I have got in trouble with implementing weighted dataset by aov function in R.
For example my dataset "data_file" has target var "Y", and four independent var named (treat, V1 ,V2, V3).
Assuming:
V1(2 groups) & treat(3 groups) --> categorical,
V2 and v3 --> continuous.
I want to check baseline comparisons of independent variables among treat groups.
I ran aov test for this purpose,example:
base_V2_aov <- aov(data_file$V2 ~ data_file$treat)
base_V2_anov <- anova(base_V2_aov)
base_V2
It worked and showed significant difference of V2 among "treat" group but other variables were non significant, then i decided to weight my data based on V2 and run aov test in weighted data.
I used mnps function in Twang package for weighting.
mnps.data <- mnps(treat ~ V2, data_file, estimand = "ATE", stop-method = "es.mean", n.trees=5000, varbose = F)
data_file$ weight <- get.weights(mnps.data, stop.method = "es.mean")
I have read in one stackoverflow answer that survey packages does not support weighting for one-way ANOVA test, but aov function does.
So i ran this code:
base_V2_aov <- aov(data_file$V2 ~ data_file$trea, weights(data_file$weight))
base_V2_anov <- anova(base_V2_aov)
print(base_V2_anov)
It showes an error:
Error: $ operator is invalid for atomic vectors
I tried :
base_V2_aov <- aov(data_file$V2 ~ data_file$trea, weights(weight))
It did not find object "weight"
I also checked this :
base_V2_aov <- aov(data_file$V2 ~ data_file$trea, weights(data_file))
It did not show error, but the results were exactly the same as without weighting(i expected to change base on significant difference without weighting)
I want to know what is the appropriate "object" for weights in aov function?
It seems you should use:
weight = your weighting variable in the aov arguments.
I replicate something like your dataset, and after using above code the results of the comparisons between groups were different which showed that weighting method had worked.
Related
I'm using a Cox Proportional Hazards (survival::coxph) model in a competing risks context- i.e. multiple event types with one endpoint for each observation. I'm having a hard time using the coxph.predict function to show an estimate of expected number of events given a supplied set of covariates and follow-up time.
Here is an example using the mgus2 dataset in the survival package:
library(survival)
#Modify data so each subject transitions only once to a state.
crdata <- mgus2
crdata$etime <- pmin(crdata$ptime, crdata$futime)
crdata$event <- ifelse(crdata$pstat==1, 1, 2*crdata$death)
crdata$event <- factor(crdata$event, 0:2, c("censor", "PCM", "death"))
cfit <- coxph(Surv(etime, event) ~ I(age/10) + sex + mspike,
id = id, crdata)
Once I fit a model, and create a "newdata" data frame, R throws an error.
I tried using a from-scratch dataframe but this results in an error suggesting that the column size or the number of rows does not mesh:
#providing both follow-up time and covariates
nd=data.frame(etime=81 ,sex= "M", age=60, mspike=1.2)
predict(cfit, newdata=nd ,type="expected")
> Data is not the same size as it was in the original fit
I get the same issue Using model.frame when extracting the same data.frame used fitting the model.
nd=model.frame(cfit)
predict(cfit,newdata=nd,type="expected")
> Data is not the same size as it was in the original fit
This results in the same error. Trying to use the original data frame to make predictions doesn't work either:
nd=crdata[1,]
predict(cfit,newdata=nd,type="expected")
> Data is not the same size as it was in the original fit
I'm wondering what I'm missing here. Thanks in advance!
I've updated my survival package from 2.7 to 3.1 and the error thrown states that "expected" predict type is not available for multistate coxph.
> predict(fit,type="expected",newdata=newdat)
Error in predict.coxphms(fit, type = "expected", newdata = newdat) :
predict method not yet available for multistate coxph
I want to plot how the estimated survival from a Cox model depends upon the value of a covariate of interest, while the rest of variables are fixed to their average values (if they are continuous variables) or lowest values for dummy. Following this example http://www.sthda.com/english/wiki/cox-proportional-hazards-model , I have construct a new data frame with three rows, one for each value of my variable of interest; and the other covariates are fixed. Among these covariates I have two factor vectors. I created the new dataset and later it is passed to survfit() via the newdata argument.
When I passed the data frame to survfit(), I obtain the following error message error in relevel.default(occupation) : 'relevel' only for factors. Where is the source of problem? If the source of problem is related to the factor vectors, how I can solve it? Below find an example of the code. Unfortunately, I cannot share the data or find a dataset that produces the same error message:
I have transformed the factor variables into integer vectors in the cox model and in the new dataset. it did not work.
I have deleated all the factor variables and it works.
