As usual, Gabor provides an elegant solution. But I hope that, in
this case, the OP provided a toy example. Otherwise, I don't see
the point of applying cut() to a vector of length 7. Why not just
use stripchart()?
Peter Ehlers
Gabor Grothendieck wrote:
> Or building on that solution but elimina
Or building on that solution but eliminating the do.call and lapply:
f <- function(x) table(cut(x, breaks = seq(0, 30, 5)))
barplot(rbind(f(x), f(y)), beside = TRUE)
On 12/7/05, Jacques VESLOT <[EMAIL PROTECTED]> wrote:
> try :
>
> barplot(do.call("rbind",lapply(list(x,y), function(x) table(cut(x
try :
barplot(do.call("rbind",lapply(list(x,y), function(x) table(cut(x,
breaks =c(0,5,10,20,25,30),beside=T)
Subhabrata a écrit :
>Hello R-users,
>
>I am new to R-commands.
>
>
>I have two sets of data:
>
>x <- c(7, 7 , 8, 9, 15, 17, 18)
>y <- c(7, 8, 9, 15, 17, 19, 20, 20, 25, 23, 22)
>
>
Hello R-users,
I am new to R-commands.
I have two sets of data:
x <- c(7, 7 , 8, 9, 15, 17, 18)
y <- c(7, 8, 9, 15, 17, 19, 20, 20, 25, 23, 22)
I have used 'cut' command to seperate them as follows
a <- cut(x, breaks =c(0,5,10,20,25,30))
b <- cut(y, breaks =c(0,5,10,20,25,30))
> table(a)
