[R] loop of quartile groups
Rui Barradas
ruipbarradas at sapo.pt
Wed Oct 17 17:23:20 CEST 2012
Hello,
There's no function cut2() but it's not very difficult to write one.
I've named your data example 'dat', it saves keystrokes.
Try the following.
dat <- structure(...etc...)
cut2 <- function(x, g = 0){
cut(x, breaks = c(-Inf, seq(min(x), max(x), length.out = g)))
}
fun <- function(x) {
ct <- cut2(x, g = 4)
factor(ct, levels = levels(ct), labels=1:4)
}
tmp <- sapply(dat[-1], fun)
colnames(tmp) <- paste("quartile", colnames(tmp), sep="_")
dat2 <- cbind(dat, tmp)
str(dat2)
Hope this helps,
Rui Barradas
Em 17-10-2012 15:23, Charles Determan Jr escreveu:
> Greetings R users,
>
> My goal is to generate quartile groups of each variable in my data set. I
> would like each experiment to have its designated group added as a
> subsequent column. I can accomplish this individually with the following
> code:
>
> brks <- with(data_variables,
>
> cut2(var2, g=4))
>
> #I don't want the actual numbers, I need a numbered group
>
> data$test1=factor(brks, labels=1:4)
>
>
> However, I cannot get a loop to work nor can I get a loop to add the
> columns with an appropriate name (ex. quartile_variable). I have tried
> multiple different ways but can't seem to get it to work. I think it would
> begin something like this:
>
>
> for(i in 11:ncol(survival_data_variables)){
> brks=as.data.frame(with(survival_data_variables,
> cut2(survival_data_variables[,i], g=4)))
>
>
> Any assistance would be sincerely appreciated. I would like the final data
> set to have the following layout:
>
>
> ID var1 var2 var3 var4 quartile var1
> quartile var2 quartile var3 quartile var4
>
>
> Here is a subset of my data to work with:
>
> structure(list(ID = c(11112L, 11811L, 12412L, 12510L, 13111L,
>
> 20209L, 20612L, 20711L, 21510L, 22012L), var1 = c(106, 107,
>
> 116, 67, 76, 146, 89, 62, 65, 116), var2 = c(0, 0, 201,
>
> 558, 526, 555, 576, 0, 531, 649), var3 = c(70.67, 81.33,
>
> 93.67, 84.33, 52, 74, 114, 101, 80.33, 91.33), var4 = c(136,
>
> 139, 142, 138, 140, 140, 136, 139, 140, 139)), .Names = c("ID",
>
> "var1", "var2", "var3", "var4"), row.names = c(NA,
>
> 10L), class = "data.frame")
>
>
> Regards,
> Charles
>
> [[alternative HTML version deleted]]
>
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