A pretty straightforward for those with intimate knowledge of R
full <- lm(hello~., hellow)
In the above specification, linear regression is being used and hello is being modeled against all variables in dataset hellow.
I have 33 variables in hellow; I wish to specify some of those as independent variable. These variables have names that carry a meaning so I really don't want to rename them to x1 x2 etc.
How can I, without having to type the individual names of the variables (since that is pretty tedious), specify a select number of variables from the whole bunch?
I tried
full <- lm(hello~hellow[,c(2,5:9)]., hellow)
but it gave me an error "Error in model.frame.default(formula = hello ~ hellow[, : invalid type (list) for variable 'hellow[, c(2, 5:9)]'
reformulate will construct a formula given the names of the variables, so something like:
(Construct data first):
set.seed(101)
hellow <- setNames(as.data.frame(matrix(rnorm(1000),ncol=10)),
c("hello",paste0("v",1:9)))
Now run the code:
ff <- reformulate(names(hellow)[c(2,5,9)],response="hello")
full <- lm(ff, data=hellow)
should work. (Works fine with this example.)
An easier solution just occurred to me; just select the columns/variables you want first:
hellow_red <- hellow[,c(1,2,5,9)]
full2 <- lm(hello~., data=hellow_red)
all.equal(coef(full),coef(full2)) ## TRUE
Related
I am trying to do an anova anaysis in R on a data set with one within factor and one between factor. The data is from an experiment to test the similarity of two testing methods. Each subject was tested in Method 1 and Method 2 (the within factor) as well as being in one of 4 different groups (the between factor). I have tried using the aov, the Anova(in car package), and the ezAnova functions. I am getting wrong values for every method I try. I am not sure where my mistake is, if its a lack of understanding of R or the Anova itself. I included the code I used that I feel should be working. I have tried a ton of variations of this hoping to stumble on the answer. This set of data is balanced but I have a lot of similar data sets and many are unblanced. Thanks for any help you can provide.
library(car)
library(ez)
#set up data
sample_data <- data.frame(Subject=rep(1:20,2),Method=rep(c('Method1','Method2'),each=20),Level=rep(rep(c('Level1','Level2','Level3','Level4'),each=5),2))
sample_data$Result <- c(4.76,5.03,4.97,4.70,5.03,6.43,6.44,6.43,6.39,6.40,5.31,4.54,5.07,4.99,4.79,4.93,5.36,4.81,4.71,5.06,4.72,5.10,4.99,4.61,5.10,6.45,6.62,6.37,6.42,6.43,5.22,4.72,5.03,4.98,4.59,5.06,5.29,4.87,4.81,5.07)
sample_data[, 'Subject'] <- as.factor(sample_data[, 'Subject'])
#Set the contrats if needed to run type 3 sums of square for unblanaced data
#options(contrats=c("contr.sum","contr.poly"))
#With aov method as I understand it 'should' work
anova_aov <- aov(Result ~ Method*Level + Error(Subject/Method),data=test_data)
print(summary(anova_aov))
#ezAnova method,
anova_ez = ezANOVA(data=sample_data, wid=Subject, dv = Result, within = Method, between=Level, detailed = TRUE, type=3)
print(anova_ez)
Also, the values I should be getting as output by SAS
SAS Anova
Actually, your R code is correct in both cases. Running these data through SPSS yielded the same result. SAS, like SPSS, seems to require that the levels of the within factor appear in separate columns. You will end up with 20 rows instead of 40. An arrangmement like the one below might give you the desired result in SAS:
Subject Level Method1 Method2
I understand that in the following
aa <- sapply(c("BMI","KOL"),function(x) as.formula(paste('Surv(BL_AGE,CVD_AGE,INCIDENT_CVD) ~', paste(colnames(s)[c(21,259,330,380)], collapse='+'))))
I am missing x
but i really don't understand how and where to insert it to be correct.
Thank you for any help.
Making this an answer instead of a comment due to amount of text.
If I understand you correctly, you're trying to iterate over a list of variables, which you want to add (each in turn) to a set of independent variables in a survival model. The issue in the code you gave is that you don't give x a place. There are several approaches to do so.
The first one is very similar to what you're doing, and creates the formulas. I demonstrate this using the 'cancer' dataset:
library(survival)
data(cancer)
myvars <- c("meal.cal","wt.loss")
a1 <- sapply(myvars,function(x){
as.formula(sprintf("Surv(time, status)~age+sex+%s",x))
}
)
#then we can fit our models
lapply(a1,function(x){coxph(formula=x,data=cancer)})
In my opinion, this is a bit convoluted and can be done in one step:
models <- lapply(myvars, function(x){
form <- as.formula(sprintf("Surv(time, status)~age+sex+%s",x))
fit <- coxph(formula=form, data=cancer)
return(fit)
})
Using the code you started with, we can simply add 'x' to the vector of dependent variables. However, this is not very readable code and I'm always a bit nervous about feeding column indices to models. You might be safer using variable names instead.
