I am Having a little problem doing a Levene test in R. I does not get any output value, only NaN. Anyone know what the problem might be?
Have used the code:
with(Test,levene.test(Sample1,Sample2,location="median"))
The problem
Best regards
The levene.test function assumes the data are in a single vector. The second argument is a grouping variable.
Concatenate your data using the c() function: data=c(Sample1, Sample2). Construct a vector of group names like gp = rep('Gp1','Gp2', each=240). Then, call the function as follows: levene.test(data, gp, location='median').
This can also be done directly:
levene.test(c(Sample1, Sample2), rep('Gp1', 'Gp2', each=240)), location='median')
Related
I have a dataset "res.sav" that I read in via haven. It contains 20 columns, called "Genes1_Acc4", "Genes2_Acc4" etc. I am trying to find a correlation coefficient between those and another column called "Condition". I want to separately list all coefficients.
I created two functions, cor.condition.cols and cor.func to do that. The first iterates through the filenames and works just fine. The second was supposed to give me my correlations which didn't work at all. I also created a new "cor.condition.Genes" which I would like to fill with the correlations, ideally as a matrix or dataframe.
I have tried to iterate through the columns with two functions. However, when I try to pass it, I get the error: "NAs introduced by conversion". This wouldn't be the end of the world (I tried also suppressWarning()). But the bigger problem I have that it seems like my function does not convert said columns into the numeric type I need for my cor() function. I receive the "y must be numeric" error when trying to run the cor() function. I tried to put several arguments within and without '' or "" without success.
When I ran str(cor.condition.cols) I only receive character strings, which makes me think that my function somehow messes up with the as.numeric function. Any suggestions of how else I could iter through these columns and transfer them?
Thanks guys :)
cor.condition.cols <- lapply(1:20, function(x){paste0("res$Genes", x, "_Acc4")})
#save acc_4 columns as numeric columns and calculate correlations
res <- (as.numeric("cor.condition.cols"))
cor.func <- function(x){
cor(res$Condition, x, use="complete.obs", method="pearson")
}
cor.condition.Genes <- cor.func(cor.condition.cols)
You can do:
cor.condition.cols <- paste0("Genes", 1:20, "_Acc4")
res2 <- as.numeric(as.matrix(res[cor.condition.cols]))
cor.condition.Genes <- cor(res2, res$Condition, use="complete.obs", method="pearson")
eventually the short variant:
cor.condition.cols <- paste0("Genes", 1:20, "_Acc4")
cor.condition.Genes <- cor(res[cor.condition.cols], res$Condition, use="complete.obs")
Here is an example with other data:
cor(iris[-(4:5)], iris[[4]])
I am a second year M.Sc student and I am running into a bit of a snag running my statistics.
I am trying to run a contingency table and Fishers test and I keep getting an error.
Error in fisher.test(GAL4UAS) : if 'x' is not a matrix, 'y' must be given
If anyone can see what I have done wrong/may be missing I would really appreciate it?
This is the code:
setwd("/Users/Pria/Desktop/Data Analysis/")
GAL4UAS <-- data.frame(Yes=c(20,21,19),No=c(10,9,11))
GAL4UAS <- lapply(GAL4UAS, abs)
fisher.test(GAL4UAS)
fisher.test(GAL4UAS[c(1,2)])
fisher.test(GAL4UAS[c(1,3)])
fisher.test() is anticipating a matrix as an input and not a data frame. Try putting your data into a matrix. One option among several would be:
m <- matrix(c(20,21,19,10,9,11),nrow = 3,ncol=2,byrow=FALSE)
fisher.test(m)
When you apply the abs() using lapply the output is a list and not a data.frame. The apply function returns the output in a matrix format which is expected in the fisher.test(). So maybe you can try this:
GAL4UAS <- data.frame(Yes=c(20,21,19),No=c(10,9,11))
GAL4UAS <- apply(GAL4UAS, abs, MARGIN=c(1,2))
fisher.test(GAL4UAS)
I have created some simple functions in R that take data and return variables. I'd appreciate instruction on taking this to another level using the code below as illustration: There's a data frame 'df' which contains variables DOFW, CDFME (there are others, but two will suffice for this example. I'd like to pass to the function the name of the variable and have it build the barplot. Basically I want to create a function that performs the actions from z<- through abline and call that function for each variable. Thank you.
