Displaying 3 results from an estimated 3 matches for "ind0".
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2009 Apr 23
2
Two 3D cones in one graph
...-seq(-astar,astar,len=50)
ystar<-seq(-bstar,bstar,len=50)
g<-expand.grid(x=xstar,y=ystar)
p1<-2*g$x*mu[1]/a**2+2*g$y*mu[2]/b**2
p2<-(g$x**2/a**2+g$y**2/b**2)
p3<-mu[1]**2/a**2+mu[2]**2/b**2-1
q<-(p1+sqrt(p1**2-4*p2*p3))/(2*p2)
z<-sqrt(1-(q*g$x)**2-(q*g$y)**2)
zstar<-(z/q)
ind0<-!(q<1)
g$z<-zstar
sc<-matrix(c(rep(c(-1,-1,-1),sum(ind0))),nrow=sum(ind0),byrow=TRUE)
gstar<-rbind(g[ind0,],sc*g[ind0,])
group<-c(rep(1,nrow(gstar)/2),rep(2,nrow(gstar)/2))
gstar$group<-group
wireframe(z~x*y|group,gstar,colorkey=TRUE,drape=TRUE,
scales=list(arrows=FALSE))
2011 Feb 07
1
multiple imputation manually
...0,8)
x2 <- 1+2*y2+ rnorm(20,0,8)
x3 <- 1+2*y3+ rnorm(20,0,8)
x4 <- 1+2*y4+ rnorm(20,0,8)
x <- c(x1,x2,x3,x4)
mcar.y <- rep(NA,80)
y.mis <- rep(NA,80)
df <- data.frame(y=y, y.mis=y.mis, mcar.y=mcar.y, x=x)
df$y.mis <- df$y
for (j in 1:80)
{
df$mcar.y <- rbinom(80,1,0.15)
}
ind0 <- which(df$mcar.y==0)
ind1 <- which(df$mcar.y==1)
if (length(ind0) > 68) {
df$mcar.y[sample(ind0, length(ind0) - 68)] <- 1
} else {
df$mcar.y[sample(ind1, 68 - length(ind0))] <- 0
}
df$y.mis[df$mcar.y==1] <- NA
This gives me data sets with missing values completely at ran...
2003 May 09
1
manipulating elements of a matrix
Dear R users:
I have the following matrix.
0 1 1
0 1 0
2 1 0
3 0 0
I would like to spread the matrix such that whenever the row sum is greater
than 1 the row is repeated the number of times given by the row sum.
Furthermore I would like to split the following cases:
0 1 1
such that it map to the following matrix
0 1 0
0 0 1
such that each row adds up to 1.
I have no problems with cases