Rosa Oliveira
2015-Sep-06 14:06 UTC
[R] split.screen to draw graphs - ggplot2 and lattice (can't slip in 4 cells)
Dear all,
I need your urgent help J
I?m na?ve, and I?m pretty sure my doubt is very simple to solve, but I?m not
getting it.
I used the following code to produce my research graphs, nonetheless, is this
problem, I do not have 6 graphs (1 ? 6),
# 3 4 5
# 2
# 6 7 8
but only 4,instead. So, I need to reformulate the code, just to
# 3 4
# 2
# 6 7
Can you please help me reformulating?
I suppose I have to change something in the split.screen code, because nowadays,
my ?third? column is blank J
I attach the code:
library(ggplot2)
library(reshape)
library(lattice)
mean.alpha1<-read.csv("graphs_mean_alpha1.csv")
mean.alpha2<-read.csv("graphs_mean_alpha2.csv")
quartz(width=10,height=5)
split.screen(figs=matrix(c(0,0.4,0.5,1,
0.4,0.7,0.5,1,
0.7,1,0.5,1,
0,0.4,0,0.5,
0.4,0.7,0,0.5,
0.7,1,0,0.5),nrow=6,byrow=TRUE),
screen=1)
screen(3)
par(mar=c(0,3.5,3,0))
# now the second set
n250<-mean.alpha1$nsample==250
matplot(x=mean.alpha1$lambda[n250],y=mean.alpha1[n250,3:5],
type="l",pch=1:3,col=c(4,2,3),xaxt="n",ylim=c(-1.2,
-0.25),main="nsample=250",ylab="", cex.main=1)
abline(h = -1, col = "gray60")
mtext(expression(paste("mean av. for ",alpha[1])),side=2,line=2,
cex.main=1)
screen(4)
par(mar=c(0,0,3,0))
# now the second set
n250<-mean.alpha1$nsample==1000
matplot(x=mean.alpha1$lambda[n1000],y=mean.alpha2[n1000,3:5],
type="l",pch=1:3,col=c(4,2,3),xaxt="n",yaxt="n",ylim=c(-1.2,
-0.25),main="nsample=1000", cex.main=1)
abline(h = -1, col = "gray60")
screen(6)
par(mar=c(3,3.5,0,0))
# now the second set
n250<-mean.alpha2$nsample==250
matplot(x=mean.alpha2$lambda[n250],y=mean.alpha2[n250,3:5],
type="l",pch=1:3,col=c(4,2,3),ylim=c(-.6,
-.35),ylab="")
abline(h = -.5, col = "gray60")
mtext(expression(paste(lambda)),side=1,line=2, cex.main=1.5)
mtext(expression(paste("mean av. for ",alpha[2])),side=2,line=2,
cex.main=1.5)
screen(7)
par(mar=c(3,0,0,0))
# now the second set
n1000<-mean.alpha2$nsample==1000
matplot(x=mean.alpha2$lambda[n1000],y=mean.alpha2[n1000,3:5],
type="l",pch=1:3,col=c(4,2,3),yaxt="n",ylim=c(-.6, -.35))
abline(h = -.5, col = "gray60")
mtext(expression(paste(lambda)),side=1,line=2, , cex.main=1.5)
screen(2)
par(mar=c(0,0,0,0))
# plot an empty plot to get the coordinates
plot(0:1,0:1,type="n",axes=FALSE)
legend(0,0.6,c("OLS", "GLS", "Reg. Cal.",
"true coefficient"),bty = "n",
lty=1:3,col=c(4,2,3,"gray60"),xpd=TRUE)
close.screen(all=TRUE)
BEST,
RO
Atenciosamente,
Rosa Oliveira
--
____________________________________________________________________________
Rosa Celeste dos Santos Oliveira,
E-mail: rosita21 at gmail.com
Tlm: +351 939355143
Linkedin: https://pt.linkedin.com/in/rosacsoliveira
____________________________________________________________________________
"Many admire, few know"
Hippocrates
Jim Lemon
2015-Sep-07 10:23 UTC
[R] split.screen to draw graphs - ggplot2 and lattice (can't slip in 4 cells)
Hi Rosa, I think all you need is the split.screen commands but this will show you where each screen number is located: split.screen(figs=matrix(c(0,0.8,0,1,0.8,1,0,1),ncol=4,byrow=TRUE)) split.screen(figs=matrix(c(0,0.5,0.5,1,0.5,1,0.5,1,0,0.5,0,0.5,0.5,1,0,0.5), ncol=4,byrow=TRUE),screen=1) screen(3) plot(0:1,0:1,type="n",xlab="",ylab="",axes=FALSE) box() text(0.5,0.5,3) screen(4) plot(0:1,0:1,type="n",xlab="",ylab="",axes=FALSE) box() text(0.5,0.5,4) screen(5) plot(0:1,0:1,type="n",xlab="",ylab="",axes=FALSE) box() text(0.5,0.5,5) screen(6) plot(0:1,0:1,type="n",xlab="",ylab="",axes=FALSE) box() text(0.5,0.5,6) screen(2) plot(0:1,0:1,type="n",xlab="",ylab="",axes=FALSE) box() text(0.5,0.5,2) close.screen(all=TRUE) I used width=10,height=6 in the device call. Jim On Mon, Sep 7, 2015 at 12:06 AM, Rosa Oliveira <rosita21 at gmail.com> wrote:> Dear all, > > > > I need your urgent help J > > > > I?m na?ve, and I?m pretty sure my doubt is very simple to solve, but I?m > not getting it. > > > > I used the following code to produce my research graphs, nonetheless, is > this problem, I do not have 6 graphs (1 ? 