df<-data.frame(y.train,x.train)
z<-ddply(df,~DOFW,summarise,median=median(PCTCHG))
row.names(z)<-z$DOFW
z$DOFW<-NULL
barplot(t(as.matrix(z)),main="Median Return by DOFW",xlab="DOFW")
abline(h=median(y.train))
z<-ddply(df,~CDFME,summarise,median=median(PCTCHG))
row.names(z)<-z$CDFME
z$CDFME<-NULL
barplot(t(as.matrix(z)),main="Median Return by CDFME",xlab="CDFME")
abline(h=median(y.train))
Actually you don't really need non-standard evaluation since ddply allows you to pass a string for the variable name rather than a formula. Here's how you could do that.
#sample data
dd<-data.frame(a=rep(1:5, 2), b=2:11, c=runif(10))
#define function
library(plyr)
myplot<-function(coln) {
z<-ddply(dd, coln, summarise, median=median(c))
barplot(z[,2], main=paste("Median Return By", coln))
abline(h=median(dd$c))
}
#make plots
myplot("a")
myplot("b")
and the easiest way to get the names of the columns as a character vector is names(dd).
I'm trying to run a regression for every zipcode in my dataset and save the coefficients to a data frame but I'm having trouble.
Whenever I run the code below, I get a data frame called "coefficients" containing every zip code but with the intercept and coefficient for every zipcode being equal to the results of the simple regression lm(Sealed$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 $squ..) and that will help.
For each of 100 data sets, I am using lm() to generate 7 different equations and would like to extract and compare the p-values and adjusted R-squared values.
Kindly assume that lm() is in fact the best regression technique possible for this scenario.
In searching the web I've found a number of useful examples for how to create a function that will extract this information and write it elsewhere, however, my code uses paste() to label each of the functions by the data source, and I can't figure out how to include these unique pasted names in the function I create.
Here's a mini-example:
temp <- data.frame(labels=rep(1:10),LogPre= rnorm(10))
temp$labels2<-temp$labels^2
testrun<-c("XX")
for (i in testrun)
{
assign(paste(i,"test",sep=""),lm(temp$LogPre~temp$labels))
assign(paste(i,"test2",sep=""),lm(temp$LogPre~temp$labels2))
}
I would then like to extract the coefficients of each equation
But the following doesn't work:
summary(paste(i,"test",sep="")$coefficients)
and neither does this:
coef(summary(paste(i,"test",sep="")))
Both generating the error :$ operator is invalid for atomic vectors
EVEN THOUGH
summary(XXtest)$coefficients
and
coef(summary(XXtest))
work just fine.
How can I use paste() within summary() to allow me to do this for AAtest, AAtest2, ABtest, ABtest2, etc.
Thanks!
Hard to tell exactly what your purpose is, but some kind of apply loop may do what you want in a simpler way. Perhaps something like this?
temp <- data.frame(labels=rep(1:10),LogPre= rnorm(10))
temp$labels2<-temp$labels^2
testrun<-c("XX")
names(testrun) <- testrun
out <- lapply(testrun, function(i) {
list(test1=lm(temp$LogPre~temp$labels),
test2=lm(temp$LogPre~temp$labels2))
})
Then to get all the p-values for the slopes you could do:
> sapply(out, function(i) sapply(i, function(x) coef(summary(x))[2,4]))
XX
test1 0.02392516
test2 0.02389790
Just using paste results in a character string, not the object with that name. You need to tell R to get the object with that name by using get.
summary(get(paste(i,"test",sep="")))$coefficients