Searched stackoverflow multiple times for a solution to this question. One of the fundamental issues I am encountering is that I am unable to predict the interaction effects of certain model averaged coefficients on the response. I am currently running an occupancy model, where I would like to visualize the effect of certain model averaged coefficients on the response (0/1)
Here's what a model averaged object from `MuMIn' looks like:
c
Call:
model.avg(object = top_clim_elev[[8]], fit = TRUE)
Component models:
‘1/2/3/4/5/6/7/8/9/10/11/12/13’ ‘1/2/3/4/5/6/7/8/9/10/11/13’
‘1/2/3/4/5/6/7/8/9/10/11/13/15’ ‘1/2/3/4/5/6/7/8/9/10/11/12/13/15’
‘1/2/3/4/5/6/7/8/9/10/11/12/13/15/16’ ‘1/2/3/4/5/6/7/8/9/10/11/12/13/16’
‘1/2/3/4/5/6/7/8/9/10/11/12/13/14’ ‘1/2/3/4/5/6/7/8/9/10/11/13/14’
‘1/2/3/4/5/6/7/8/9/10/11/13/14/15’ ‘1/2/3/4/5/6/7/8/9/10/11/12/13/14/15’
‘1/2/3/4/5/6/7/8/9/10/11/12/13/14/15/16’ ‘1/2/3/4/5/6/7/8/9/10/11/12/13/14/16’
Coefficients:
psi(Int) psi(alt.y) psi(bio_17.y) psi(bio_18.y) psi(bio_4.y) psi(prec_interannual.y)
full -3.10525 -0.05192473 -0.2068835 -1.482401 0.6015917 0.2626282
subset -3.10525 -0.05192473 -0.2068835 -1.482401 0.6015917 0.4324259
psi(alt.y:bio_17.y) p(Int) p(duration_minutes) p(effort_distance_km) p(expertise)
full -0.7490456 -1.245292 0.411762 -0.05852994 0.3375549
subset -0.7490456 -1.245292 0.411762 -0.05852994 0.3375549
p(julian_date) p(min_obs_started) p(number_observers) p(protocol_typeTraveling)
full -0.04548198 -0.01558561 -0.02327381 0.3527979
subset -0.04548198 -0.01558561 -0.02327381 0.3527979
psi(alt.y:bio_4.y) psi(alt.y:prec_interannual.y) psi(alt.y:bio_18.y)
full 0.09419262 0.05598506 0.01681933
subset 0.24061807 0.26320150 0.06110500
I considered wrote the coefficients out of the full model as a dataframe, along with the standard error and upper and lower CI.
For example:
I would like to use the coefficient estimates for alt.y:bio_17.y and predict it's effect on the response as a function of the moderator alt.y. I have tried multiple ways to do the same, but would like to achieve the 'predict' function in R without having the need to store a model object. Further, even if I store a model averaged object in R, as shown above - I am unable to use the same via packages like sjPlot or others in R.
Related
My question is closely related to this previous one: Simulation-based hypothesis testing on spatial point pattern hyperframes using "envelope" function in spatstat
I have obtained an mppm object by fitting a model on several independent datasets using the mppmfunction from the R package spatstat. How can I study its envelope to compare it to my observations ?
I fitted my model as such:
data <- listof(NMJ1,NMJ2,NMJ3)
data <- hyperframe(X=1:3, Points=data)
model <- mppm(Points ~marks*sqrt(x^2+y^2), data)
where NMJ1, NMJ2, and NMJ3 are marked ppp and are independent realizations of the same experiment.
