I'm trying to replicate some Stata results in R and am having a lot of trouble. Specifically, I want to recover the same eigenvalues as Stata does in exploratory factor analysis. To provide a specific example, the factor help in Stata uses bg2 data (something about physician costs) and gives you the following results:
webuse bg2
factor bg2cost1-bg2cost6
(obs=568)
Factor analysis/correlation Number of obs = 568
Method: principal factors Retained factors = 3
Rotation: (unrotated) Number of params = 15
--------------------------------------------------------------------------
Factor | Eigenvalue Difference Proportion Cumulative
-------------+------------------------------------------------------------
Factor1 | 0.85389 0.31282 1.0310 1.0310
Factor2 | 0.54107 0.51786 0.6533 1.6844
Factor3 | 0.02321 0.17288 0.0280 1.7124
Factor4 | -0.14967 0.03951 -0.1807 1.5317
Factor5 | -0.18918 0.06197 -0.2284 1.3033
Factor6 | -0.25115 . -0.3033 1.0000
--------------------------------------------------------------------------
LR test: independent vs. saturated: chi2(15) = 269.07 Prob>chi2 = 0.0000
I'm interested in the eigenvalues in the first column of the table. When I use the same data in R, I get the following results:
bg2 = read.dta("bg2.dta")
eigen(cor(bg2)
$values
[1] 1.7110112 1.4036760 1.0600963 0.8609456 0.7164879 0.6642889 0.5834942
As you can see, these values are quite different from Stata's results. It is likely that the two programs are using different means of calculating the eigenvalues, but I've tried a wide variety of different methods of extracting the eigenvalues, including most (if not all) of the options in R commands fa, factanal, principal, and maybe some other R commands. I simply cannot extract the same eigenvalues as Stata. I've also read through Stata's manual to try and figure out exactly what method Stata uses, but couldn't figure it out with enough specificity.
I'd love any help! Please let me know if you need any additional information to answer the question.
I would advise against carrying out a factor analysis on all the variables in the bg2 data as one of the variables is clinid, which is an arbitrary identifier 1..568 and carries no information, except by accident.
Sensibly or not, you are not using the same data, as you worked on the 6 cost variables in Stata and those PLUS the identifier in R.
Another way to notice that would be to spot that you got 6 eigenvalues in one case and 7 in the other.
Nevertheless the important principle is that eigen(cor(bg2)) is just going to give you the eigenvalues from a principal component analysis based on the correlation matrix. So you can verify that pca in Stata would match what you report from R.
So far, so clear.
But your larger question remains. I don't know how to mimic Stata's (default) factor analysis in R. You may need a factor analysis expert, if any hang around here.
In short, PCA is not equal to principal axis method factor analysis.
Different methods of calculating eigenvalues are not the issue here. I'd bet that given the same matrix Stata and R match up well in reporting eigenvalues. The point is that different techniques mean different eigenvalues in principle.
P.S. I am not an R person, but I think what you call R commands are strictly R functions. In turn I am open to correction on that small point.
Related
I am applying an Exploratory factor analysis on a dataset using the factanal() package in R. After applying Scree Test I found out that 2 factors need to be retained from 20 features.
Trying to find what this uniqueness represents, I found the following from here
"A high uniqueness for a variable usually means it doesn’t fit neatly into our factors. ..... If we subtract the uniquenesses from 1, we get a quantity called the communality. The communality is the proportion of variance of the ith variable contributed by the m common factors. ......In general, we’d like to see low uniquenesses or high communalities, depending on what your statistical program returns."
I understand that if the value of uniquenesses is high, it could be better represented in a single factor. But what is a good threshold for this uniquenesses measure? All of my features show a value greater than 0.3, and most of them range from 0.3 to 0.7. Does the following mean that my factor analysis doesn't work well on my data? I have tried rotation, the results are not very different. What else should I try then?
You can partition an indicator variable's variance into its...
Uniqueness (h2): Variance that is not explained by the common factors
Communality (1-h2): The variance that is explained by the common factors
Which values can be considered "good" depends on your context. You should look for examples in your application domain to know what you can expect. In the package psych, you can find some examples from psychology:
library(psych)
m0 <- fa(Thurstone.33,2,rotate="none",fm="mle")
m0
m0$loadings
When you run the code, you can see that the communalities are around 0.6. The absolute factor loadings of the unrotated solution vary between 0.27 and 0.85.