6), > > > > # 3 4 5 > > # 2 > > # 6 7 8 > > > > > > but only 4,instead. So, I need to reformulate the code, just to > > > > # 3 4 > > # 2 > > # 6 7 > > > > Can you please help me reformulating? > > I suppose I have to change something in the split.screen code, because > nowadays, my ?third? column is blank J > > > > > > I attach the code: > > > > library(ggplot2) > > library(reshape) > > library(lattice) > > > > mean.alpha1<-read.csv("graphs_mean_alpha1.csv") > > mean.alpha2<-read.csv("graphs_mean_alpha2.csv") > > quartz(width=10,height=5) > > split.screen(figs=matrix(c(0,0.4,0.5,1, > > 0.4,0.7,0.5,1, > > 0.7,1,0.5,1, > > 0,0.4,0,0.5, > > 0.4,0.7,0,0.5, > > 0.7,1,0,0.5),nrow=6,byrow=TRUE), > > screen=1) > > > > > > > > screen(3) > > par(mar=c(0,3.5,3,0)) > > # now the second set > > n250<-mean.alpha1$nsample==250 > > matplot(x=mean.alpha1$lambda[n250],y=mean.alpha1[n250,3:5], > > type="l",pch=1:3,col=c(4,2,3),xaxt="n",ylim=c(-1.2, > -0.25),main="nsample=250",ylab="", cex.main=1) > > abline(h = -1, col = "gray60") > > mtext(expression(paste("mean av. for ",alpha[1])),side=2,line=2, > cex.main=1) > > > > > > > > screen(4) > > par(mar=c(0,0,3,0)) > > # now the second set > > n250<-mean.alpha1$nsample==1000 > > matplot(x=mean.alpha1$lambda[n1000],y=mean.alpha2[n1000,3:5], > > type="l",pch=1:3,col=c(4,2,3),xaxt="n",yaxt="n",ylim=c(-1.2, > -0.25),main="nsample=1000", cex.main=1) > > abline(h = -1, col = "gray60") > > > > > > > > screen(6) > > par(mar=c(3,3.5,0,0)) > > # now the second set > > n250<-mean.alpha2$nsample==250 > > matplot(x=mean.alpha2$lambda[n250],y=mean.alpha2[n250,3:5], > > type="l",pch=1:3,col=c(4,2,3),ylim=c(-.6, -.35),ylab="") > > abline(h = -.5, col = "gray60") > > mtext(expression(paste(lambda)),side=1,line=2, cex.main=1.5) > > mtext(expression(paste("mean av. for ",alpha[2])),side=2,line=2, > cex.main=1.5) > > > > > > > > screen(7) > > par(mar=c(3,0,0,0)) > > # now the second set > > n1000<-mean.alpha2$nsample==1000 > > matplot(x=mean.alpha2$lambda[n1000],y=mean.alpha2[n1000,3:5], > > type="l",pch=1:3,col=c(4,2,3),yaxt="n",ylim=c(-.6, -.35)) > > abline(h = -.5, col = "gray60") > > mtext(expression(paste(lambda)),side=1,line=2, , cex.main=1.5) > > > > > > screen(2) > > par(mar=c(0,0,0,0)) > > # plot an empty plot to get the coordinates > > plot(0:1,0:1,type="n",axes=FALSE) > > legend(0,0.6,c("OLS", "GLS", "Reg. Cal.", "true coefficient"),bty = "n", > lty=1:3,col=c(4,2,3,"gray60"),xpd=TRUE) > > > > > > close.screen(all=TRUE) > > > > > > BEST, > > RO > > > > > > Atenciosamente, > Rosa Oliveira > > -- > > ____________________________________________________________________________ > > > Rosa Celeste dos Santos Oliveira, > > E-mail: rosita21 at gmail.com > Tlm: +351 939355143 > Linkedin: https://pt.linkedin.com/in/rosacsoliveira > > ____________________________________________________________________________ > "Many admire, few know" > Hippocrates > > ______________________________________________ > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.[[alternative HTML version deleted]]