I am attempting to run a chi sqare analysis on the data frame (called "habitat.re") below however im having difficulty as I've gotten it to read the data but its giving the wrong results, when i prompt it with $expected it returns 18 different colums when there should be 3 (one for each site).
All the tourorials ive been able to find have the data as a table, however i've not been able to convert it correctly myself.
The chisq.test function is intended to work with two variables, or columns in this case. If you want to compare all three of your columns, then I suspect you would want to compare 1-2, 2-3, and 3-3, e.g.
chisq.test(x=habitat.re$Gidgee, y=habitat.re$`Ian's Place`)
chisq.test(x=habitat.re$`Ian's Place`, y=habitat.re$`Saw Mulga`)
chisq.test(x=habitat.re$Gidgee, y=habitat.re$`Saw Mulga`)
Actually, just typing in the above should reveal much useful information directly to the R console, something like this:
data: habitat.re$Gidgee and y=habitat.re$`Ian's Place`
X-squared = 5.5569, df = 1, p-value = 0.01841
A sufficiently low p-value might indicate that the two columns are in fact dependent.
Pearson's Chi-Squared Test requires a data frame to be made into a matrix table containing only the variables you need as numerical values. N.B. my data frame is called "habitat.re"
habitat.df<-data.matrix(habitat.re, rownames.force = NA)# convert to matrix table
habitat.df<- habitat.df[,-c(1,2,3)] # delete first 3 columns
rownames(habitat.df) <- habitat.re$COMMON.NAME #pull names from original
chisq.test(habitat.df) #do chisquare test
chisq.test(habitat.df)$expected #return predicted values
The following are images of my data frames
habitat.re
habitat.df
Related
I am trying to run a survival analysis on a set of data I have collected. In this data frame (m3), each row is a new patient and each column is a mutation I have identified. I have made a binary data table to indicate whether each patient is positive or negative for the mutation. I can run a survfit function for each column(mutation), but I have hundreds and want to loop through them. I have written the following code, but don't think it is correct (nothing is being output).
for (i in m3[,2:256]) {survdiff(Surv(m3$Overall.Survival, m3$Status) ~ i,
data = m3)}
Once I gather this data I want to make a table with each mutation (column) as a row and put the p-value from this survfit object as the column.
I'm not sure why I don't have any output for the for loop and even more so how to generate the new data frame. I believe I would be subsetting it.
I've got a 1000x1000 matrix consisting of a random distribution of the letters a - z, and I need to be able to plot the data in a rank abundance distribution plot; however I'm having a lot of trouble with it due to a) it all being in character format, b) it being as a matrix and not a vector (though I have changed it to a vector in one attempt to sort it), and c) I seem to have no idea how to summarise the data so that I get species abundance, let alone then be able to rank it.
My code for the matrix is:
##Create Species Vector
species.b<-letters[1:26]
#Matrix creation (Random)
neutral.matrix2<- matrix(sample(species.b,10000,replace=TRUE),
nrow=1000,
ncol=1000)
##Turn Matrix into Vector
neutral.b<-as.character(neutral.matrix2)
##Loop
lo.op <- 2
neutral.v3 <- neutral.matrix2
neutral.c<-as.character(neutral.v3)
repeat {
neutral.v3[sample(length(neutral.v3),1)]<-as.character(sample(neutral.c,1))
neutral.c<-as.character(neutral.v3)
lo.op <- lo.op+1
if(lo.op > 10000) {
break
}
}
Which creates a matrix, 1000x1000, then replaces 10,000 elements randomly (I think, I don't know how to check it until I can check the species abundances/rank distribution).
I've run it a couple of times to get neutral.v2, neutral.v3, and neutral.b, neutral.c, so I should theoretically have two matrices/vectors that I can plot and compare - I just have no idea how to do so on a purely character dataset.
I also created a matrix of the two vectors:
abundance.matrix<-matrix(c(neutral.vb,neutral.vc),
nrow=1000000,
ncol=2)
As a later requirement is for sites, and each repeat of my code (neutral.v2 to neutral.v11 eventually) could be considered a separate site for this; however this didn't change the fact that I have no idea how to treat the character data set in the first place.
I think I need to calculate the abundance of each species in the matrix/vectors, then run it through either radfit (vegan) or some form of the rankabundance/rankabun plot (biodiversityR). However the requirements for those functions:
rankabundance(x,y="",factor="",level,digits=1,t=qt(0.975,df=n-1))
x Community data frame with sites as rows, species as columns and species abundance
as cell values.
y Environmental data frame.
factor Variable of the environment
aren't available in the data I have, as for all intents and purposes I just have a "map" of 1,000,000 species locations, and no idea how to analyse it at all.
Any help would be appreciated: I don't feel like I've explained it very well though, so sorry about that!.
I'm not sure exactly what you want, but this will summarise the data and make it into a data.frame for rankabundance
counts <- as.data.frame(as.list(table(neutral.matrix2)))
BiodiversityR::rankabundance(counts)
Dear Friends I would appreciate if someone can help me in some question in R.
I have a data frame with 8 variables, lets say (v1,v2,...,v8).I would like to produce groups of datasets based on all possible combinations of these variables. that is, with a set of 8 variables I am able to produce 2^8-1=63 subsets of variables like {v1},{v2},...,{v8}, {v1,v2},....,{v1,v2,v3},....,{v1,v2,...,v8}
my goal is to produce specific statistic based on these groupings and then compare which subset produces a better statistic. my problem is how can I produce these combinations.
thanks in advance
You need the function combn. It creates all the combinations of a vector that you provide it. For instance, in your example:
names(yourdataframe) <- c("V1","V2","V3","V4","V5","V6","V7","V8")
varnames <- names(yourdataframe)
combn(x = varnames,m = 3)
This gives you all permutations of V1-V8 taken 3 at a time.