aa <- sapply(c("BMI","KOL"),function(x) as.formula(paste('Surv(BL_AGE,CVD_AGE,INCIDENT_CVD) ~', paste(c(x,colnames(s)[c(21,259,330,380)]), collapse='+'))))
I'm trying to write a function I can apply to a string vector or list instead of writing a loop. My goal is to run a regression for different endogenous variables and save the resulting tables. Since experienced R users tell us we should learn the apply functions, I want to give it a try. Here is my attempt:
Broken Example:
library(ExtremeBounds)
Data <- data.frame(var1=rbinom(30,1,0.2),var2=rbinom(30,1,0.2),var3=rnorm(30),var4=rnorm(30),var5=rnorm(30),var6=rnorm(30))
spec1 <- list(y=c("var1"),freevars=("var3"),doubtvars=c("var4","var5"))
spec2 <- list(y=c("var2"),freevars=("var4"),doubtvars=c("var3","var5","var6"))
specs <- c("spec1","spec2")
myfunction <- function(x){
eba <- eba(data=Data, y=x$y,
free=x$freevars,
doubtful=x$doubtvars,
reg.fun=glm, k=1, vif=7, draws=50, se.fun = se.robust, weights = "lri", family = binomial(logit))
output <- eba$bounds
output <- output[,-(3:7)]
}
lapply(specs,myfunction)
Which gives me an error that makes me guess that R does not understand when x should be "spec1" or "spec2". Also, I don't quite understand what lapply would try to collect here. Could you provide me with some best practice/hints how to communicate such things to R?
error: Error in x$y : $ operator is invalid for atomic vectors
Working example:
Here is a working example for spec1 without using apply that shows what I'm trying to do. I want to loop this example through 7 specs but I'm trying to get away from loops. The output does not have to be saved as a csv, a list of all outputs or any other collection would be great!
eba <- eba(data=Data, y=spec1$y,
free=spec1$freevars,
doubtful=spec1$doubtvars,
reg.fun=glm, k=1, vif=7, draws=50, se.fun = se.robust, weights = "lri", family = binomial(logit))
output <- eba$bounds
output <- output[,-(3:7)]
write.csv(output, "./Results/eba_pmr.csv")
Following the comments of #user20650, the solution is quite simple:
In the lapply command, use lapply(mget(specs),myfunction) which gets the names of the list elements of specs instead of the lists themselves.
Alternatively, one could define specs as a list: specs <- list(spec1,spec2) but that has the downside that the lapply command will return a list where the different specifications are numbered. The first version keeps the names of the specifications (spec1 and spec2) which which makes work with the resulting list much easier.
I have a graph structure, determined from another method, and I want to do parameter learning. The bnlearn methods, however, seem to do parameter learning directly on the dataset (strictly in a dataframe). I have two questions: how do I do parameter learning from an igraph or graphNEL structure with bnlearn?
Second question: I am getting a check.data() error when I try to do parameter learning using my dataset. Their example code works, and I can't understand why my dataset does not. See their code below and a reproducible example, below.
Here is their example code:
require(bnlearn)
require(Rgraphviz)
data(learning.test)
bn <- naive.bayes(learning.test, "A")
pred <- predict(bn, learning.test)
table(pred, learning.test[,"A"])
My reproducible example (errors on naive.bayes() call):
require(bnlearn, Rgraphviz)
data <- data <- matrix(sample.int(200, 61*252, TRUE), nrow=252, ncol=61)
data <- as.data.frame(matrix(as.numeric(as.matrix(data)), ncol=ncol(data),
byrow=TRUE))
bn <- naive.bayes(data, names(data)[1])
Error message:
Error in check.data(data, allowed.types = discrete.data.types) :
valid data types are:
* all variables must be unordered factors.
* all variables must be ordered factors.
* variables can be either ordered or unordered factors.
I do not think this error comes from detecting integers, because when I cast my data to a dataframe, I first cast it to numeric, because other methods in bnlearn require numeric or factored data. This dataset IS count data, but I want to use the method assuming I am using continuous datasets. Does this make sense?
I'm trying to run a regression for every zipcode in my dataset and save the coefficients to a data frame but I'm having trouble.
Whenever I run the code below, I get a data frame called "coefficients" containing every zip code but with the intercept and coefficient for every zipcode being equal to the results of the simple regression lm(Sealed$hhincome ~ Sealed$square_footage).
When I run the code as indicated in Ranmath's example at the link below, everything works as expected. I'm new to R after many years with STATA, so any help would be greatly appreciated :)
R extract regression coefficients from multiply regression via lapply command
library(plyr)
Sealed <- read.csv("~/Desktop/SEALED.csv")
x <- function(df) {
lm(Sealed$hhincome ~ Sealed$square_footage)
}
regressions <- dlply(Sealed, .(Sealed$zipcode), x)
coefficients <- ldply(regressions, coef)
Because dlply takes a ... argument that allows additional arguments to be passed to the function, you can make things even simpler:
dlply(Sealed,.(zipcode),lm,formula=hhincome~square_footage)
The first two arguments to lm are formula and data. Since formula is specified here, lm will pick up the next argument it is given (the relevant zipcode-specific chunk of Sealed) as the data argument ...
You are applying the function:
x <- function(df) {
lm(Sealed$hhincome ~ Sealed$square_footage)
}
to each subset of your data, so we shouldn't be surprised that the output each time is exactly
lm(Sealed$hhincome ~ Sealed$square_footage)
right? Try replacing Sealed with df inside your function. That way you're referring to the variables in each individual piece passed to the function, not the whole variable in the data frame Sealed.
The issue is not with plyr but rather in the definition of the function. You are calling a function, but not doing anything with the variable.
As an analogy,
myFun <- function(x) {
3 * 7
}
> myFun(2)
[1] 21
> myFun(578)
[1] 21
If you run this function on different values of x, it will still give you 21, no matter what x is. That is, there is no reference to x within the function. In my silly example, the correction is obvious; in your function above, the confusion is understandable. The $hhincome and $square_footage should conceivably serve as variables.
But you want your x to vary over what comes before the $. As #Joran correctly pointed out, swap sealed$hhincome with df$hhincome (and same for $squ..) and that will help.