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)
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 have a dataset like this:
contingency_table<-tibble::tibble(
x1_not_happy = c(1,4),
x1_happy = c(19,31),
x2_not_happy = c(1,4),
x2_happy= c(19,28),
x3_not_happy=c(14,21),
X3_happy=c(0,9),
x4_not_happy=c(3,13),
X4_happy=c(17,22)
)
in fact, there are many other variables that come from a poll aplied in two different years.
Then, I apply a Fisher test in each 2X2 contingency matrix, using this code:
matrix1_prueba <- contingency_table[1:2,1:2]
matrix2_prueba<- contingency_table[1:2,3:4]
fisher1<-fisher.test(matrix1_prueba,alternative="two.sided",conf.level=0.9)
fisher2<-fisher.test(matrix2_prueba,alternative="two.sided",conf.level=0.9)
I would like to run this task using a short code by mean of a function or a loop. The output must be a vector with the p_values of each questions.
Thanks,
Frederick
So this was a bit of fun to do. The main thing that you need to recognize is that you want combinations of your data. There are a number of functions in R that can do that for you. The main workhorse is combn() Link
So in the language of the problem, we want all combinations of your tibble taken 2 at a time link2
From there, you just need to do some looping structure to get your tests to work, and extract the p-values from the object.
list_tables <- lapply(combn(contingency_table,2,simplify=F), fisher.test)
unlist(lapply(list_tables, `[`, 'p.value'))
This should produce your answer.
EDIT
Given the updated requirements for just adjacement data.frame columns, the following modifications should work.
full_list <- combn(contingency_table,2,simplify=F)
full_list <- full_list[sapply(
full_list, function(x) all(startsWith(names(x), substr(names(x)[1], 1,2))))]
full_list <- lapply(full_list, fisher.test)
unlist(lapply(full_list, `[`, 'p.value'))
This is approximately the same code as before, but now we have to find the subsets of the data that have the same question prefix name. This only works if the prefixes are exactly the same (X3 != x3). I think this is a better solution than trying to work with column indexes, and without the guarantee of always being next to one another. The sapply code does just that. The final output should be what you need for the problem.
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')
I've been working on a project for a little bit for a homework assignment and I've been stuck on a logistical problem for a while now.
What I have at the moment is a list that returns 10000 values in the format:
[[10000]]
X-squared
0.1867083
(This is the 10000th value of the list)
What I really would like is to just have the chi-squared value alone so I can do things like create a histogram of the values.
Is there any way I can do this? I'm fine with repeating the test from the start if necessary.
My current code is:
nsims = 10000
for (i in 1:nsims) {cancer.cells <- c(rep("M",24),rep("B",13))
malig[i] <- sum(sample(cancer.cells,21)=="M")}
benign = 21 - malig
rbenign = 13 - benign
rmalig = 24 - malig
for (i in 1:nsims) {test = cbind(c(rbenign[i],benign[i]),c(rmalig[i],malig[i]))
cancerchi[i] = chisq.test(test,correct=FALSE) }
It gives me all I need, I just cannot perform follow-up analysis on it such as creating a histogram.
Thanks for taking the time to read this!
I'll provide an answer at the suggestion of #Dr. Mike.
hist requires a vector as input. The reason that hist(cancerchi) will not work is because cancerchi is a list, not a vector.
There a several ways to convert cancerchi, from a list into a format that hist can work with. Here are 3 ways:
hist(as.data.frame(unlist(cancerchi)))
Note that if you do not reassign cancerchi it will still be a list and cannot be passed directly to hist.
# i.e
class(cancerchi)
hist(cancerchi) # will still give you an error
If you reassign, it can be another type of object:
(class(cancerchi2 <- unlist(cancerchi)))
(class(cancerchi3 <- as.data.frame(unlist(cancerchi))))
# using the ldply function in the plyr package
library(plyr)
(class(cancerchi4 <- ldply(cancerchi)))
these new objects can be passed to hist directly
hist(cancerchi2)
hist(cancerchi3[,1]) # specify column because cancerchi3 is a data frame, not a vector
hist(cancerchi4[,1]) # specify column because cancerchi4 is a data frame, not a vector
A little extra information: other useful commands for looking at your objects include str and attributes.
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