However, the envelope function does not accept inputs of type mppm:
> envelope(model, Kcross.inhom, nsim=10)
Error in UseMethod("envelope") :
no applicable method for 'envelope' applied to an object of class "c('mppm', 'list')"
The answer provided to the previously mentioned question indicates how to plot global envelopes for each pattern, and to use the product rule for multiple testing. However, my fitted model implies that my 3 ppp objects are statistically equivalent, and are independent realizations of the same experiment (ie no different covariates between them). I would thus like to obtain one single plot comparing my fitted model to my 3 datasets. The following code:
gamma= 1 - 0.95^(1/3)
nsims=round(1/gamma-1)
sims <- simulate(model, nsim=2*nsims)
SIMS <- list()
for(i in 1:nrow(sims)) SIMS[[i]] <- as.solist(sims[i,,drop=TRUE])
Hplus <- cbind(data, hyperframe(Sims=SIMS))
EE1 <- with(Hplus, envelope(Points, Kcross.inhom, nsim=nsims, simulate=Sims))
pool(EE1[1],EE1[2],EE1[3])
leads to the following error:
Error in pool.envelope(`1` = list(r = c(0, 0.78125, 1.5625, 2.34375, 3.125, :
Arguments 2 and 3 do not belong to the class “envelope”
Wrong type of subset index. Use
pool(EE1[[1]], EE1[[2]], EE1[[3]])
or just
pool(EE1)
These would have given an error message that the envelope commands should have been called with savefuns=TRUE. So you just need to change that step as well.
However, statistically this procedure makes little sense. You have already fitted a model, which allows for rigorous statistical inference using anova.mppm and other tools. Instead of this, you are generating simulated data from the fitted model and performing a Monte Carlo test, with all the fraught issues of multiple testing and low power. There are additional problems with this approach - for example, even if the model is "the same" for each row of the hyperframe, the patterns are not statistically equivalent unless the windows of the point patterns are identical, and so on.
I am having trouble getting the Brier Score for my Machine Learning Predictive models. The outcome "y" was categorical (1 or 0). Predictors are a mix of continuous and categorical variables.
I have created four models with different predictors, I will call them "model_1"-"model_4" here (except predictors, other parameters are the same). Example code of my model is:
Model_1=rfsrc(y~ ., data=TrainTest, ntree=1000,
mtry=30, nodesize=1, nsplit=1,
na.action="na.impute", nimpute=3,seed=10,
importance=T)
When I run the "Model_1" function in R, I got the results:
My question was how can I get the predicted possibility for those 412 people? And how to find the observed probability for each person? Do I need to calculate by hand? I found the function BrierScore() in "DescTools" package.
But I tried "BrierScore(Model_1)", it gives me no results.
codes I added:
library(scoring)
library(DescTools)
BrierScore(Raw_SB)
class(TrainTest$VL_supress03)
TrainTest$VL_supress03_nu<-as.numeric(as.character(TrainTest$VL_supress03))
class(TrainTest$VL_supress03_nu)
prediction_Raw_SB = predict(Raw_SB, TrainTest)
BrierScore(prediction_Raw_SB, as.numeric(TrainTest$VL_supress03) - 1)
BrierScore(prediction_Raw_SB, as.numeric(as.character(TrainTest$VL_supress03)) - 1)
BrierScore(prediction_Raw_SB, TrainTest$VL_supress03_nu - 1)
I tried some codes: have so many error messages:
One assumption I am making about your approach is that you want to compute the BrierScore on the data you train your model on (which is usually not the correct approach, google train-test split if you need more info there).
In general, therefore you should reflect on whether your approach is correct there.
The BrierScore method in DescTools only has a defined method for glm models, otherwise, it expects as input a vector of predicted probabilities and a vector of true values (see ?BrierScore).
What you would need to do though is to predict on your data using:
prediction = predict(model_1, TrainTest, na.action="na.impute")
and then compute the brier score using
BrierScore(as.numeric(TrainTest$y) - 1, prediction$predicted[, 1L])
(Note, that we transform TrainTest$y into a numeric vector of 0's and 1's in order to compute the brier score.)
Note: The randomForestSRC package also prints a normalized brier score when you call print(prediction).
In general, using one of the available workbenches for machine learning in R (mlr3, tidymodels, caret) might simplify this approach for you and prevent a lot of errors in this direction. This is a really good practice, especially if you are less experienced in ML as it can prevent many errors.
