问题
I'm trying to run an MCA on a datatable using FactoMineR. It contains only 0/1 numerical columns, and its size is 200.000 * 20.
require(FactoMineR)
result <- MCA(data[, colnames, with=F], ncp = 3)
I get the following error :
Error in which(unlist(lapply(listModa, is.numeric))) : argument to 'which' is not logical
I didn't really know what to do with this error. Then I tried to turn every column to character, and everything worked. I thought it could be useful to someone else, and that maybe someone would be able to explain the error to me ;)
Cheers
回答1:
It's difficult to tell without further input, but what you can do is:
- Find the function where the error occurred (via
traceback()
), Set a breakpoint and debug it:
trace(tab.disjonctif, browser)
I did the following (offline) to find the name of tab.disjonctif
:
- Found the package on the CRAN mirror on GitHub
- Search for that particular expression that gives the error
回答2:
Are the classes of your variables character or factor?I was having this problem. My solution was to change al variables to factor.
#my data.frame was "aux.da"
i=0
while(i < ncol(aux.da)){
i=i+1 aux.da[,i] = as.factor(aux.da[,i])
}
回答3:
I just started to learn R yesterday, but the error comes from the fact that the MCA is for categorical data
, so that's why your data cannot be numeric. Then to be more precise, before the MCA a "tableau disjonctif" (sorry i don't know the word in english : Complete disjunctive matrix) is created.
So FactomineR is using this function :
https://github.com/cran/FactoMineR/blob/master/R/tab.disjonctif.R
Where i think it's looking for categorical values that can be matched to a numerical value (like Y = 1, N = 0).
For others ; be careful : for R categorical data is related to factor
type, so even if you have characters you could get this error
.
回答4:
Same problem as well and changing to factor did not solve my answer either, because I had put every variable as supplementary.
What I did first was transform all my numeric data to factor :
Xfac = factor(X[,1], ordered = TRUE)
for (i in 2:29){
tfac = factor(X[,i], ordered = TRUE)
Xfac = data.frame(Xfac, tfac)
}
colnames(Xfac)=labels(X[1,])
Still, it would not work. But my 2nd problem was that I included EVERY factor as supplementary variable ! So these :
MCA(Xfac, quanti.sup = c(1:29), graph=TRUE)
MCA(Xfac, quali.sup = c(1:29), graph=TRUE)
Would generate the same error, but this one works :
MCA(Xfac, graph=TRUE)
Not transforming the data to factors also generated the problem.
I posted the same answer to a related topic : https://stackoverflow.com/a/40737335/7193352
来源:https://stackoverflow.com/questions/34266186/what-does-argument-to-which-is-not-logical-mean-in-factominer-mca