Yep it is faster but it is wrong! :-) Maybe good for two columns only...
----- Original Message -----
From: "Liaw, Andy" <[EMAIL PROTECTED]>
To: "'Jim Brennan'" <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Monday, July 19, 2004 9:45 PM
Subject: RE: [R] How to compare X1 = X2 = ... = Xn?
> 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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