See e.g. this chapter in the mlr3 book for more information.
For reference, here is some very similar code using the mlr3 package, automatically also taking care of train-test splits.
data(breast, package = "randomForestSRC") # with target variable "status"
library(mlr3)
library(mlr3extralearners)
task = TaskClassif$new(id = "breast", backend = breast, target = "status")
algo = lrn("classif.rfsrc", na.action = "na.impute", predict_type = "prob")
resample(task, algo, rsmp("holdout", ratio = 0.8))$score(msr("classif.bbrier"))
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.
library(Sunclarco)
library(MASS)
library(survival)
library(SPREDA)
library(SurvCorr)
library(doBy)
#Dataset
diabetes=data("diabetes")
data1=subset(diabetes,select=c("LASER","TRT_EYE","AGE_DX","ADULT","TIME1","STATUS1"))
data2=subset(diabetes,select=c("LASER","TRT_EYE","AGE_DX","ADULT","TIME2","STATUS2"))
#Adding variable which identify cluster
data1$CLUSTER<- rep(1,197)
data2$CLUSTER<- rep(2,197)
#Renaming the variable so that that we hve uniformity in the common items in the data
names(data1)[5] <- "TIME"
names(data1)[6] <- "STATUS"
names(data2)[5] <- "TIME"
names(data2)[6] <- "STATUS"
#merge the files
Total_data=rbind(data1,data2)
# Re arranging the database
diabete_full=orderBy(~LASER+TRT_EYE+AGE_DX,data=Total_data)
diabete_full
#using Sunclarco package for Clayton a nd Gumbel
Clayton_1step <- SunclarcoModel(data=diabete_full,time="TIME",status="STATUS",
clusters="CLUSTER",covariates=c("LASER","TRT_EYE","ADULT"),
stage=1,copula="Clayton",marginal="Weibull")
summary(Clayton_1step)
# Estimates StandardErrors
#lambda 0.01072631 0.005818201
#rho 0.79887565 0.058942208
#theta 0.10224445 0.090585891
#beta_LASER 0.16780224 0.157652947
#beta_TRT_EYE 0.24580489 0.162333369
#beta_ADULT 0.09324001 0.158931463
# Estimate StandardError
#Kendall's Tau 0.04863585 0.04099436
Clayton_2step <- SunclarcoModel(data=diabete_full,time="TIME",status="STATUS",
clusters="CLUSTER",covariates=c("LASER","TRT_EYE","ADULT"),
stage=2,copula="Clayton",marginal="Weibull")
summary(Clayton_1step)
# Estimates StandardErrors
#lambda 0.01131751 0.003140733
#rho 0.79947406 0.012428824
#beta_LASER 0.14244235 0.041845100
#beta_TRT_EYE 0.27246433 0.298184235
#beta_ADULT 0.06151645 0.253617142
#theta 0.18393973 0.151048024
# Estimate StandardError
#Kendall's Tau 0.08422381 0.06333791
Gumbel_1step <- SunclarcoModel(data=diabete_full,time="TIME",status="STATUS",
clusters="CLUSTER",covariates=c("LASER","TRT_EYE","ADULT"),
stage=1,copula="GH",marginal="Weibull")
# Estimates StandardErrors
#lambda 0.01794495 0.01594843
#rho 0.70636113 0.10313853
#theta 0.87030690 0.11085344
#beta_LASER 0.15191936 0.14187943
#beta_TRT_EYE 0.21469814 0.14736381
#beta_ADULT 0.08284557 0.14214373
# Estimate StandardError
#Kendall's Tau 0.1296931 0.1108534
Gumbel_2step <- SunclarcoModel(data=diabete_full,time="TIME",status="STATUS",
clusters="CLUSTER",covariates=c("LASER","TRT_EYE","ADULT"),
stage=2,copula="GH",marginal="Weibull")
Am required to fit copula models in R for different copula classes particularly the Gaussian, FGM,Pluckett and possibly Frank (if i still have time). The data am using is Diabetes data available in R through the package Survival and Survcorr.