An absolute value of 0.4 is often used as an arbitrary cutoff for acceptable factor loadings in psychological domains.
I use from code R for table Tests of Between-Subjects Effects calculate in spss but results are diffrent and really I don't know that why are the reasons? , please help me . thanka alot
fit2222<-manova(cbind(postrunning,postbalance,postbilateral,poststrngth,postupperlimb,postresponse
,postvisumotor,postdexterity)
~ group+prerunning+ prebalance+prebilateralcoordination + prestrength + preupperlimbcoordination
+preresponse+ previsumotor+predexterity ,data=Data)
summary.aov(fit2222)
input in R
'''
input in spss
It's impossible to give a complete answer to this, or at least exactly what you should be doing differently, but it's clear from the output you're showing that you're not comparing the same results in the two packages. Note the numerator degrees of freedom (df) values, which are all 1 in R and 2 in SPSS.
I am not sure if this is a right place to ask a question like this, but Im not sure where to ask this.
I am currently doing some research on data and have been asked to find the intraclass correlation of the observations within patients. In the data, some patients have 2 observations, some only have 1 and I have an ID variable to assign each observation to the corresponding patient.
I have come across the ICC package in R, which calculates the intraclass correlation coefficient, but there are 2 commands available: ICCbare and ICCbareF.
I do not understand what is the difference between them as they do give completely different ICC values on the same variables. For example, on the same variable, x:
ICCbare(ID,x) gave me a value of -0.01035216
ICCbareF(ID,x) gave me a value of 0.475403
The second one using ICCbareF is almost the same as the estimated correlation I get when using random effects models.
So I am just confused and would like to understand the algorithm behind them so I could explain them in my research. I know one is to be used when the data is balanced and there are no NA values.
In the description it says that it is either calculated by hand or using ANOVA - what are they?
By: https://www.rdocumentation.org/packages/ICC/versions/2.3.0/topics/ICCbare
ICCbare can be used on balanced or unbalanced datasets with NAs. ICCbareF is similar, however ICCbareF should not be used with unbalanced datasets.
Perhaps this is a philosophical question rather than a programming question, but here goes...
In R, is there some package or method that will let you deal with "less than"s as a concept?
Backstory: I have some data which, for privacy reasons, is given as <5 for small numbers (representing integers 1, 2, 3 or 4, in fact). I'd like to do some simple arithmetic on this data (adding, subtracting, averaging, etc.) but obviously I need to find some way to deal with these <5s conceptually. I could replace them all with NAs, sure, but of course that's throwing away potentially useful information, and I would like to avoid that if possible.
Some examples of what I mean:
a <- c(2,3,8)
b <- c(<5,<5,8)
mean(a)
> 4.3333
mean(b)
> 3.3333 -> 5.3333
If you are interested in the values at the bounds, I would take each dataset and split it into two datasets; one with all <5s set to 1 and one with all <5s set to 4.
a <- c(2,3,8)
b1 <- c(1,1,8)
b2 <- c(4,4,8)
mean(a)
# 4.333333
mean(b1)
# 3.3333
mean(b2)
# 5.3333
Following #hedgedandlevered proposal, but he's wrong wrt normal and/or uniform. You ask for integer numbers, so you have to use discrete distributions, like Poisson, binomial (including negative one), geometric etc
In statistics "less than" data is known as "left censored" https://en.wikipedia.org/wiki/Censoring_(statistics), searching on "censored data" might help.
My favoured approach to analysing such data is maximum likelihood https://en.wikipedia.org/wiki/Maximum_likelihood. There are a number of R packages for maximum likelihood estimation, I like the survival package https://cran.r-project.org/web/packages/survival/index.html but there are others, e.g. fitdistrplus https://cran.r-project.org/web/packages/fitdistrplus/index.html which "provides functions for fitting univariate distributions to different types of data (continuous censored or non-censored data and discrete data) and allowing different estimation methods (maximum likelihood, moment matching, quantile matching and maximum goodness-of-t estimation)".
You will have to specify (assume?) the form of the distribution of the data; you say it is integer so maybe a Poisson [related] distribution may be appropriate.
