I have a linear model1<-lm(divorce_rate~marriage_rate+median_age+population) for which the leverage plot shows an outlier at 28 (State variable id for "Nevada"). I'd like to specify a model without Nevada in the dataset. I tried the following but got stuck.
data<-read.dta("census.dta")
attach(data)
data1<-data.frame(pop,divorce,marriage,popurban,medage,divrate,marrate)
attach(data1)
model1<-lm(divrate~marrate+medage+pop,data=data1)
summary(model1)
layout(matrix(1:4,2,2))
plot(model1)
dfbetaPlots(lm(divrate~marrate+medage+pop),id.n=50)
vif(model1)
dataNV<-data[!data$state == "Nevada",]
attach(dataNV)
model3<-lm(divrate~marrate+medage+pop,data=dataNV)
The last line of the above code gives me
Error in model.frame.default(formula = divrate ~ marrate + medage + pop, :
variable lengths differ (found for 'medage')
I suspect that you have some glitch in your code such that you have attach()ed copies that are still lying around in your environment -- that's why it's really best practice not to use attach(). The following code works for me:
library(foreign)
## best not to call data 'data'
mydata <- read.dta("http://www.stata-press.com/data/r8/census.dta")
I didn't find divrate or marrate in the data set: I'm going to speculate that you want the per capita rates:
## best practice to use a new name rather than transforming 'in place'
mydata2 <- transform(mydata,marrate=marriage/pop,divrate=divorce/pop)
model1 <- lm(divrate~marrate+medage+pop,data=mydata2)
library(car)
plot(model1)
dfbetaPlots(model1)
This works fine for me in a clean session:
dataNV <- subset(mydata2,state != "Nevada")
## update() may be nice to avoid repeating details of the
## model specification (not really necessary in this case)
model3 <- update(model1,data=dataNV)
Or you can use the subset argument:
model4 <- update(model1,subset=(state != "Nevada"))
Related
I'm running into some problems while running plm regressions in my panel database. Basically, I have to take out a year from my base and also all observations from some variable that are zero. I tried to make a reproducible example using a dataset from AER package.
require (AER)
library (AER)
require(plm)
library("plm")
data("Grunfeld", package = "AER")
View(Grunfeld)
#Here I randomize some observations of the third variable (capital) as zero, to reproduce my dataset
for (i in 1:220) {
x <- rnorm(10,0,1)
if (mean(x) >=0) {
Grunfeld[i,3] <- 0
}
}
View(Grunfeld)
panel <- Grunfeld
#First Method
#This is how I was originally manipulating my data and running my regression
panel <- Grunfeld
dd <-pdata.frame(panel, index = c('firm', 'year'))
dd <- dd[dd$year!=1935, ]
dd <- dd[dd$capital !=0, ]
ols_model_2 <- plm(log(value) ~ (capital), data=dd)
summary(ols_model_2)
#However, I couuldn't plot the variables of the datasets in graphs, because they weren't vectors. So I tried another way:
#Second Method
panel <- panel[panel$year!= 1935, ]
panel <- panel[panel$capital != 0,]
ols_model <- plm(log(value) ~ log(capital), data=panel, index = c('firm','year'))
summary(ols_model)
#But this gave extremely different results for the ols regression!
In my understanding, both approaches sould have yielded the same outputs in the OLS regression. Now I'm afraid my entire analysis is wrong, because I was doing it like the first way. Could anyone explain me what is happening?
Thanks in advance!
You are a running two different models. I am not sure why you would expect results to be the same.
Your first model is:
ols_model_2 <- plm(log(value) ~ (capital), data=dd)
While the second is:
ols_model <- plm(log(value) ~ log(capital), data=panel, index = c('firm','year'))
As you see from the summary of the models, both are "Oneway (individual) effect Within Model". In the first one you dont specify the index, since dd is a pdata.frame object. In the second you do specify the index, because panel is a simple data.frame. However this makes no difference at all.
The difference is using the log of capital or capital without log.
As a side note, leaving out 0 observations is often very problematic. If you do that, make sure you also try alternative ways of dealing with zero, and see how much your results change. You can get started here https://stats.stackexchange.com/questions/1444/how-should-i-transform-non-negative-data-including-zeros
I'm trying to train a RF model in R, but when i try to define the model:
rf <- randomForest(labs ~ .,data=as.matrix(dd.train))
It gives me the error:
Error in randomForest.default(m, y, ...) :
Can not handle categorical predictors with more than 53 categories.
Any idea what could it be?
And no, before you say "You have some categoric variable with more than 53 categories". No, all variables but labs are numeric.
