest::coeftest(model), conf.int = TRUE)
I would like to analysis my data based on the gradient boosted model.
On the other hand, as my data is a kind of cohort, I have a trouble understanding the result of this model.
Here's my code. Analysis was performed based on the example data.
install.packages("randomForestSRC")
install.packages("gbm")
install.packages("survival")
library(randomForestSRC)
library(gbm)
library(survival)
data(pbc, package="randomForestSRC")
data <- na.omit(pbc)
set.seed(9512)
train <- sample(1:nrow(data), round(nrow(data)*0.7))
data.train <- data[train, ]
data.test <- data[-train, ]
set.seed(9741)
gbm <- gbm(Surv(days, status)~.,
data.train,
interaction.depth=2,
shrinkage=0.01,
n.trees=500,
distribution="coxph")
summary(gbm)
set.seed(9741)
gbm.pred <- predict.gbm(gbm,
n.trees=500,
newdata=data.test,
type="response")
As I read the package documnet, "gbm.pred" is the result of cox's partial likelihood.
set.seed(9741)
lambda0 = basehaz.gbm(t=data.test$days,
delta=data.test$status,
t.eval=sort(data.test$days),
cumulative = FALSE,
f.x=gbm.pred,
smooth=T)
hazard=lambda0*exp(gbm.pred)
In this code, lambda0 is a baseline hazard fuction.
So, according to formula: h(t/x)=lambda0(t)*exp(f(x))
"hazard" is hazard function.
However, what I've wanted to calculte was the "survival function".
Because, I would like to compare the outcome of original data (data$status) to the prediction result (survival function).
Please let me know how to calculate survival function.
Thank you
Actually, the returns is cumulative baseline hazard function(integral part: \int^t\lambda(z)dz), and survival function can be computed as below:
s(t|X)=exp{-e^f(X)\int^t\lambda(z)dz}
f(X) is prediction of gbm, which is equal to log-hazard proportion.
I think this tutorial about gbm-based survival analysis would help to u!
https://github.com/liupei101/Tutorial-Machine-Learning-Based-Survival-Analysis/blob/master/Tutorial_Survival_GBM.ipynb
I'm trying to teach myself a bit about modeling time series for 'counts' data. I found a pretty simple model, the Autoregressive Conditional Poisson model (ACP) (Heinen 2003), that has an accompanying R package {acp}. I'm having trouble finding information about how to construct n-step-ahead prediction intervals for predictions made from an ACP model. Inconveniently, forecast doesn't work with these ACP objects. Any thoughts on how to construct these?
Additionally, when using predict() with an ACP model, you have to include an argument, newydata, that is a data frame of the values you want to predict...? Maybe I'm misinterpreting this, but it seems like you need to already have y when predicting yhat. Why?
Below I copy/pasted the example code from the {acp} package.
library(acp)
data(polio)
trend=(1:168/168)
cos12=cos((2*pi*(1:168))/12)
sin12=sin((2*pi*(1:168))/12)
cos6=cos((2*pi*(1:168))/6)
sin6=sin((2*pi*(1:168))/6)
#Autoregressive Conditional Poisson Model with explaning covariates
polio_data<-data.frame(polio, trend , cos12, sin12, cos6, sin6)
mod1 <- acp(polio~-1+trend+cos12+sin12+cos6+sin6,data=polio_data, p = 1 ,q = 2)
summary(mod1)
#Static out-of-sample fit example
train<-data.frame(polio_data[c(1: 119),])
mod1t <- acp(polio~-1+trend+cos12+sin12+cos6+sin6,data=train, p = 1 ,q = 2)
xpolio_data<-data.frame(trend , cos12, sin12, cos6, sin6)
test<-xpolio_data[c(120:nrow(xpolio_data)),]
yfor<-polio_data[120:nrow(polio_data),1]
predict(mod1t,yfor,test)
#Autoregressive Conditional Poisson Model without explaning covariates
polio_data<-data.frame(polio)
mod2 <- acp(polio~-1,data=polio_data, p = 3 ,q = 1)
summary(mod2)
The second argument in the predict() command is the vector of observed y values that confuses me.
Thanks!