# [R] How to compare X1 = X2 = ... = Xn?

Liaw, Andy andy_liaw at merck.com
Tue Jul 20 03:45:34 CEST 2004

```Not so fast:

> x <- matrix(rep(c(2, 1, 3), 2), nr=2, byrow=TRUE)
> x
[,1] [,2] [,3]
[1,]    2    1    3
[2,]    2    1    3
> rowSums(x) / ncol(x) == x[,1]
[1] TRUE TRUE

Andy

> From: Jim Brennan
>
> This similar method may  be quicker
>  x1\$new <- 1*(rowSums(x1)/ncol(x1)==x1[,1])
> Learning lots from these type questions!
>
> Jim
>
> From: "Gabor Grothendieck"
>
> > Berton Gunter <gunter.berton <at> gene.com> writes:
> >
> > :
> > : How about:
> > :
> > : X<-as.matrix(yourframe)
> > : apply(X,2, '==',X[,1])%*%rep(1,ncol(X)) == ncol(x)
> > :
> > : avoiding the rowwise apply overhead?
> >
> > Following up on your idea we can use rowSums instead of matrix
> multiplication
> > to speed it up even more:
> >
> > R> x <- data.frame(X1 = c(1.5, 1.5, 2.5, 4.5),
> > + X2 = c(4.5, 1.5, 2.5, 5.5), X3 = c(2.5, 1.5, 2.5, 2.5))
> > R> set.seed(1)
> > R> x1 <- x2 <- x[sample(4,100000,rep=T),]
> >
> > R> gc();system.time({x1\$new <- (rowSums(x1==x1[,1])==ncol(x))+0})
> >           used (Mb) gc trigger (Mb)
> > Ncells  634654 17.0    1590760 42.5
> > Vcells 1017322  7.8    3820120 29.2
> > [1] 0.48 0.00 0.48   NA   NA
> >
> > R> gc(); system.time({X <- as.matrix(x2); x2\$new <- c(apply(X,2,
> '==',X[,1])%*%
> > rep(1,ncol(X)) == ncol(x))+0})
> >           used (Mb) gc trigger (Mb)
> > Ncells  634668 17.0    1590760 42.5
> > Vcells 1517333 11.6    3820120 29.2
> > [1] 1.39 0.03 1.50   NA   NA
> >
> > R> all.equal(x1,x2)
> > [1] TRUE
> >
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>

```