Tim Biegeleisen: Read the last line of my question and you will see why is not the same as the one you are linking!
Edited to address followup from OP
I believe using as.matrix in this case implicitly creates factors. It is also not necessary for this packages. You can keep it as a data frame, but will need to make sure that any unused factor levels are dropped by using droplevels (or something similar). There are many reasons an unused factor may be in your data set, but a common one is a dropped observation.
Below is a quick example that reproduces your error:
library('randomForest')
#making a toy data frame
x <- data.frame('one' = c(1,1,1,1,1,seq(50) ),
'two' = c(seq(54),NA),
'three' = seq(55),
'four' = seq(55) )
x$one <- as.factor(x$one)
x <- na.omit(x) #getting rid of an NA. Note this removes the whole row.
randomForest(one ~., data = as.matrix(x)) #your first error
randomForest(one ~., data = x) #your second error
x <- droplevels(x)
randomForest(one ~., data = x) #OK
I'm fairly new to using the caret library and it's causing me some problems. Any
help/advice would be appreciated. My situations are as follows:
I'm trying to run a general linear model on some data and, when I run it
through the confusionMatrix, I get 'the data and reference factors must have
the same number of levels'. I know what this error means (I've run into it before), but I've double and triple checked my data manipulation and it all looks correct (I'm using the right variables in the right places), so I'm not sure why the two values in the confusionMatrix are disagreeing. I've run almost the exact same code for a different variable and it works fine.
I went through every variable and everything was balanced until I got to the
confusionMatrix predict. I discovered this by doing the following:
a <- table(testing2$hold1yes0no)
a[1]+a[2]
1543
b <- table(predict(modelFit,trainTR2))
dim(b)
[1] 1538
Those two values shouldn't disagree. Where are the missing 5 rows?
My code is below:
set.seed(2382)
inTrain2 <- createDataPartition(y=HOLD$hold1yes0no, p = 0.6, list = FALSE)
training2 <- HOLD[inTrain2,]
testing2 <- HOLD[-inTrain2,]
preProc2 <- preProcess(training2[-c(1,2,3,4,5,6,7,8,9)], method="BoxCox")
trainPC2 <- predict(preProc2, training2[-c(1,2,3,4,5,6,7,8,9)])
trainTR2 <- predict(preProc2, testing2[-c(1,2,3,4,5,6,7,8,9)])
modelFit <- train(training2$hold1yes0no ~ ., method ="glm", data = trainPC2)
confusionMatrix(testing2$hold1yes0no, predict(modelFit,trainTR2))
I'm not sure as I don't know your data structure, but I wonder if this is due to the way you set up your modelFit, using the formula method. In this case, you are specifying y = training2$hold1yes0no and x = everything else. Perhaps you should try:
modelFit <- train(trainPC2, training2$hold1yes0no, method="glm")
Which specifies y = training2$hold1yes0no and x = trainPC2.
I am learning how to use various random forest packages and coded up the following from example code:
library(party)
library(randomForest)
set.seed(415)
#I'll try to reproduce this with a public data set; in the mean time here's the existing code
data = read.csv(data_location, sep = ',')
test = data[1:65] #basically data w/o the "answers"
m = sample(1:(nrow(factor)),nrow(factor)/2,replace=FALSE)
o = sample(1:(nrow(data)),nrow(data)/2,replace=FALSE)
train2 = data[m,]
train3 = data[o,]
#random forest implementation
fit.rf <- randomForest(train2[,66] ~., data=train2, importance=TRUE, ntree=10000)
Prediction.rf <- predict(fit.rf, test) #to see if the predictions are accurate -- but it errors out unless I give it all data[1:66]
#cforest implementation
fit.cf <- cforest(train3[,66]~., data=train3, controls=cforest_unbiased(ntree=10000, mtry=10))
Prediction.cf <- predict(fit.cf, test, OOB=TRUE) #to see if the predictions are accurate -- but it errors out unless I give it all data[1:66]
Data[,66] is the is the target factor I'm trying to predict, but it seems that by using "~ ." to solve for it is causing the formula to use the factor in the prediction model itself.
How do I solve for the dimension I want on high-ish dimensionality data, without having to spell out exactly which dimensions to use in the formula (so I don't end up with some sort of cforest(data[,66] ~ data[,1] + data[,2] + data[,3}... etc.?
EDIT:
On a high level, I believe one basically
loads full data
breaks it down to several subsets to prevent overfitting
trains via subset data
generates a fitting formula so one can predict values of target (in my case data[,66]) given data[1:65].
so my PROBLEM is now if I give it a new set of test data, let’s say test = data{1:65], it now says “Error in eval(expr, envir, enclos) :” where it is expecting data[,66]. I want to basically predict data[,66] given the rest of the data!