I have tried to implement this strategy, but it did not work: Plotting predicted survival curves for continuous covariates in ggplot
fit <- coxph(Surv(entry, exit, event == 1) ~ status_plot +
exp_national + relevel(occupation, 5) + age + gender + EDUCATION , data = data)
data_rank <- with(data,
data.frame(status_plot = c(1,2,3), # factor vector of interest
exp_national=rep(mean(exp_national, na.rm = TRUE), 3),
occupation = c(5,5,5), # factor with 6 categories, number 5 is the category of reference in the cox model
age=rep(mean(age, na.rm = TRUE), 3),
gender = c(1,1,1),
EDUCATION=rep(mean(EDUCATION, na.rm = TRUE), 3) ))
surv.fin <- survfit(fit, newdata=data_rank) # this produces the error
Looking at the code it appears you probably attempted to take the mean of a factor. So do post at least str(data) as an edit to the body of your question. You should also realize that you can give a single value to a column in a data.frame call and have it recycled to the correct length, you all the meanss could be entered as a single item rather thanrep`-ng.
I am trying to make sure that all my features of type factors are represented fully (in terms of all possible factor levels) both in my tree object and in my test set for prediction.
for (j in 1:length(predictors)){
if (is.factor(Test[,j])){
ct [[names(predictors)[j]]] <- union(ct$xlevels[[names(predictors)[j]]], levels(Test[,c(names(predictors)[j])]))
}
}
however, for object ct (ctree from package party) I can't seem to understand how to access the features' factor levels, as I am getting an error
Error in ct$xlevels : $ operator not defined for this S4 class
I had this problem countless times and today I come up with a little hack that should make not needed to fix levels' discrepancy in factors.
Just make the model on the whole dataset (train + test) giving zero weight to test observations. This way the ctree model will not drop factor levels.
a <- ctree(Y ~ ., DF[train.IDs,]) %>% predict(newdata = DF) # Would trigger error if the data passed to predict would not match the train data levels
b <- ctree(Y ~ ., weights = as.numeric((1:nrow(DF) %in% train.IDs)), data = DF) %>% predict(newdata = DF) # passing the IDs as 0-1 in the weights instead of subsetting the data solves it
mean(a == b) # test that predictions are equals, should be 1
Tell me if it works as expected!
Using R 3.2.0 with caret 6.0-41 and randomForest 4.6-10 on a 64-bit Linux machine.
When trying to use the predict() method on a randomForest object trained with the train() function from the caret package using a formula, the function returns an error.
When training via randomForest() and/or using x= and y= rather than a formula, it all runs smoothly.
Here is a working example:
library(randomForest)
library(caret)
data(imports85)
imp85 <- imports85[, c("stroke", "price", "fuelType", "numOfDoors")]
imp85 <- imp85[complete.cases(imp85), ]
imp85[] <- lapply(imp85, function(x) if (is.factor(x)) x[,drop=TRUE] else x) ## Drop empty levels for factors.
modRf1 <- randomForest(numOfDoors~., data=imp85)
caretRf <- train( numOfDoors~., data=imp85, method = "rf" )
modRf2 <- caretRf$finalModel
modRf3 <- randomForest(x=imp85[,c("stroke", "price", "fuelType")], y=imp85[, "numOfDoors"])
caretRf <- train(x=imp85[,c("stroke", "price", "fuelType")], y=imp85[, "numOfDoors"], method = "rf")
modRf4 <- caretRf$finalModel
p1 <- predict(modRf1, newdata=imp85)
p2 <- predict(modRf2, newdata=imp85)
p3 <- predict(modRf3, newdata=imp85)
p4 <- predict(modRf4, newdata=imp85)
Among the last 4 lines, only the second one p2 <- predict(modRf2, newdata=imp85) returns the following error:
Error in predict.randomForest(modRf2, newdata = imp85) :
variables in the training data missing in newdata
It seems that the reason for this error is that the predict.randomForest method uses rownames(object$importance) to determine the name of the variables used to train the random forest object. And when looking at
rownames(modRf1$importance)
rownames(modRf2$importance)
rownames(modRf3$importance)
rownames(modRf4$importance)
We see:
[1] "stroke" "price" "fuelType"
[1] "stroke" "price" "fuelTypegas"
[1] "stroke" "price" "fuelType"
[1] "stroke" "price" "fuelType"
So somehow, when using the caret train() function with a formula changes the name of the (factor) variables in the importance field of the randomForest object.
Is it really an inconsistency between the formula and and non-formula version of the caret train() function? Or am I missing something?
First, almost never use the $finalModel object for prediction. Use predict.train. This is one good example of why.
There is some inconsistency between how some functions (including randomForest and train) handle dummy variables. Most functions in R that use the formula method will convert factor predictors to dummy variables because their models require numerical representations of the data. The exceptions to this are tree- and rule-based models (that can split on categorical predictors), naive Bayes, and a few others.
So randomForest will not create dummy variables when you use randomForest(y ~ ., data = dat) but train (and most others) will using a call like train(y ~ ., data = dat).
The error occurs because fuelType is a factor. The dummy variables created by train don't have the same names so predict.randomForest can't find them.