I'll use data.table instead of data.frame;
I'll include an extraneous variable for robustness.
This will get you your subsetted data frames:
nn<-8L
dt<-setnames(as.data.table(cbind(1:100,matrix(rnorm(100*nn),ncol=nn))),
c("id",paste0("V",1:nn)))
#should be a smarter (read: more easily generalized) way to produce this,
# but it's eluding me for now...
#basically, this generates the indices to include when subsetting
x<-cbind(rep(c(0,1),each=128),
rep(rep(c(0,1),each=64),2),
rep(rep(c(0,1),each=32),4),
rep(rep(c(0,1),each=16),8),
rep(rep(c(0,1),each=8),16),
rep(rep(c(0,1),each=4),32),
rep(rep(c(0,1),each=2),64),
rep(c(0,1),128)) *
t(matrix(rep(1:nn),2^nn,nrow=nn))
#now get the correct column names for each subset
# by subscripting the nonzero elements
incl<-lapply(1:(2^nn),function(y){paste0("V",1:nn)[x[y,][x[y,]!=0]]})
#now subset the data.table for each subset
ans<-lapply(1:(2^nn),function(y){dt[,incl[[y]],with=F]})
You said you wanted some statistics from each subset, in which case it may be more useful to instead specify the last line as:
ans2<-lapply(1:(2^nn),function(y){unlist(dt[,incl[[y]],with=F])})
#exclude the first row, which is null
means<-lapply(2:(2^nn),function(y){mean(ans2[[y]])})
My question is very similar to this one here , but I still can't solve my problem and thus would like to get little bit more help to make it clear. The original dataframe "ddf" looks like:
CONC <- c(0.15,0.52,0.45,0.29,0.42,0.36,0.22,0.12,0.27,0.14)
SPP <- c(rep('A',3),rep('B',3),rep('C',4))
LENGTH <- c(390,254,380,434,478,367,267,333,444,411)
ddf <- as.data.frame(cbind(CONC,SPECIES,LENGTH))
the regression model is constructed based on Species:
model <- dlply(ddf,.(SPP), lm, formula = CONC ~ LENGTH)
the regression model works fine and returns individual models for each species.
What I am going to get is the residual and expected value of 'Length' variable in terms of each models (corresponding to different species) and I want those data could be added into my original dataset ddf as new columns. so the new dataset should looks like:
SPP LENGTH CONC EXPECTED RESIDUAL
Firstly, I use the following code to get the expected value:
model_pre <- lapply(model,function(x)predict(x,data = ddf))
I loom there might be some mistakes in the above code, but it actually works! The result comes with two columns ( predicated value and species). My first question is whether I could believe this result of above code? (Does R fully understand what I am aiming to do, getting expected value of "length" in terms of different model?)
Then i used the following code to attach those data to ddf:
ddf_new <- cbind(ddf, model_pre)
This code works fine as well. But the problem comes here. It seems like R just attach the model_pre result directly to the original dataframe, since the result of model_pre is not sorted the same as the original ddf and thus is obviously wrong(justifying by the species column in original dataframe and model_pre).
I was using resid() and similar lapply, cbind code to get residual and attach it to original ddf. Same problem comes.
Therefore, how can I attach those result correctly in terms of length by species? (please let me know if you confuse what I am trying to explain here)
Any help would be greatly appreciated!
There are several problems with your code, you refer to columns SPP and Conc., but columns by those names don't exist in your data frame.
Your predicted values are made on the entire dataset, not just the subset corresponding to that model (this may be intended, but seems strange with the later usage).
When you cbind a data frame to a list of data frames, does it really cbind the individual data frames?
Now to more helpful suggestions.
Why use dlply at all here? You could just fit a model with interactions that effectively fits a different regression line to each species:
fit <- lm(CONC ~ SPECIES * LENGTH, data= ddf)
fitted(fit)
predict(fit)
ddf$Pred <- fitted(fit)
ddf$Resid <- ddf$CONC - ddf$Pred
Or if there is some other reason to really use dlply and the problem is combining 2 data frame that have different ordering then either use merge or reorder the data frames to match first (see functions like ordor, sort.list, and match).
I am new to R and trying to use wilcox.test on my data : I have a dataframe 36021X246 with rownames as probeIDs and the last row is a label which indicates which group the samples belong to - "control" for the first 140 and "treated" for the last 106.
I would greatly appreciate knowing how to define the two groups when I perform the test....I am unable to find much information on the "formula" argument online except that -
"formula
a formula of the form lhs ~ rhs where lhs is a numeric variable giving the data values and rhs a factor with two levels giving the corresponding groups."
If someone could explain what lhs~rhs means and how to define this formula I would really appreciate it.
Thanks!
R typically assumes that each row is a case and the columns are associated variables. If the cases from both your samples occur in the same data frame, one column would be an indicator variable for sample membership. Let's call is IndSample. The Wilcoxon is a univariate test, so you would have another column containing the response values you are testing on. Let's call it Y. You then write
wilcox.test(y ~ IndSample, data=MyData, .....)
and the rest of your parameters for the test: is it two-sided? Do you want an exact statistic? (Probably not, in your case.)
It looks to me as if your data is on its side. That's problematic with a data frame, since you can't just pull out a row from a data frame, the way you would with a matrix.
You need to grab the last row and turn it into a factor - something like
factor(c(MyData[lastrow,]))
Then pull out the row that contains your response:
Y <- as.numeric(c(MyData[ResponseRow,]))
Then do the wilcoxon.
However, I am not sure that I have properly understood your situation. That seems to be a very large data matrix for a modest wilcoxon test.