Treat them as a certain probability distribution of your choosing, and replace them with actual randomly generated numbers. All equal to 2.5, normal-like distribution capped at 0 and 5, uniform on [0,5] are all options
I deal with similar data regularly. I strongly dislike any of the suggestions of replacing the <5 values with a particular number. Consider the following two cases:
c(<5,<5,<5,<5,<5,<5,<5,<5,6,12,18)
c(<5,6,12,18)
The problem comes when you try to do arithmetic with these.
I think a solution to your issue is to think of the values as factors (in the R sense. You can bin the values above 5 too if that helps, for example
c(<5,<5,<5,<5,<5,<5,<5,<5,5-9,10-14,15-19)
c(<5,5-9,10-14,15-19)
Now, you still wouldn't do arithmetic on these, but your summary statistics (histograms/proportion tables/etc...) would make more sense.
I'm working on a very big data-set.(csv)
The data set is composed from both numeric and categorical columns.
One of the columns is my "target column" , meaning i want to use the other columns to determine which value (out of 3 possible known values) is likely to be in the "target column". In the end check my classification vs the real data.
My question:
I'm using R.
I am trying to find a way to select the subset of features which give the best classifiation.
going over all the subsets is impossible.
Does anyone know an algorithm or can think of a way do it on R?
This seems to be a classification problem. Without knowing the amount of covariates you have for your target, can't be sure, but wouldn't a neural network solve your problem?
You could use the nnet package, which uses a Feed-forward neural network and works with multiple classes. Having categorical columns is not a problem since you could just use factors.
Without a datasample I can only explain it just a bit, but mainly using the function:
newNet<-nnet(targetColumn~ . ,data=yourDataset, subset=yourDataSubset [..and more values]..)
You obtain a trained neural net. What is also important here is the size of the hidden layer which is a tricky thing to get right. As a rule of thumb it should be roughly 2/3 of the amount of imputs + amount of outputs (3 in your case).
Then with:
myPrediction <- predict(newNet, newdata=yourDataset(with the other subset))
You obtain the predicted values. About how to evaluate them, I use the ROCR package but currently only supports binary classification, I guess a google search will show some help.
If you are adamant about eliminate some of the covariates, using the cor() function may help you to identify the less caracteristic ones.
Edit for a step by step guide:
Lets say we have this dataframe:
str(df)
'data.frame': 5 obs. of 3 variables:
$ a: num 1 2 3 4 5
$ b: num 1 1.5 2 2.5 3
$ c: Factor w/ 3 levels "blue","red","yellow": 2 2 1 2 3
The column c has 3 levels, that is, 3 type of values it can take. This is something done by default by a dataframe when a column has strings instead of numerical values.
Now, using the columns a and b we want to predict which value c is going to be. Using a neural network. The nnet package is simple enough for this example. If you don't have it installed, use:
install.packages("nnet")
Then, to load it:
require(nnet)
after this, lets train the neural network with a sample of the data, for that, the function
portion<-sample(1:nrow(df),0.7*nrow(df))
will store in portion, 70% of the rows from the dataframe. Now, let's train that net! I recommend you to check the documentation for the nnet package with ?nnet for a deeper knowledge. Using only basics:
myNet<-nnet( c~ a+b,data=df,subset=portion,size=1)
c~ a+b is the formula for the prediction. You want to predict the column c using the columns a and b
data= means the data origin, in this case, the dataframe df
subset= self explanatory
size= the size of the hidden layer, as I said, use about 2/3 of the total columns(a+b) + total outputs(1)
We have trained net now, lets use it.
Using predict you will use the trained net for new values.
newPredictedValues<-predict(myNet,newdata=df[-portion,])
After that, newPredictedValues will have the predictions.
Since you have both numerical and categorical data, then you may try SVM.
I am using SVM and KNN on my numerical data and I also tried to apply DNN. DNN is pretty slow for training especially big data in R. KNN does not need to be trained, but is used for numerical data. And the following is what I am using. Maybe you can have a look at it.
#Train the model
y_train<-data[,1] #first col is response variable
x_train<-subset(data,select=-1)
train_df<-data.frame(x=x_train,y=y_train)
svm_model<-svm(y~.,data=train_df,type="C")
#Test
y_test<-testdata[,1]
x_test<-subset(testdata,select=-1)
pred<-predict(svm_model,newdata = x_test)
svm_t<-table(pred,y_test)
sum(diag(svm_t))/sum(svm_t) #accuracy