Using the non-formula method with train will pass the factor predictors to randomForest and everything will work.
TL;DR
Use the non-formula method with train if you want the same levels or use predict.train
There can be two reasons why you get this error.
1. The categories of the categorical variables in the train and test sets don't match. To check that, you can run something like the following.
Well, first of all, it is good practice to keep the independent variables/features in a list. Say that list is "vars". And say, you separated "Data" into "Train" and "Test". Let's go:
for (v in vars){
if (class(Data[,v]) == 'factor'){
print(v)
# print(levels(Train[,v]))
# print(levels(Test[,v]))
print(all.equal(levels(Train[,v]) , levels(Test[,v])))
}
}
Once you find the non-matching categorical variables, you can go back, and impose the categories of Test data onto Train data, and then re-build your model. In a loop similar to above, for each nonMatchingVar, you can do
levels(Test$nonMatchingVar) <- levels(Train$nonMatchingVar)
2. A silly one. If you accidentally leave the dependent variable in the set of independent variables, you may run into this error message. I have done that mistake. Solution: Just be more careful.
Another way is to explicitly code the testing data using model.matrix, e.g.
p2 <- predict(modRf2, newdata=model.matrix(~., imp85))
I have set of data that I've analyzed with a generalized linear model that has three categorical factors in 3-way interaction (factorA, factorB, factorC) and a fourth continuous factor (factorD) that is simply added in the model. I am trying to obtain a set of Tukey letter groups (ie, compact letter display) from the model but haven't found a way to include the interaction successfully. I'm not interested in including factorD, just the three in the interaction.
I have gotten the Tukey-adjusted pairwise comparisons with this:
lsmeans(my.glm, factorA*factorB*factorC)
But I was not able to figure out how to produce a compact letters display from that. It can be done with multcomp package but I could only find ways to do it with main effects with that package, not interactions.
So then I tried the agricolae package, as this post (https://stats.stackexchange.com/questions/31547/how-to-obtain-the-results-of-a-tukey-hsd-post-hoc-test-in-a-table-showing-groupe) discusses that that should work. However, following the instructions in that answer led to a non-functional response from HSD.test. Specifically, I could get the main effects tests to work fine, e.g. HSD.test(my.glm,"factorA") but I could not get the interactions to work. I tried this:
intxns<-with(my.data, interaction(factorA,factorB,factorC))
HSD.test(my.glm,"intxns",group=TRUE)
But a get an error that indicates the HSD.test function didn't recognize "intxns" as a valid object, it looks like this (also, I checked the intxns object and it looks good and the number of rows matched the number of residuals of my glm):
Name: inxtns
factorA factorB factorC factorD
I get that same error if I just put nonsense into the factor field in the HSD.test function call. I checked the inxtns object and it looks good and the number of rows matched the number of residu
The agricolae notes don't actually cover the use of interactions in HSD.test, but I assume it can work.
Does anyone know how to get HSD.test to work with interactions? Or is there any other function you've gotten to work to produce compact letter displays for a glm with interactions?
I've been working on this for a number of days now and haven't been able to find a solution, hopefully I'm not missing something obvious.
Thanks!
I don't know how you've specified your glm model, but for HSD.test, it's looking to match the particular treatment name with the same name specified in the glm formula as well as the data frame. This is why your main effect, factorA will work, but not the 3-way interaction. For multiple comparison tests on interactions, I find it easiest to generate the interactions separately and add them to the data frame as additional columns. The glm model can then be specified using the new variables which code for the interaction.
For example,
set.seed(42)
glm.dat <- data.frame(y = rnorm(1000), factorA = sample(letters[1:2],
size = 1000, replace = TRUE),
factorB = sample(letters[1:2], size = 1000, replace = TRUE),
factorC = sample(letters[1:2], size = 1000, replace = TRUE))
# Generate interactions explicitly and add them to the data.frame
glm.dat$factorAB <- with(glm.dat, interaction(factorA, factorB))
glm.dat$factorAC <- with(glm.dat, interaction(factorA, factorC))
glm.dat$factorBC <- with(glm.dat, interaction(factorB, factorC))
glm.dat$factorABC <- with(glm.dat, interaction(factorA, factorB, factorC))
# General linear model
glm.mod <- glm(y ~ factorA + factorB + factorC + factorAB + factorAC +
factorBC + factorABC, family = 'gaussian', data = glm.dat)
# Multiple comparison test
library(agricolae)
comp <- HSD.test(glm.mod, trt = "factorABC", group = TRUE)
giving
comp$groups giving
trt means M
1 a.a.a 0.070052189 a
2 a.b.b 0.035684571 a
3 b.a.a 0.020517535 a
4 b.b.b -0.008153257 a
5 a.b.a -0.036136140 a
6 a.a.b -0.078891136 a
7 b.a.b -0.080845419 a
8 b.b.a -0.115808772 a