I think that if the response is in train3 then it will be used as a feature.
I believe this is more like what you want:
crtl <- cforest_unbiased(ntree=1000, mtry=3)
mod <- cforest(iris[,5] ~ ., data = iris[,-5], controls=crtl)
I have a function that returns an lm object. I want to produce predicted values based on some new data. The new data is a data.frame in the exact format as the data passed to the lm function, except that the response has been removed (since we're predicting, not training). I would expect to execute the following, but get an error:
predict( model , newdata )
"Error in eval(expr, envir, enclos) : object 'ModelResponse' not found"
In my case, ModelResponse was the name of the response column in the data I originally trained on. So just for kicks, I tried to insert NA reponse:
newdata$ModelResponse = NA
predict( model , newdata )
Error in terms.default(object, data = data) : no terms component nor attribute
Highly frustrating! R's notion of models/regression doesn't match mine: 1. I train a model with some data and get a model object. 2. I can score new data from any environment/function/frame/etc. so long as I input data into the model object that "looks like" the data I trained on (i.e. same column names). This is a standard black-box paradigm.
So here are my questions:
1. What concept(s) am I missing here?
2. How do I get my scenario to work?
3. How can I get model object to be portable? str(model) shows me that the model object saved the original data it trained on! So the model object is massive. I want my model to be portable to any function/environment/etc. and only contain the data it needs to score.
In the absence of str() on either the model or the data offered to the model, here's my guess regarding this error message:
predict( model , newdata )
"Error in eval(expr, envir, enclos) : object 'ModelResponse' not found"
I guess that you made a model object named "model" and that your outcome variable (the left-hand-side of the formula( in the original call to lm was named "ModelResponse" and that you then named a column in newdata by the same name. But what you should have done was leave out the "ModelResponse" columns (because that is what you are predicting) and put in the "Model_Predictor1", Model_Predictor2", etc. ... i.e. all the names on the right-hand-side of the formula given to lm()
The coef() function will allow you to extract the information needed to make the model portable.
mod.coef <- coef(model)
mod.coef
Since you expressed interest in the rms/Hmisc package combo Function, here it is using the help-example from ols and comparing the output with an extracted function and the rms Predict method. Note the capitals, since these are designed to work with the package equivalents of lm and glm(..., family="binomial") and coxph, which in rms become ols, lrm, and cph.
> set.seed(1)
> x1 <- runif(200)
> x2 <- sample(0:3, 200, TRUE)
> distance <- (x1 + x2/3 + rnorm(200))^2
> d <- datadist(x1,x2)
> options(datadist="d") # No d -> no summary, plot without giving all details
>
>
> f <- ols(sqrt(distance) ~ rcs(x1,4) + scored(x2), x=TRUE)
>
> Function(f)
function(x1 = 0.50549065,x2 = 1) {0.50497361+1.0737604* x1-
0.79398383*pmax(x1-0.083887788,0)^3+ 1.4392827*pmax(x1-0.38792825,0)^3-
0.38627901*pmax(x1-0.65115162,0)^3-0.25901986*pmax(x1-0.92736774,0)^3+
0.06374433*x2+ 0.60885222*(x2==2)+0.38971577*(x2==3) }
<environment: 0x11b4568e8>
> ols.fun <- Function(f)
> pred1 <- Predict(f, x1=1, x2=3)
> pred1
x1 x2 yhat lower upper
1 1 3 1.862754 1.386107 2.339401
Response variable (y): sqrt(distance)
Limits are 0.95 confidence limits
# The "yhat" is the same as one produces with the extracted function
> ols.fun(x1=1, x2=3)
[1] 1.862754
(I have learned through experience that the restricted cubic-spline fit functions coming from rms need to have spaces and carriage returns added to improve readability. )
Thinking long-term, you should probably take a look at the caret package. Many or most modeling functions work with data frames and matrices, others have a preference, and there may be other variations of their expectations. It's important to quickly get your head around each, but if you want a single wrapper that will simplify life for you, making the intricacies into a "black box", then caret is as close as you can get.
As a disclaimer: I do not use caret, as I don't think modeling should be a be a black box. I've had more than a few emails to maintainers of modeling packages resulting from looking into their code and seeing something amiss. Wrapping that in another layer would not serve my interests. So, in the very long-run, avoid caret and develop an enjoyment for dissecting what's going into and out of the different modeling functions. :)