Its my thesis am working on and its a study for the exploratory purposes to see how does copula class serves different purposes as in results they lead to having different results on the same. I found a package Sunclarco in Rstudio which i was able to fit Clayton and Gumbel copula class but its not available yet for the other classes.
The challenge am facing is that since i have censored data which has to be incorporated in likelihood estimation then it becomes harder fro me to write a syntax since as I don't have a strong programming background. In addition, i have to incorporate the covariates present in programming and see their impact on the association if it present or not. However, taking to my promoter he gave me insights on how to approach the syntax writing for this puzzle which goes as follows
• ******First of all, forget about the likelihood function. We only work with the log-likelihood function. In this way, you do not need to take the product of the contributions over each of the observations, but can take the sum of the log-contributions over the different observations.
• Next, since we have a balanced design, we can use the regular data frame structure in which we have for each cluster only one row in the data frame. The different variables such as the lifetimes, the indicators and all the covariates are the columns in this data frame.
• Due to the bivariate setting, there are only 4 possible ways to give a contribution to the log-likelihood function: both uncensored, both censored, first uncensored and second censored, or first censored and second uncensored. Well, to create the loglikelihood function, you create a new variable in your data frame in which you put the correct contribution of the log-likelihood based on which individual in the couple is censored. When you take the sum of this variable, you have the value of the log-likelihood function.
• Since this function depends on parameters, you can use any optimizer, like optim or nlm to get your optimal values. By careful here, optim and nlm look for the minimum of a function, not a maximum. This is easy solved since the minimum of a function -f is the same as the maximum of a function f.
• Since you have for each copula function, the different expressions for the derivatives, it should be possible to get the likelihood functions now.******
Am still struggling to find a way as for each copula class each of the likelihood changes as the generator function is also unique for the respective copula since it needs to be adapted during estimation. Lastly, I should run analysis for both one and two steps of copula estimations as i will use to compare results.
if someone could help me to figure it out then I will be eternally grateful. Even if for just one copula class e.g. Gaussian then I will figure it the rest based on the one that am requesting to be assisted since I tried everything and still i have nothing to show up for and now i feel time is running out to get answers by myself.
I have created a loop to fit a non-linear model to six data points by participants (each participant has 6 data points). The first model is a one parameter model. Here is the code for that model that works great. The time variable is defined. The participant variable is the id variable. The data is in long form (one row for each datapoint of each participant).
Here is the loop code with 1 parameter that works:
1_p_model <- dlply(discounting_long, .(Participant), function(discounting_long) {wrapnls(indiff ~ 1/(1+k*time), data = discounting_long, start = c(k=0))})
However, when I try to fit a two parameter model, I get this error "Error: singular gradient matrix at initial parameter estimates" while still using the wrapnls function. I realize that the model is likely over parameterized, that is why I am trying to use wrapnls instead of just nls (or nlsList). Some in my field insist on seeing both model fits. I thought that the wrapnls model avoids the problem of 0 or near-0 residuals. Here is my code that does not work. The start values and limits are standard in the field for this model.
2_p_model <- dlply(discounting_long, .(Participant), function(discounting_long) {nlxb(indiff ~ 1/(1+k*time^s), data = discounting_long, lower = c (s = 0), start = c(k=0, s=.99), upper = c(s=1))})
I realize that I could use nlxb (which does give me the correct parameter values for each participant) but that function does not give predictive values or residuals of each data point (at least I don't think it does) which I would like to compute AIC values.
I am also open to other solutions for running a loop through the data by participants.
You mention at the end that 'nlxb doesn't give you residuals', but it does. If your result from your call to nlxbis called fit then the residuals are in fit$resid. So you can get the fitted values using just by adding them to the original data. Honestly I don't know why nlxb hasn't been made to work with the predict() function, but at least there's a way to get the predicted values.